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5., , 6., , 7., , 8., , 9., , 10., , (b) The projection of the particle on any one of, the diameters executes S.H.M., , The amplitude and the time period in a S.H.M. is, 0.5 cm and 0.4 sec respectively. If the initial, phase is  / 2 radian, then the equation of S.H.M., will be, (a) y = 0 . 5 sin 5t, , (b) y = 0 .5 sin 4t, , (c) y = 0 . 5 sin 2 .5t, , (d) y = 0 . 5 cos 5t, , (c) The projection of the particle on any of the, diameters executes S.H.M., (d) None of the above, , (a) 2nt, , (b) , , A particle is executing simple harmonic motion, with a period of T seconds and amplitude a metre., a, The shortest time it takes to reach a point, m, 2, , (c) 2nt + , , (d) 2t, , from its mean position in seconds is [EAMCET (Med.) 2000], , The equation of S.H.M. is y = a sin(2nt +  ) , then its, phase at time t is, [DPMT 2001], , A particle is oscillating according to the, equation X = 7 cos 0.5t , where t is in second. The, point moves from the position of equilibrium to, maximum displacement in time, (a) 4.0 sec, , (b) 2.0 sec, , (c) 1.0 sec, , (d) 0.5 sec, , 13., , 14., , (c) T/8, , (d) T/16, , A simple, motion is represented by, [CPMTharmonic, 1989], ., The, amplitude of the S.H.M., F(t) = 10 sin (20 t + 0.5), [DPMT 1998; CBSE PMT 2000; MH CET 2001], , A simple harmonic oscillator has an amplitude a, (a) a = 30, and time period T. The time required by it to, travel from x = a to x = a / 2 is[CBSE PMT 1992; SCRA 1996; BHU, (c) 1997], a = 10, (a) T / 6, , (b) T / 4, , (c) T / 3, , (d) T / 2, , 15., , Which of the following expressions represent, simple harmonic motion, [Roorkee 1999], (a) x = A sin( t +  ), , (b) x = B cos( t +  ), , (c) x = A tan( t +  ), , (d) x = A sin  t cos  t, , 16., , A 1.00  10 −20 kg particle is vibrating with simple, , (a) 1.59 mm, , (b) 1.00 m, , (c) 10 m, , (d) None of these, , 17., , (b) Only the direction of motion of the particle at, time t, , A particle is moving with constant angular, velocity along the circumference of a circle., Which of the following statements is true, (a) The particle so moving executes S.H.M., , Which of the following equation, represent a simple harmonic motion, (a) y = a sin  t, , (b) y = a cos  t, , (c) y = a sin t + b cos  t, , (d) y = a tan  t, , does, , not, , A particle in S.H.M. is described by the, displacement function x (t) = a cos(t +  ) . If the, initial (t = 0 ) position of the particle is 1 cm and its, initial velocity is  cm/s . The angular frequency of, , (c) 2 cm, , (a) Only the position of the particle at time t, , (d) Neither the position of the particle nor its, direction of motion at time t, , (d) a = 5, , (Med.) 1999] (b), (a) 1 [AMU, cm, , The phase (at a time t) of a particle in simple, harmonic motion tells, [AMU (Engg.) 1999], , (c) Both the position and direction of motion of, the particle at time t, , (b) a = 20, , the particle is  rad / s , then it’s amplitude is, , maximum displacement of the particle is, , 12., , (b) T/4, , is, , harmonic motion with a period of 1 .00  10 −5 sec, and a maximum speed of 1 .00  10 3 m/s . The, , 11., , (a) T, , 18., , 2 cm, , (d) 2.5 cm, , A particle executes a simple harmonic motion of, time period T. Find the time taken by the particle, to go directly from its mean position to half the, amplitude, [UPSEAT 2002], (a) T / 2, , (b) T / 4, , (c) T / 8, , (d) T / 12, , A particle executing simple harmonic motion along, y-axis has its motion described by the equation, y = A sin( t) + B . The amplitude of the simple, harmonic motion is, [Orissa JEE 2003], , (a) A, , (b) B, [AMU (Engg.) 1999], , (c) A + B, , (d), , A+B

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19., , A particle executing S.H.M. of amplitude 4 cm and, T = 4 sec. The time taken by it to move from, positive extreme position to half the amplitude is, (a) 1 sec, (c) 2/3 sec, , 20., , 25., , (a) Inertia only, [BHU 1995], , (b) Elasticity as well as inertia, , (b) 1/3 sec, (d), , (c) Elasticity, inertia and an external force, , 3 / 2 sec, , Which one of the following is a simple harmonic, motion, , A system exhibiting S.H.M. must possess[KCET 1994], , (d) Elasticity only, 26., , [CBSE PMT 1994], , (a) Wave moving through a string fixed at both, ends, (b) Earth spinning about its own axis, , , , If x = a sin t +  and x  = a cos t , then what is, 6, , the phase difference between the two waves[RPET 1996], (a)  / 3, , (b)  / 6, , (c)  / 2, , (d) , , Velocity of Simple Harmonic Motion, , (c) Ball bouncing between two rigid vertical walls, , (a) Periodic and simple harmonic, , A simple pendulum performs simple harmonic, motion about X = 0 with an amplitude A and time, A, period T. The speed of the pendulum at X =, 2, will be, [MP PMT 1987], , (b) Periodic but not simple harmonic, , (a), , (d) Particle moving in a circle with uniform speed, 21., , 1., , A particle is moving in a circle with uniform, speed. Its motion is, [CPMT 1978; CBSE PMT 2005], , A 3, T, , (c) A periodic, (d) None of the above, 22., , Two simple harmonic motions are represented by, , , y1 = 0 . 1 sin  100  t +  and, the, equations, 3, , , 2T, , (d), , T, , 3 2 A, T, , A body is executing simple harmonic motion with, an angular frequency 2rad / s . The velocity of the, body at 20 mm displacement, when the amplitude, , of particle 1 with respect to the velocity of, particle 2 is, [AIEEE 2005], , of motion is 60 mm, is, , (c), , 24., , 2., , A 3, , A, , y 2 = 0.1 cos  t. The phase difference of the velocity, , −, (a), 3, , 23., , (c), , (b), , −, 6, , CPMT 1999], , , (b), 6, (d), , , 3, , 3., , Two particles are executing S.H.M. The equation, of, their, motion, are, , T , 3T , , y1 = 10 sin  t +, . What is, , y 2 = 25 sin   t +, , 4 , 4 , , , the ratio of their amplitude [DCE 1996], (a) 1 : 1, , (b) 2 : 5, , (c) 1 : 2, , (d) None of these, , The periodic time of a body executing simple, harmonic motion is 3 sec. After how much interval, from time t = 0, its displacement will be half of, its amplitude, [BHU 1998], , [, , (a) 40 mm /s, , (b) 60 mm / s, , (c) 113 mm / s, , (d) 120 mm / s, , A body of mass 5 gm is executing S.H.M. about a, point with amplitude 10 cm. Its maximum velocity, is 100 cm/sec. Its velocity will be 50 cm/sec at a, , 4., , distance, , [CPMT 1976], , (a) 5, , (b) 5 2, , (c) 5 3, , (d) 10 2, , A simple harmonic oscillator has a period of 0.01, sec and an amplitude of 0.2 m. The magnitude of, the velocity in m sec −1 at the centre of oscillation, is [JIPMER 1997], , (a), , 1, sec, 8, , (b), , 1, sec, 6, , (a) 20 , , (b) 100, , (c), , 1, sec, 4, , (d), , 1, sec, 3, , (c) 40, , (d) 100 

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5., , A particle executes S.H.M. with a period of 6, second and amplitude of 3 cm. Its maximum speed, , 11., , If the displacement of a particle executing SHM is, given by y = 0 .30 sin(220 t + 0 .64 ) in metre, then the, , in cm/sec is, , frequency and maximum velocity of the particle is[AFMC 1, [AIIMS 1982], , 6., , 7., , (a)  / 2, , (b) , , (c) 2, , (d) 3, , 12., , (a) 35 Hz, 66 m / s, , (b) 45 Hz, 66 m / s, , (c) 58 Hz, 113 m / s, , (d) 35 Hz, 132 m / s, , The maximum velocity and the maximum, acceleration of a body moving in a simple, , A particle is executing S.H.M. If its amplitude is 2, , harmonic oscillator are 2 m/s and 4 m /s 2 . Then, , m and periodic time 2 seconds, then the maximum, velocity of the particle will be, , angular velocity will be, , [MP PMT 1985], [Pb. PMT 1998; MH CET 1999, 2003], , 2 m / s, , (a)  m / s, , (b), , (c) 2 m / s, , (d) 4 m / s, , A S.H.M. has amplitude ‘a’ and time period T. The, , 13., , (a) 3 rad/sec, , (b) 0.5 rad/sec, , (c) 1 rad/sec, , (d) 2 rad/sec, , If a particle under S.H.M. has time period 0.1 sec, and amplitude 2  10 −3 m . It has maximum velocity, , maximum velocity will be, , [RPET 2000], , [MP PMT 1985; CPMT 1997; UPSEAT 1999], , 4a, (a), T, , (a), , 2a, (b), T, , (c), 2a, (d), T, , a, (c) 2, T, 8., , 14., , A body is executing S.H.M. When its displacement, , (b), , m/s, , (d) None of these, , 26, , m/s, , A particle executing simple harmonic motion has, an amplitude of 6 cm. Its acceleration at a, , The maximum speed of the particle is [EAMCET (Engg.) 200, corresponding velocity of the body is 10 cm/sec, (a)MP, 8 PET, cm/s1995], (b) 12 cm/s, and 8 cm/sec. Then the time period of the body is [CPMT 1991;, (c) 16 cm/s, (d) 24 cm/s, (a) 2 sec, (b)  / 2 sec, 15. A particle executes simple harmonic motion with, (c)  sec, (d) 3 / 2 sec, an amplitude of 4 cm. At the mean position the, velocity of the particle is 10 cm/s. The distance of, A particle has simple harmonic motion. The, the particle from the mean position when its, speed becomes 5 cm/s is, , , , equation of its motion is x = 5 sin 4 t −  , where x, 6, , is its displacement. If the displacement of the, particle is 3 units, then it velocity is, , [EAMCET (Med.) 2000], , (a), , [MP PMT 1994], , (a), , 2, 3, , (b), , 5, 6, , 3 cm, , (c) 2( 3 ) cm, , 5 cm, , (d) 2( 5 ) cm, , If a simple pendulum oscillates with an amplitude, of 50 mm and time period of 2 sec, then its, maximum velocity is, [AIIMS 1998; MH CET 2000; DPMT 2000], , [EAMCET 2001], , (d) 16, , 16., , (b), , Two particles P and Q start from origin and, execute Simple Harmonic Motion along X-axis, with same amplitude but with periods 3 seconds, and 6 seconds respectively. The ratio of the, velocities of P and Q when they meet is, , (c) 20, 10., , , 30, , , , m/s, , distance of 2 cm from the mean position is 8 cm /s 2 ., , from the mean position is 4 cm and 5 cm, the, , 9., , , 25, , (a) 0.10 m / s, , (b) 0.15 m / s, , (a) 1 : 2, , (b) 2 : 1, , (c) 0.8 m / s, , (d) 0.26 m / s, , (c) 2 : 3, , (d) 3 : 2

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17., , A particle is performing simple harmonic motion, with amplitude A and angular velocity . The, ratio of maximum velocity to maximum, acceleration is, [Kerala (Med.) 2002], (a) , (c), , 18., , 24., , by, , (b) 1/, , 2, , [BCECE 2005], , (d) A, , The angular velocities of three bodies in simple, harmonic motion are 1, 2 , 3, with their, respective amplitudes as A1 , A2 , A3 . If all the three, bodies have same mass and velocity, then, (a) A11 = A22 = A33, , (b) A11 = A22 = A33, 2, , 2, , 25., 2, , (c) A1 21 = A2 22 = A3 23 (d) A1 212 = A2 22 2 = A 2, 19., , The velocity of a particle performing simple, harmonic motion, when it passes through its, mean position is, , 26., , (b), , 3, 6, , (c) 100, , (d), , , 6, , 2002], The [BHU, displacement, equation of a particle is, x = 3 sin 2t + 4 cos 2t. The amplitude and maximum, velocity will be respectively, , (a) 5, 10, , (b) 3, 2, , (c) 4, 2, , (d) 3, 4, , (a) Infinity, , (b) Zero, , (c) Minimum, , (d) Maximum, , (a) 4v, , The velocity of a particle in simple harmonic, motion at displacement y from mean position is, [BCECE 2003; RPMT 2003], , 21., , (a) 300, , Velocity at mean position of a particle executing, S.H.M. is v, they velocity of the particle at a, distance equal to half of the amplitude, , [MH CET (Med.) 2002; BCECE 2004], , 20., , The maximum velocity of a simple harmonic, , , motion represented by y = 3 sin  100 t +  is given, 6, , , (a)  a 2 + y 2, , (b)  a 2 − y 2, , (c) y, , (d)  2 a 2 − y 2, , (c), 27., , A particle is executing the motion x = A cos( t −  ) ., The maximum velocity of the particle is, , 22., , 23., , (b) A , , (c) A sin , , (d) None of these, , A particle executing simple harmonic motion with, amplitude of 0.1 m. At a certain instant when its, displacement is 0.02 m, its acceleration is 0.5 m/s2., The maximum velocity of the particle is (in m/s), , (d), , (a), , , 4, , (b), , (c), , , , , (d), , 1., , (a) Constant period, , (c) 0.5, , (d) 0.25, , (b) Constant acceleration, , The amplitude of a particle executing SHM is 4, cm. At the mean position the speed of the particle, is 16 cm/sec. The distance of the particle from the, mean position at which the speed of the particle, , (b), , (c) 1 cm, , (d) 2 cm, , 3 cm, , , , (c) Proportionality between acceleration, displacement from equilibrium position, , and, , (d) Proportionality between restoring force and, displacement from equilibrium position, 2., , (a) 2 3 cm, , 2, , Which of the following is a necessary and, [MP PET, 2003] for S.H.M., sufficient, condition, , (b) 0.05, , [Pb. PET 2003], , , 2, , Acceleration of Simple Harmonic Motion, , (a) 0.01, , becomes 8 3 cm / s, will be, , 3, v, 4, , The instantaneous displacement of a simple, , , pendulum oscillator is given by x = A cos  t +  ., 4, , Its speed will be maximum at time, , [BHU 2003; CPMT 2004], , (a) A cos , , 3, v, 2, , (b) 2v, , If a hole is bored along the diameter of the earth, and a stone is dropped into hole, (a) The stone reaches the centre of the earth and, stops there, , [

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(b) The stone reaches the other side of the earth, and stops there, , 8., , (c) The stone executes simple harmonic motion, about the centre of the earth, , [MP PMT 1997; AIIMS 1999; Kerala PMT 2005], , (d) The stone reaches the other side of the earth, and escapes into space, 3., , The acceleration of a particle in S.H.M. is[MP PMT 1993], , (b) Always constant, , the, , (b) The restoring force, towards a fixed point, , of, , (d) Maximum at the equilibrium position, The displacement of a particle moving in S.H.M., at any instant is given by y = a sin t . The, acceleration after time t =, , T, 4, , 9., , [MP PET 1984], , (a) a, , (b) −a, , (c) a 2, , (d) − a 2, , always, , is maximum, , always, directed, at, , the, , A particle of mass 10 grams is executing simple, harmonic motion with an amplitude of 0.5 m and, periodic time of ( / 5) seconds. The maximum, , value of the force acting on the particle is[MP PET 1999; M, , is (where T is the, , time period), , particle, , (d) The acceleration of the particle is maximum at, the equilibrium position, , (c) Maximum at the extreme position, , 5., , (a) The total energy of, remains the same, , (c) The restoring force, extreme positions, , (a) Always zero, , 4., , For a particle executing simple harmonic motion,, which of the following statements is not correct, , 10., , The amplitude of a particle executing S.H.M. with, frequency of 60 Hz is 0.01 m. The maximum value, of the acceleration of the particle is, , (a) 25 N, , (b) 5 N, , (c) 2.5 N, , (d) 0.5 N, , The displacement of an oscillating particle varies, with time (in seconds) according to the equation y, (cm) = sin, ,  t, , 1,  +  . The maximum acceleration, 2 2 3, , of the particle is approximately, , [DPMT 1998; CBSE PMT 1999; AFMC 2001;, , (a) 5.21cm / s 2, , (b) 3.62cm / s 2, , (c) 1.81cm / s 2, , (d) 0.62cm / s 2, , Pb. PMT 2001; Pb. PET 2001, 02; CPMT 1993, 95, 04;, RPMT 2005; MP PMT 2005], , (a) 144  2m /sec 2, (c), 6., , 7., , 144, , 2, , m /sec 2, , (b) 144 m /sec 2, , 11., , (d) 288  2m /sec 2, , A particle moving along the x-axis executes, simple harmonic motion, then the force acting on, it is given by, [CBSE PMT 1994], , A small body of mass 0.10 kg is executing S.H.M., of amplitude 1.0 m and period 0.20 sec. The, maximum force acting on it is, (a) 98.596 N, , (b) 985.96 N, , (c) 100.2 N, , (d) 76.23 N, , (a) – A Kx, , (b) A cos (Kx), , (c) A exp (– Kx), , (d) A Kx, , Where A and K are positive constants, 12., , A body executing simple harmonic motion has a, maximum acceleration equal to 24 metres /sec 2 and, , A body is vibrating in simple harmonic motion, with an amplitude of 0.06 m and frequency of 15, Hz. The velocity and acceleration of body is, (a) 5.65 m/s and 5 . 32  10 2 m /s 2, , maximum velocity equal to 16 metres /sec . The, (b) 6.82 m/s and 7 . 62  10 2 m /s 2, , amplitude of the simple harmonic motion is, , (c) 8.91 m/s and 8 . 21  10 2 m /s 2, , [MP PMT 1995; DPMT 2002; RPET 2003; Pb. PET 2004], , (a), , 32, metres, 3, , (b), , 3, metres, 32, , (c), , 1024, metres, 9, , (d), , 64, metres, 9, , (d) 9.82 m/s and 9 . 03  10 2 m /s 2, 13., , A particle executes harmonic motion with an, angular velocity and maximum acceleration of 3.5

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rad/sec and 7.5 m/s2 respectively. The amplitude, of oscillation is, , 20., , [AIIMS 1999; Pb. PET 1999], , 14., , (a) 0.28 m, , (b) 0.36 m, , (c) 0.53 m, , (d) 0.61 m, , [Kerala PET 2005], , What is the maximum acceleration experienced by, , 15., , (a), , A 0.10 kg block oscillates back and forth along a, horizontal surface. Its displacement from the, origin is given by: x = (10 cm)cos[(10 rad/s) t +  /2 rad] ., the block, , [AMU (Engg.) 2000], , (a) 10 m /s 2, , (b) 10  m /s 2, , (c), , 10 , m /s 2, 2, , (d), , (c), , 1, , (b) 2 3, , 2 3, 2, , (d), , 3, 21., , 3, 2, , In simple harmonic motion, the ratio of, acceleration of the particle to its displacement at, any time is a measure of, [UPSEAT 2001], , 10 , m /s 2, 3, , (a) Spring constant, , (b) Angular frequency, , (c) (Angular frequency)2, , In S.H.M. maximum acceleration is at, [RPET 2001; BVP 2003], , (a) Amplitude, , A particle executes linear simple harmonic motion, with an amplitude of 2 cm. When the particle is at, 1 cm from the mean position the magnitude of its, velocity is equal to that of its acceleration. Then, its time period in seconds is, , (d), , Restoring force, , Energy of Simple Harmonic Motion, , (b) Equilibrium, , (c) Acceleration is constant (d), , None of these, , 1., , The total energy of a particle executing S.H.M. is, proportional to, , 16., , A particle is executing simple harmonic motion, [CPMT 1974, 78; EAMCET 1994; RPET 1999;, with an amplitude of 0.02 metre and frequency 50, MP, PMT 2001; Pb. PMT 2002; MH CET 2002], Hz. The maximum acceleration of the particle is [MP PET 2001], (a) Displacement from equilibrium position, (a) 100 m /s 2, (b) 100  2 m /s 2, (b) Frequency of oscillation, (c) 100 m /s 2, (d) 200  2 m /s 2, (c) Velocity in equilibrium position, , 17., , Acceleration of a particle, executing SHM, at it’s, mean position is, [MH CET (Med.) 2002], , 18., , (a) Infinity, , (b) Varies, , (c) Maximum, , (d) Zero, , (d) Square of amplitude of motion, 2., , Which one of the following statements is true for, the speed v and the acceleration a of a particle, executing simple harmonic motion, , (a)  A, (c) , 3., , (c) When v is zero, a is zero, (d) When v is maximum, a is zero, 19., , What is the maximum acceleration of the particle,  t, , doing the SHM y = 2 sin  +   where 2 is in cm[DCE 2003], 2, , , (a), (c), , , 2, , , 4, , cm / s 2, , (b), , cm / s 2, , (d), , 2, 2, , , 4, , cm / s 2, , cm / s 2, , (b) Zero, , [CBSE PMT 2004], , (a) When v is maximum, a is maximum, (b) Value of a is zero, whatever may be the value, of v, , A particle executes simple harmonic motion, along a straight line with an amplitude A. The, potential energy is maximum when the, displacement is, [CPMT 1982], , 4., , A, 2, , (d) , , A, 2, , A particle is vibrating in a simple harmonic, motion with an amplitude of 4 cm. At what, displacement from the equilibrium position, is its, energy half potential and half kinetic[NCERT 1984; MNR 19, RPMT 1995; DCE 2000; UPSEAT 2000], , (a) 1 cm, , (b), , 2 cm, , (c) 3 cm, , (d) 2 2 cm, , For a particle executing simple harmonic motion,, the kinetic energy K is given by K = K o cos 2 t . The, maximum value of potential energy is

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(a) K 0, (c), 5., , 6., , [CBSE PMT 1993; EAMCET (Engg.) 1995;, , (b) Zero, , MP PMT 1994, 2000; MP PET 1995, 96, 2002], , K0, 2, , (d) Not obtainable, , The potential energy of a particle with, displacement X is U(X). The motion is simple, harmonic, when (K is a positive constant), , KX 2, (a) U = −, 2, , (b) U = KX 2, , (c) U = K, , (d) U = KX, , 11., , 12., , MH CET 1997, 99; AFMC 1999; CPMT 2000], , 7., , 2, , 13., , E, 2, , (b), , (b) 3 .75  2 ergs, , (c) 375  2 ergs, , (d) 0.375  2ergs, , When the displacement is half the amplitude, the, ratio of potential energy to the total energy is, , (d), , 3, E, 4, , (a) 18 J, , (b) 10 J, , (c) 12 J, , (d) 2.5 J, , 14., , (a) X 2 2 /(a 2 − X 2 2 ), , (b) X 2 /(a 2 − X 2 ), , (c) (a − X  ) / X , , (d) (a − X ) / X, , 2, , 2, , (c) 1, , (d), , 1, 8, , The P.E. of a particle executing SHM at a distance, x from its equilibrium position is, , 2, , 2, , 2, , When the potential energy of a particle executing, simple harmonic motion is one-fourth of its, maximum value during the oscillation, the, displacement of the particle from the equilibrium, position in terms of its amplitude a is, , (a), , 1, m2x 2, 2, , (b), , (c), , 1, m  2 (a 2 − x 2 ), 2, , (d) Zero, , 1, m  2a2, 2, , A vertical, system executes simple, [DPMTmass-spring, 2001], harmonic oscillations with a period of 2 s. A, , (a) Velocity, (b) Potential energy, , kinetic energy is T and potential energy is V, then, the ratio of T to V is, [CBSE PMT 1991], , 2, , 1, 4, , quantity of this system which exhibits simple, harmonic variation with a period of 1 s is, , The angular velocity and the amplitude of a, simple pendulum is  and a respectively. At a, , 2, , (b), , [Roorkee 1992; CPMT 1997; RPMT 1999], , The potential energy of a particle executing, S.H.M. is 2.5 J, when its displacement is half of, amplitude. The total energy of the particle be, , 2, , 1, 2, , E, 4, , displacement X from the mean position if its, , 10., , (a) 37 .5 2 ergs, , (a), , a 2, (d), 3, , a, , 3E, (c), 4, , 9., , A particle, mass 10 gm is describing S.H.M., [CPMTof, 1982], , [CPMT 1999; JIPMER 2000; Kerala PET 2002], , [RPMT 1994, 96; CBSE PMT 1995; JIPMER 2002], , 8., , (d) 2a / 3, , (b) a 2, , The total energy of the body executing S.H.M. is, E. Then the kinetic energy when the displacement, is half of the amplitude, is, , (a), , (c) a / 2, , at 5 cm from its equilibrium position is, , [MP PMT 1987; CPMT 1990; DPMT 1996;, , (c), , (b) a / 3, , along a straight line with period of 2 sec and, amplitude of 10 cm. Its kinetic energy when it is, , The kinetic energy and potential energy of a, particle executing simple harmonic motion will be, equal, when displacement (amplitude = a) is, , a, (a), 2, , (a) a / 4, , (c) Phase difference between acceleration and, displacement, (d) Difference, , between, , kinetic, , energy, , and, , potential energy, 15., , For any S.H.M., amplitude is 6 cm. If, instantaneous potential energy is half the total, energy then distance of particle from its mean, position is, [RPET 2000], (a) 3 cm, , (b) 4.2 cm

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(c) 5.8 cm, 16., , 22., , (d) 6 cm, , A body of mass 1 kg is executing simple harmonic, motion. Its displacement y (cm ) at t seconds is, given by y = 6 sin(100 t +  /4) . Its maximum kinetic, energy is, , When a mass M is attached to the spring of force, constant k, then the spring stretches by l. If the, mass oscillates with amplitude l, what will be, maximum potential energy stored in the spring, (a), , kl, 2, , (b) 2kl, , (c), , 1, Mgl, 2, , (d) Mgl, , [EAMCET (Engg.) 2000], , 17., , 18., , (a) 6 J, , (b) 18 J, , (c) 24 J, , (d) 36 J, , 23., , A particle is executing simple harmonic motion, with frequency f. The frequency at which its, kinetic energy change into potential energy is, (a) f/2, , (b) f, , (c) 2 f, , (d) 4 f, , There is a body having mass m and performing, S.H.M. with amplitude a. There is a restoring, force F = −Kx , where x is the displacement. The, total energy of body depends upon, , [MP PET 2000], , 24., , (b), , 1, E, 4, , (c), , 1, E, 2, , (d), , 2, E, 3, , (b) K, a, , (c) K, a, x, , (d) K, a, v, , (a) P.E. is maximum when x = 0, (b) K.E. is maximum when x = 0, , The total energy of a particle executing S.H.M. is, 80 J. What is the potential energy when the, particle is at a distance of 3/4 of amplitude from, the mean position, (a) 60 J, , (b) 10 J, , (c) 40 J, , (d) 45 J, , (c) T.E. is zero when x = 0, (d) K.E. is maximum when x is maximum, 25., , In a simple harmonic oscillator, at the mean, position, [AIEEE 2002], , (a) Kinetic energy is minimum, potential energy is, maximum, (b) Both kinetic, maximum, , and, , potential, , energies, , (d) Both kinetic, minimum, , and, , potential, , energies, , 26., , are, , (c) Kinetic energy is maximum, potential energy is, minimum, , 21., , 1, E, 8, , (a) K, x, , [Kerala (Engg.) 2001], , 20., , (a), , A body executes simple harmonic motion. The, potential energy (P.E.), the kinetic energy (K.E.), and total energy (T.E.) are measured as a function, of displacement x. Which of the following, statements is true, [AIEEE 2003], , [CBSE PMT 2001], , 19., , The potential energy of a simple harmonic, oscillator when the particle is half way to its end, point is (where E is the total energy), , 27., , are, , (b) + a, , (c)  a, , a, (d) , 4, , (a) <E> = <U>, , (b) <E> = 2<U>, , (c) <E> = – 2<U>, , (d) <E>= – <U>, , The total energy of a particle, executing simple, harmonic motion is, [AIEEE 2004], (a)  x, , (b)  x 2, , (c) Independent of x, , (d)  x 1 / 2, , The kinetic energy of a particle executing S.H.M., is 16 J when it is at its mean position. If the mass, of the particle is 0.32 kg, then what is the, maximum velocity of the particle, [MH CET 2004], , Displacement between maximum potential energy, position and maximum kinetic energy position for, a particle executing S.H.M. is, (a) – a, , If <E> and <U> denote the average kinetic and, the average potential energies respectively of, mass describing a simple harmonic motion, over, one period, then the correct relation is, , (a) 5 m / s, [CBSE PMT 2002], , (c) 10 m / s, 28., , (b) 15 m / s, (d) 20 m / s, , Consider the following statements. The total, energy of a particle executing simple harmonic, motion depends on its

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(1) Amplitude (2) Period, Of these statements, , (3), , Displacement, , (c) Periodic and simple harmonic, (d) Periodic with variable time period, , [RPMT 2001; BCECE 2005], , (a) (1) and (2) are correct, , 34., , (b) (2) and (3) are correct, , displacement y its potential energy is E 2 . The, , (c) (1) and (3) are correct, , potential energy E at displacement (x + y ) is [EAMCET 200, , (d) (1), (2) and (3) are correct, 29., , A body is executing Simple Harmonic Motion. At a, displacement x its potential energy is E1 and at a, , (a), , A particle starts simple harmonic motion from the, mean position. Its amplitude is a and total energy, E. At one instant its kinetic energy is 3E / 4. Its, displacement at that instant is, , E = E1 − E2, , (b) a / 2, , a, , (c), , (d) E = E1 + E2, , (c) E = E1 + E2, , Time Period and Frequency, , [Kerala PET 2005], , (a) a / 2, , E = E1 + E2, , (b), , 1., , (d) a / 3, , 3/2, , A particle moves such that its acceleration a is, given by a = −bx , where x is the displacement, from equilibrium position and b is a constant. The, period of oscillation is, [NCERT 1984; CPMT 1991; MP PMT 1994;, , 30., , A particle executes simple harmonic motion with, a frequency f . The frequency with which its, , MNR 1995; UPSEAT 2000], , (a) 2 b, kinetic energy oscillates is [IIT JEE 1973, 87; Manipal MEE 1995;, MP PET 1997; DCE 1997; DCE 1999; UPSEAT 2000;, RPET 2002; RPMT 2004; BHU 2005], , 31., , 32., , (a) f / 2, , (b) f, , (c) 2 f, , (d) 4 f, , (c), 2., , (b), , (c), , 9, E, 16, , (d) None of these, , A particle of mass m is hanging vertically by an, ideal spring of force constant K. If the mass is, made to oscillate vertically, its total energy is, , (c), 3., , (a) Not periodic, (b) Periodic but not simple harmonic, , equation, , of, , motion, , , b, of, , a, , particle, , is, , [RPMT 2004], , 2, , (b) 2K, (d) 2 K, , A tunnel has been dug through the centre of the, earth and a ball is released in it. It will reach the, other end of the tunnel after, [CPMT 1978; RPET 1999], , (a) 84.6 minutes, (b) 42.3 minutes, (c) 1 day, , (c) Minimum at mean position, , A body is moving in a room with a velocity of 20, m / s perpendicular to the two walls separated by, 5 meters. There is no friction and the collisions, with the walls are elastic. The motion of the body, is, [MP PMT 1999], , (d) 2, , K, , (b) Maximum at mean position, , 33., , 2, b, , 2, (a), K, , (a) Maximum at extreme position, , (d) Same at all position, , b, , d y, + Ky = 0 , where K is positive constant. The, dt 2, time period of the motion is given by, , 3, E, 4, , E, 2, , 2, , 2, , The amplitude of a particle executing SHM is, made three-fourth keeping its time period, constant. Its total energy will be, (a), , The, , (b), , (d) Will not reach the other end, 4., , The maximum speed of a particle executing, S.H.M. is 1m / s and its maximum acceleration is, 1.57 m / sec 2 . The time period of the particle will be, [DPMT 2002], , (a), , 1, sec, 1 . 57, , (c) 2 sec, , (b) 1.57 sec, (d) 4 sec

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5., , The motion of a particle executing S.H.M. is given, by x = 0 .01 sin 100  (t + .05 ) , where x is in metres and, , 12., , 6., , (b) 0.02 sec, , (c) 0.1 sec, , (b) 0.2 sec, , [CPMT, 1990], (b) All, motion having same time period are S.H.M., , (c) In S.H.M. total energy is proportional to, square of amplitude, (d) Phase constant of S.H.M. depends upon initial, conditions, , The kinetic energy of a particle executing S.H.M., is 16 J when it is in its mean position. If the, amplitude of oscillations is 25 cm and the mass of, the particle is 5.12 kg, the time period of its, , 13., , A particle in SHM is described by the, displacement equation x (t) = A cos(t +  ). If the, initial (t = 0) position of the particle is 1 cm and, its initial velocity is  cm/s, what is its, amplitude? The angular frequency of the particle, is s −1, , oscillation is, [Haryana CEE 1996; AFMC 1998], , (a), , , sec, 5, , (b) 2 sec, , (c) 20  sec, 7., , [DPMT 2004], , (d) 5 sec, , The acceleration of a particle performing S.H.M., is 12 cm /sec, , 2, , at a distance of 3 cm from the mean, , 14., , position. Its time period is [MP PET 1996; MP PMT 1997], , 8., , (a) 0.5 sec, , (b) 1.0 sec, , (c) 2.0 sec, , (d) 3.14 sec, , To make the frequency double of an oscillator, we, have to, , 15., , (b) Half the mass, , 2 cm, , (c) 2 cm, , (d) 2.5 cm, , A particle executes SHM in a line 4 cm long. Its, velocity when passing through the centre of line, is 12 cm/s. The period will be, (a) 2.047 s, , (b) 1.047 s, , (c) 3.047 s, , (d) 0.047 s, , (c) Quadruple the mass, , The frequency of the motion will be [UPSEAT 2004], , (d) Reduce the mass to one-fourth, , (a) 0.5 Hz, , , , What is constant in S.H.M., (a) Restoring force, , (b) Kinetic energy, , (c) Potential energy, , (d) Periodic time, , Hz, [UPSEAT, (c), 1999], 2, , 16., , If a simple harmonic oscillator has got a, displacement of 0.02 m and acceleration equal to, 2 . 0 ms −2 at any time, the angular frequency of the, oscillator is equal to, [CBSE PMT 1992; RPMT 1996], , 11., , (b), , , , x = 0 .01 cos   t + , 4, , , (a) Double the mass, , 10., , (a) 1 cm, , The displacement x (in metre) of a particle in,, simple harmonic motion is related to time t (in, seconds) as, , [CPMT 1999], , 9., , [, , (a) All S.H.M.’s have fixed time period, , time is in seconds. The time period is, (a) 0.01 sec, , Mark the wrong statement, , (a) 10 rad s −1, , (b) 0 .1 rad s −1, , (c) 100 rad s −1, , (d) 1 rad s −1, , The equation of a simple harmonic motion is, X = 0 .34 cos( 3000 t + 0 .74 ) where X and t are in mm, and sec. The frequency of motion is, (a) 3000, , (b) 3000 /2, , (c) 0.74 /2, , (d) 3000 /, , 17., , (b) 1.0 Hz, (d)  Hz, , A simple harmonic wave having an amplitude a, and time period T is represented by the equation, y = 5 sin  (t + 4 )m . Then the value of amplitude (a) in, (m) and time period (T) in second are [Pb. PET 2004], (a) a = 10, T = 2, , (b) a = 5, T = 1, , (c) a = 10 , T = 1, , (d) a = 5, T = 2, , A particle executing simple harmonic motion of, amplitude 5 cm has maximum speed of 31.4 cm/s., The frequency of its oscillation is, (a) 3 Hz, , (b) 2 Hz, , (c) 4 Hz, , (d) 1 Hz, , 18.[Kerala, The (Engg.), displacement, 2002] x (in metres) of a particle, performing simple harmonic motion is related to

Page 11 :
, , time t (in seconds) as x = 0 . 05 cos  4  t +  . The, 4, , frequency of the motion will be, (a) 0.5 Hz, (c) 1.5 Hz, , (c) 2 sec, 6., , (b) 1.0 Hz, (d) 2.0 Hz, , (d), , 1, sec, 2, , [MP PMT/PET, A simple pendulum, is set 1998], up in a trolley which, moves to the right with an acceleration a on a, horizontal plane. Then the thread of the pendulum, , in the mean position makes an angle  with the, , Simple Pendulum, 1., , vertical, , The period of a simple pendulum is doubled, when, , (a) tan −1, , a, in the forward direction, g, , (b) tan −1, , a, in the backward direction, g, , (c) tan −1, , g, in the backward direction, a, , (d) tan −1, , g, in the forward direction, a, , [CPMT 1974; MNR 1980; AFMC 1995; Pb. PET/PMT 2002], , (a) Its length is doubled, (b) The mass of the bob is doubled, (c) Its length is made four times, (d) The mass of the bob and the length of the, pendulum are doubled, 2., , The period of oscillation of a simple pendulum of, constant length at earth surface is T. Its period, inside a mine is, , 7., , 3., , (b) Less than T, , (c) Equal to T, , (d) Cannot be compared, , small, , A simple pendulum is made of a body which is a, , (b) Inversely proportional to the square root of, the acceleration due to gravity, , hollow sphere containing mercury suspended by, means of a wire. If a little mercury is drained off,, , (c) Dependent on the mass, size and material of, the bob, (d) Independent of the amplitude, , [NCERT 1972; BHU 1979], , (a) Remains unchanged, , 8., , The time period of a second's pendulum is 2 sec., The spherical bob which is empty from inside has, a mass of 50 gm. This is now replaced by another, , (b) Increase, , solid bob of same radius but having different, mass of 100 gm. The new time period will be, , (c) Decrease, (d) Become erratic, , 5., , for, , (a) Directly proportional to square root of the, length of the pendulum, , the period of pendulum will, , 4., , Which of the following statements is not true ? In, the case of a simple pendulum, amplitudes the period of oscillation is, , [CPMT 1973; DPMT 2001], , (a) Greater than T, , [CPMT 1983], , A pendulum suspended from the ceiling of a train, has a period T, when the train is at rest. When the, train is accelerating with a uniform acceleration, a, the period of oscillation will[NCERT 1980; CPMT 1997]9., (a) Increase, , (b) Decrease, , (c) Remain unaffected, , (d) Become infinite, , The mass and diameter of a planet are twice those, of earth. The period of oscillation of pendulum on, this planet will be (If it is a second's pendulum on, earth), , (a) 4 sec, , (b) 1 sec, , (c) 2 sec, , (d) 8 sec, , A man measures the period of a simple pendulum, inside a stationary lift and finds it to be T sec. If, the lift accelerates upwards with an acceleration, g / 4 , then the period of the pendulum will be, (a) T, (b), , [IIT 1973; DCE 2002], , (a), , 1, 2, , sec, , (b) 2 2 sec, , (c), , T, 4, , Pendulum, , 2T, 5, , (d) 2T 5, , Lift, , [

Page 12 :
10., , A simple pendulum is suspended from the roof of, a trolley which moves in a horizontal direction, with an acceleration a, then the time period is, given by T = 2, , 17., , is H. [BHU, If the1997], acceleration due to gravity is ‘g’, then, the velocity of the bob as it passes through B is, , (b) g − a, , (c) g + a, , (d), , g +a, 2, , [CBSE PMT 1995; DPMT 1995; Pb. PMT 1996], 2, , A, , A second's pendulum is placed in a space, laboratory orbiting around the earth at a height, , (b) 2 3 sec, , (c) 4 sec, , (d) Infinite, , 18., , (c) 2mE, 13., , 2mE, , (d) mE 2, , The length of the second pendulum on the surface, of earth is 1 m. The length of seconds pendulum, on the surface of moon, where g is 1/6th value of, g on the surface of earth, is, , (c) 2 gH, , (d) Zero, , Identify correct statement among the following, , (c) As the length of a simple pendulum is, increased, the maximum velocity of its bob, during its oscillation will also decreases, (d) The fractional change in the time period of a, pendulum, on changing the temperature is independent, of the length of the pendulum, , [CPMT 1971], , 14., , 15., , (a) 1 / 6 m, , (b) 6 m, , (c) 1 / 36 m, , (d) 36 m, , If the length of second's pendulum is decreased by, 2%, how many seconds it will lose per day, (a) 3927 sec, , (b) 3727 sec, , (c) 3427 sec, , (d) 864 sec, , The period of simple pendulum is measured as T, in a stationary lift. If the lift moves upwards with, an acceleration of 5 g, the period will be, (a) The same, , 16., , (b) Increased by 3/5, , 19., , The bob of a pendulum of length l is pulled aside, from its equilibrium position through an angle , and then released. The bob will then pass through, its equilibrium position with a speed v, where v, equals, [Haryana CEE 1996], , [CPMT 1992], , 20., , (a), , 2 gl(1 − sin ), , (b), , 2 gl(1 + cos  ), , (c), , 2 gl(1 − cos  ), , (d), , 2 gl(1 + sin ), , A simple pendulum executing S.H.M. is falling, [MNR 1979], freely along with the support. Then, (a) Its periodic time decreases, , (c) Decreased by 2/3 times (d) None of the above, , (b) Its periodic time increases, , The length of a simple pendulum is increased by, 1%. Its time period will [MP PET 1994; RPET 2001], , (c) It does not oscillate at all, , (a) Increase by 1%, , (b) Increase by 0.5%, , 2 gH, , (b) A simple pendulum with a bob of mass M, swings with an angular amplitude of 40 o ., When its angular amplitude is 20 o , the, tension in the string is less than Mg cos 20 o ., , [MP PMT 1986], , (b), , (b), , (a) The greater the mass of a pendulum bob, the, shorter is its frequency of oscillation, , equal to, , 2E, m, , (a) mgH, , [Manipal MEE 1995], , The bob of a simple pendulum of mass m and total, energy E will have maximum linear momentum, , (a), , C, , B, , [CPMT 1989; RPMT 1995], , (a) Zero, , P, H, , 3R, where R is the radius of the earth. The time, period of the pendulum is, , 12., , (d) Increase by 2%, , A simple pendulum with a bob of mass ‘m’, oscillates from A to C and back to A such that PB, , l, , where g  is equal to, g, , (a) g, , 11., , (c) Decrease by 0.5%, , (d) None of these

Page 13 :
21., , A pendulum bob has a speed of 3 m/s at its lowest, position. The pendulum is 0.5 m long. The speed, of the bob, when the length makes an angle of, , 27., , (a) 100%, , 60 o to the vertical, will be (If g = 10 m / s 2 ), , 1, (b) m / s, 3, , (a) 3m / s, 1, m/s, (c), 2, , 22., , (c) 300%, 28., , 29., , 23., , (b) 12 sec, , (c) 8 sec, , (d) 4 sec, , (d) 400%, , The length of a seconds pendulum is, , [RPET 2000], , (a) 99.8 cm, , (b) 99 cm, , (c) 100 cm, , (d) None of these, , The time period of a simple pendulum in a lift, descending with constant acceleration g is [DCE 1998; MP P, (a) T = 2, , [CBSE PMT 1999; DPMT 1999], , (a) 16 sec, , (b) 200%, , [MP PET 1996], , (d) 2m / s, , The time period of a simple pendulum is 2 sec. If, its length is increased 4 times, then its period, becomes, , If the length of simple pendulum is increased by, 300%, then the time period will be increased by[RPMT 199, , l, g, , (c) Zero, , (b) T = 2, , l, 2g, , (d) Infinite, , 30. A chimpanzee swinging on a swing in a sitting, position, stands up suddenly, the time period will, If the metal bob of a simple pendulum is replaced, by a wooden bob, then its time period will[AIIMS 1998, 99], [KCET (Engg./Med.) 2000; AIEEE 2002; DPMT, 2004], , (a) Increase, (b) Decrease, , (a) Become infinite, , (b) Remain same, , (c) Remain the same, , (c) Increase, , (d) Decrease, , (d) First increase then decrease, 24., , (c), , 26., , l, = constant, T, , l, T2, , = constant, , (d), , [EAMCET, (Med.), 1995], pendulum, of length, one, metre is, , l2, = constant, T, , (b), , (a), , l2, = constant, T2, , A pendulum has time period T. If it is taken on to, another planet having acceleration due to gravity, half and mass 9 times that of the earth then its, time period on the other planet will be, (a), , T, , (b) T, , (c) T, , 1/3, , (d), , 2T, , A simple pendulum is executing simple harmonic, motion with a time period T. If the length of the, pendulum is increased by 21%, the percentage, increase in the time period of the pendulum of, increased length is, [BHU 1994, 96; Pb. PMT 1995; AFMC 2001;, AIIMS 2001; AIEEE 2003], , (a) 10%, , (b) 21%, , (c) 30%, , (d) 50%, , The acceleration due to gravity at a place is, ,  2 m /sec 2 . Then the time period of a simple, , In a simple pendulum, the period of oscillation T, is related to length of the pendulum l as, (a), , 25., , 31., , 2, , , , sec, , (c) 2 sec, 32., , (b) 2 sec, (d)  sec, , A plate oscillated with time period ‘T’. Suddenly,, another plate put on the first plate, then time, period, [CMEET Bihar 1995], , 33., , [AIEEE 2002], , (a) Will decrease, , (b) Will increase, , (c) Will be same, , (d) None of these, , A simple pendulum of length l has a brass bob, attached at its lower end. Its period is T. If a steel, bob of same size, having density x times that of, brass, replaces the brass bob and its length is, changed so that period becomes 2T, then new, length is, [MP PMT 2002], (a) 2 l, , (b) 4 l, , (c) 4 l x, , (d), , 4l, x

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34., , In a seconds pendulum, mass of bob is 30 gm. If it, is replaced by 90 gm mass. Then its time period, will, , 41., , [Orissa PMT 2001], , 35., , (a) 1 sec, , (b) 2 sec, , (c) 4 sec, , (d) 3 sec, , (b) Decreases, , (c) Remains unchanged (d) Becomes infinite, 36., , (a) Zero, , (b) 0.57 m / s, , (c) 0.212 m / s, , (d) 0.32 m / s, , 42., , The time period of a simple pendulum when it is, made to oscillate on the surface of moon, (a) Increases, , The amplitude of an oscillating simple pendulum, is 10cm and its period is 4 sec. Its speed after 1, sec after it passes its equilibrium position, is, , A simple pendulum consisting of a ball of mass m, tied to a thread of length l is made to swing on a, circular arc of angle  in a vertical plane. At the, [J & K CET 2004], end of this arc, another ball of mass m is placed at, rest. The momentum transferred to this ball at, rest by the swinging ball is, [NCERT 1977], , A simple pendulum is attached to the roof of a lift., If time period of oscillation, when the lift is, stationary is T. Then frequency of oscillation,, when the lift falls freely, will be, , (c), , [DCE 2002], , 37., , 43., , (a) Zero, , (b) T, , (c) 1/T, , (d) None of these, , A simple pendulum, suspended from the ceiling of, a stationary van, has time period T. If the van, starts moving with a uniform velocity the period, of the pendulum will be, [RPMT 2003], , 38., , (a) Less than T, , (b) Equal to 2T, , (c) Greater than T, , (d) Unchanged, , 39., , 44., , If the length of the simple pendulum is increased, by 44%, then what is the change in time period of, pendulum, [MH CET 2004; UPSEAT 2005], , (a) 22%, , (b) 20%, , (c) 33%, , (d) 44%, , (b) m , , (a) Zero, , 45., , m, l, , l, g, , (d), , m, l, 2, l, g, , A simple pendulum hangs from the ceiling of a, car. If the car accelerates with a uniform, acceleration, the frequency of the simple, pendulum will, [Pb. PMT 2000], (a) Increase, , (b) Decrease, , (c) Become infinite, , (d) Remain constant, , The periodic time of a simple pendulum of length, 1 m and amplitude 2 cm is 5 seconds. If the, amplitude is made 4 cm, its periodic time in, seconds will be, [MP PMT 1985], (a) 2.5, , (b) 5, , (c) 10, , (d) 5 2, , The ratio of frequencies of two pendulums are 2 :, 3, then their length are in ratio, (a), , 2/3, , (b), , 3/2, , (c) 4 / 9, (d) 9 / 4, To show that a simple pendulum executes simple, harmonic motion, it is necessary to assume that[CPMT 46., 2001] Two pendulums begin to swing simultaneously. If, the ratio of the frequency of oscillations of the, (a) Length of the pendulum is small, two is 7 : 8, then the ratio of lengths of the two, pendulums will be, (b) Mass of the pendulum is small, [J & K CET 2005], , (c) Amplitude of oscillation is small, , 40., , g, l, , (d) Acceleration due to gravity is small, , (a) 7 : 8, , (b) 8 : 7, , The height of a swing changes during its motion, from 0.1 m to 2.5 m. The minimum velocity of a, boy who swings in this swing is, , (c) 49 : 64, , (d) 64 : 49, , (a) 5.4 m / s, , (b) 4.95 m / s, , (c) 3.14 m / s, , (d) Zero, , 47., , A simple pendulum hanging from the ceiling of a, [CPMTlift, 1997], stationary, has a time period T1. When the lift, moves downward with constant velocity, the time, period is T2, then

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[Orissa JEE 2005], , 48., , (a) Decrease the length 2 times, , (a) T2 is infinity, , (b) T2  T1, , (b) Decrease the length 4 times, , (c) T2  T1, , (d) T2 = T1, , (c) Increase the length 2 times, , If the length of a pendulum is made 9 times and, mass of the bob is made 4 times then the value of, time period becomes, , (d) Increase the length 4 times, 54., , [BHU 2005], , 49., , (a) 3T, , (b) 3/2T, , (c) 4T, , (d) 2T, , Length of a simple pendulum is l and its maximum, angular displacement is , then its maximum K.E., is, [RPMT 1995; BHU 2003], , A simple pendulum is taken from the equator to, the pole. Its period, [Kerala (PET/PMT) 2005], 55., , (a) Decreases, , (a) mgl sin , , (b) mgl (1 + sin  ), , (c) mgl (1 + cos  ), , (d) mgl (1 − cos  ), , The velocity of simple pendulum is maximum at, [RPMT 2004], , (b) Increases, , 50., , (c) Remains the same, , (a) Extremes, , (b) Half displacement, , (d) Decreases and then increases, , (c) Mean position, , (d) Every where, , A pendulum of length 2m lift at P. When it reaches, , 56., , Q, it losses 10% of its total energy due to air, resistance. The velocity at Q is, (a) 6 m/sec, (b) 1 m/sec, , P, 2m, , 57., , (c) 2 m/sec, (d) 8 m/sec, 51., , Q, , There is a simple pendulum hanging from the, ceiling of a lift. When the lift is stand still, the, time period of the pendulum is T. If the resultant, acceleration becomes g / 4 , then the new time, period of the pendulum is, , 52., , (b) 0.25 T, , (c) 2 T, , (d) 4 T, , The period of a simple pendulum measured inside, a stationary lift is found to be T. If the lift starts, , [DCE 1998], (a) Increasing, amplitude, , (b)Constant amplitude, , (c) Decreasing amplitude, , (d) First (c) then (a), , The time period of a simple pendulum of length L, as measured in an elevator descending with, g, acceleration, is, 3, [CPMT 2000], , (a) 2, , [DCE 2004], , (a) 0.8 T, , A simple pendulum is vibrating in an evacuated, chamber, it will oscillate with, ,  3L , , (b)  ,  g , , 3L, g, ,  3L , , (c) 2 ,  2g , 58., , 2L, 3g, , (d) 2, , If a body is released into a tunnel dug across the, diameter of earth, it executes simple harmonic, motion with time period, [CPMT 1999], , accelerating upwards with acceleration of, 2 Re, R, g / 3, then the time period of the pendulum is [RPMT 2000; DPMT, 2000,, = 2 03]e, (a) T, (b) T = 2, g, g, T, T, (a), (b), 3, Re, 3, (c) T = 2, (d) T = 2 seconds, 2g, 3, (c), (d) 3 T, T, 59. What is the velocity of the bob of a simple, 2, 53., , Time period of a simple pendulum will be double,, if we, [MH CET 2003], , pendulum at its mean position, if it is able to rise, to vertical height of 10cm (g = 9.8 m/s2), (a) 2.2 m/s, , A, , B, M

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(b) 1.8 m/s, (c) 1.4 m/s,  m, (a) 2 ,  k1k 2, , (d) 0.6 m/s, 60., , A simple pendulum has time period T. The bob is, given negative charge and surface below it is, given positive charge. The new time period will, ,  m, (c) 2 ,  k1 − k 2, , be [AFMC 2004], , 61., , (a) Less than T, , (b) Greater than T, , (c) Equal to T, , (d) Infinite, , 3., , What effect occurs on the frequency of a, pendulum if it is taken from the earth surface to, deep into a mine, , , , , , , A spring has a certain mass suspended from it, and its period for vertical oscillation is T. The, spring is now cut into two equal halves and the, same mass is suspended from one of the halves., The period of vertical oscillation is now, , (a), , (a) Increases, (c), , (b) Decreases, 4., , T, 2, , T, , (b), , 2, (d) 2T, , 2T, , Two masses m 1 and m 2 are suspended together, by a massless spring of constant k. When the, masses are in equilibrium, m 1 is removed without, , (d) None of these, , disturbing the system. Then, frequency of oscillation of m 2 is, , Spring Pendulum, 1., ,  m, (d) 2 ,  k1 + k 2, , , , , , , [MP PET 1995], , [AFMC 2005], , (c) First increases then decrease, , k , (b) 2 m  1 ,  k2 , , , , , , , the, , angular, , Two bodies M and N of equal masses are, suspended from two separate massless springs of, force constants k1 and k2 respectively. If the two, bodies oscillate vertically such that their, maximum velocities are equal, the ratio of the, amplitude M to that of N is, [IIT-JEE 1988; MP PET 1997, 2001; MP PMT 1997;, , 5., , (a), , k, m1, , (b), , k, m2, , (c), , k, m1 + m 2, , (d), , k, m 1m 2, , In arrangement given in figure, if the block of, mass m is displaced, the frequency is given by, , BHU 1998; Pb. PMT 1998; MH CET 2000, 03; AIEEE 2003], , [BHU 1994; Pb. PET 2001], , k, (a) 1, k2, , (b), , k1, k2, , A, , B, m, , K1, , k, (c) 2, k1, 2., , (d), , k2, k1, , (a) n =, , 1, 2, ,  k1 − k 2 , , ,  m , , (b) n =, , 1, 2, ,  k1 + k 2 , , ,  m , , (c) n =, , 1, 2, ,  m, , k +k, 2,  1, , (d) n =, , 1, 2, ,  m, , k −k, 2,  1, , A mass m is suspended by means of two coiled, spring which have the same length in unstretched, condition as in figure. Their force constant are k1, and k2 respectively. When set into vertical, vibrations, the period will be [MP PMT 2001], A, , B, , k1, , k2, , m, , 6., , K2, , , , , , , , , , , , Two identical spring of constant K are connected, in series and parallel as shown in figure. A mass, m is suspended from them. The ratio of their, , frequencies of vertical oscillations will be [MP PET 1993; B, , K, K, , K, m, , K

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equal halves and a mass 2m is suspended from it., The frequency of oscillation will now become, [CPMT 1988], , 12., , 7., , (a) 2 : 1, , (b) 1 : 1, , (c) 1 : 2, , (d) 4 : 1, , suspended mass will be, [CBSE PMT 1990; Pb. PET 2002], , , , , , ,  m (K 1 + K 2 ) , , (c) T = 2 , ,  K1 K 2, , , 8., ,  m, (b) T = 2 ,  K1 + K 2, , , , , , ,  mK 1 K 2, (d) T = 2 ,  K1 + K 2, , , , , , , (b) 2n, , (c) n / 2, , (d) n(2)1 / 2, , A mass M is suspended from a light spring. An, additional mass m added displaces the spring, , further by a distance x. Now the combined mass, will oscillate on the spring with period[CPMT 1989, 1998 ; U, , A mass m is suspended from the two coupled, springs connected in series. The force constant for, springs are K 1 and K 2 . The time period of the, ,  m, (a) T = 2 ,  K1 + K 2, , (a) n, , (a) T = 2, , (mg / x(M + m)), , (b) T = 2, , ((M + m)x / mg ), , (c) T = ( / 2) (mg / x(M + m)), (d) T = 2, 13., , ((M + m) / mgx ), , In the figure, S 1 and S 2 are identical springs. The, oscillation frequency of the mass m is f . If one, , spring is removed, the frequency will become [CPMT 1971], , A spring is stretched by 0.20 m, when a mass of, 0.50 kg is suspended. When a mass of 0.25 kg is, suspended, then its period of oscillation will be, , A, , (g = 10 m / s 2 ), , 9., , 10., , S1, , (a) 0.328 sec, , (b) 0.628 sec, , (c) 0.137 sec, , (d) 1.00 sec, , A mass M is suspended from a spring of negligible, mass. The spring is pulled a little and then, released so that the mass executes simple, harmonic oscillations with a time period T. If the, mass is increased by m then the time period, m, 5 , becomes  T  . The ratio of, is, [CPMT 1991], M, 4 , , 14., , B, m, , S2, , (a) f, , (b) f  2, , (c) f  2, , (d) f / 2, , The vertical extension in a light spring by a, weight of 1 kg suspended from the wire is 9.8 cm., The period of oscillation, [CPMT 1981; MP PMT 2003], , (a) 9/16, , (b) 25/16, , (a) 20 sec, , (b) 2 sec, , (c) 4/5, , (d) 5/4, , (c) 2 / 10 sec, , (d) 200  sec, , A spring having a spring constant ‘K’ is loaded, with a mass ‘m’. The spring is cut into two equal, parts and one of these is loaded again with the, , 15., , A particle of mass 200 gm executes S.H.M. The, restoring force is provided by a spring of force, , constant 80 N / m. The time period of oscillations, same mass. The new spring constant is [NCERT 1990; KCET 1999;, is, [MP PET 1994], Kerala PMT 2004; BCECE 2004], , 11., , (a) K / 2, , (b) K, , (c) 2K, , (d) K 2, , A weightless spring which has a force constant, oscillates with frequency n when a mass m is, suspended from it. The spring is cut into two, , 16., , (a) 0.31 sec, , (b) 0.15 sec, , (c) 0.05 sec, , (d) 0.02 sec, , The length of a spring is l and its force constant is, k. When a weight W is suspended from it, its, length increases by x. If the spring is cut into two

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equal parts and put in parallel and the same, weight W is suspended from them, then the, extension will be, (a) 2 x, (c), 17., , 21., , K 2 . Both are stretched till their elastic energies, , [MP PMT 1994], , are equal. If the stretching forces are F1 and F2 ,, then F1 : F2 is, , (b) x, , x, 2, , (d), , [MP PET 2002], , x, 4, , A block is placed on a frictionless horizontal table., The mass of the block is m and springs are, attached on either side with force constants K 1, , 22., ,  K1 K 2 , (b) , ,  m (K1 + K 2 ) , ,  K1 K 2 , (c) , ,  (K1 − K 2 )m , , 18., , 1/2, ,  K 2 + K 22 , (d)  1, ,  (K1 + K 2 )m , , 19., , 20., , (c) 2 : 3, , (d) 2 : 1, , (d) K12 : K 22, , K1 : K 2, , A mass m is vertically suspended from a spring of, , spring, , 1/2, , 1/2, , 23., , [CBSE PMT 1998], , (a) n / 4, , (b) 4 n, , (c) n / 2, , (d) 2n, , If the period of oscillation of mass m suspended, from a spring is 2 sec, then the period of mass 4m, will be, [AIIMS 1998], , [Manipal MEE 1995], , (b) 1 : 2, , (c), , system if a mass 4 m is suspended from the same, , A uniform spring of force constant k is cut into, two pieces, the lengths of which are in the ratio 1, : 2. The ratio of the force constants of the shorter, and the longer pieces is, (a) 1 : 3, , (b) K2 : K1, , frequency n. What will be the frequency of the, , oscillate, then the angular frequency of oscillation, will be, [MP PMT 1994], 1/2, , (a) K1 : K2, , negligible mass; the system oscillates with a, , and K 2 . If the block is displaced a little and left to, ,  K + K2 , (a)  1, ,  m, , , The force constants of two springs are K 1 and, , 24., , (a) 1 sec, , (b) 2 sec, , (c) 3 sec, , (d) 4 sec, , Five identical springs are used in the following, , three configurations. The time periods of vertical, A mass m =100 gms is attached at the end of a, light spring which oscillates on a frictionless, oscillations in configurations (i), (ii) and (iii) are, horizontal table with an amplitude equal to 0.16, in the ratio, [AMU 1995], metre and time period equal to 2 sec. Initially the, mass is released from rest at t = 0 and, displacement x = −0.16 metre. The expression for, K, K, K, K, the displacement of the mass at any time t is [MP PMT 1995], (a) x = 0 .16 cos(t), , (b) x = − 0.16 cos(t), , (c) x = 0 .16 sin(t +  ), , (d) x = − 0.16 sin(t +  ), , m, , (i), , A block of mass m, attached to a spring of spring, constant k, oscillates on a smooth horizontal, table. The other end of the spring is fixed to a, wall. The block has a speed v when the spring is, at its natural length. Before coming to an, instantaneous rest, if the block moves a distance x, from the mean position, then, (a) x = m / k, , 1, m /k, (b) x =, v, , (c) x = v m / k, , (d) x = mv / k, , K, m, , m, , (iii, ), , (ii), (a) 1 : 2 :, , (c), 25., , 1, 2, , 1[MP PET 1996], : 2 :1, 2, , (b) 2 : 2 :, , (d) 2 :, , 1, , 1, 2, , :1, , 2, , A mass m performs oscillations of period T when, hanged by spring of force constant K. If spring is, cut in two parts and arranged in parallel and

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(c) K1 − K2, , same mass is oscillated by them, then the new, time period will be, , 30., , [CPMT 1995; RPET 1997; RPMT 2003], , (a) 2T, , (d) K1 K2 / K1 − K2, , A mass m attached to a spring oscillates every 2, sec. If the mass is increased by 2 kg, then timeperiod increases by 1 sec. The initial mass is, AIIMS 2000; MP PET 2000; DPMT 2001; Pb. PMT 2003], , (b) T, (c), , K, , T, 2, , m, m, , 31., , (b) 3.9 kg, , (c) 9.6 kg, , (d) 12.6 kg, , A mass M is suspended by two springs of force, , T, (d), 2, , 26., , (a) 1.6 kg, , constants K1 and K2 respectively as shown in the, diagram. The total elongation (stretch) of the two, springs is, , If a watch with a wound spring is taken on to the, , [MP PMT 2000; RPET 2001], , moon, it, [AFMC 1993], , 27., , (a) Runs faster, , (b) Runs slower, , (c) Does not work, , (d) Shows no change, , What will be the force constant of the spring, system shown in the figure, , (a), , Mg, K1 + K 2, Mg (K1 + K 2 ), K1 K 2, , K1, , (b), , (c), , Mg K1 K 2, K1 + K 2, , K2, , (d), , K1 + K 2, K1 K 2 Mg, , [RPET 1996; Kerala (Med./ Engg.) 2005], , (a), , K1, + K2, 2, ,  1, 1 , +, (b) , ,  2 K1 K 2 , , −1, , K1, , K1, , 1, 1, (c), +, 2 K1 K 2,  2, 1 , +, (d) , ,  K1 K1 , , 28., , 32., , (a), , 1, 2, , K, m, , (b), , 1, 2, , (K1 + K 2 )m, K1 K 2, , m, , (c) 2, , Two springs have spring constants K A and K B, and K A  KB . The work required to stretch them, by same extension will be, , 29., , The frequency of oscillation of the springs shown, in the figure will be, [AIIMS 2001; Pb. PET 2002], , K2, , −1, , (d), , (b) More in spring B, , (c) Equal in both, , (d) Noting can be said, , 33., , K, m, , K2, , K1 K 2, m (K1 + K 2 ), , m, , [RPMT 1999], , K2, , 34., (b) K1 K2 / K1 + K2, , The scale of a spring balance reading from 0 to 10, kg is 0.25 m long. A body suspended from the, balance oscillates vertically with a period of  /10, second. The mass suspended is (neglect the mass, of the spring), , The effective spring constant of two spring, system as shown in figure will be, , (a) K1 + K2, , 1, 2, , K1, , [RPMT 1999], , (a) More in spring A, , K1, , m, , [Kerala (Engg.) 2001], , (a) 10 kg, , (b) 0.98 kg, , (c) 5 kg, , (d) 20 kg, , If a spring has time period T, and is cut into n, equal parts, then the time period of each part will, be, [AIEEE 2002]

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35., , (a) T n, , (b) T/ n, , (c) nT, , (d) T, , 40., , (a), , 3, K, 4, , (b), , (c) K, 36., , effective force constant of the spring will be, , One-forth length of a spring of force constant K is, cut away. The force constant of the remaining, spring will be, [MP PET 2002], , 4, K, 3, , Infinite springs with force constant k, 2k, 4k and, 8k.... respectively are connected in series. The, , 41., , (a) 2K, , (b) k, , (c) k/2, , (d) 2048, , To make the frequency double of a spring, oscillator, we have to, [CPMT 2004; MP PMT 2005], (a) Reduce the mass to one fourth, , (d) 4 K, , A mass m is suspended separately by two, different springs of spring constant K1 and K2, gives the time-period t1 and t 2 respectively. If, , (b) Quardruple the mass, , same mass m is connected by both springs as, shown in figure then time-period t is given by the, relation, , (d), , (c) Double of mass, , 42., , The springs shown are identical. When A = 4 kg ,, the elongation of spring is 1 cm. If B = 6 kg , the, , [CBSE PMT 2002], , (a) t = t1 + t 2, (b) t =, , Half of the mass, , elongation produced by it is, , t1 .t2, t1 + t2, , K, , [, , K, , (c) t 2 = t1 2 + t 2 2, (d) t −2 = t1 −2 + t 2 −2, 37., , m, A, , Two springs of force constants K and 2K are, connected to a mass as shown below. The, frequency of oscillation of the mass is [RPMT 1996; DCE 2000; AIIMS 2003], K, , B, , 2K, m, , 38., , K, , (a) (1 / 2 ) (K/m ), , (b) (1 / 2 ) (2 K /m ), , (c) (1 / 2 ) (3 K/m ), , (d) (1 / 2 ) (m /K ), , 43., , (a) 4 cm, , (b) 3 cm, , (c) 2 cm, , (d) 1 cm, , When a body of mass 1.0 kg is suspended from a, certain light spring hanging vertically, its length, increases by 5 cm. By suspending 2.0 kg block to, the spring and if the block is pulled through 10 cm, , Two springs of constant k 1 and k 2 are joined in, , and released the maximum velocity in it in m/s is, , series. The effective spring constant of the, combination is given by, , : (Acceleration due to gravity = 10 m /s 2 ), , [CBSE PMT 2004], , (a), , k 1k 2, , (c) k1 + k 2, 39., , (a) 0.5, , (b) 1, , (d) k1k 2 /(k1 + k 2 ), , (c) 2, , (d) 4, , A particle at the end of a spring executes simple, harmonic motion with a period t1 , while the, corresponding period for another spring is t 2 . If, the period of oscillation with the two springs in, series is T, then, [AIEEE 2004], (b) T 2 = t12 + t 22, , (a) T = t1 + t2, (c) T, , −1, , =, , t1−1, , +, , t 2−1, , [EAMCET 2003], , (b) (k1 + k 2 ) / 2, , (d) T, , −2, , =, , t1−2, , +, , t 2−2, , 44., , Two springs with spring constants K1 = 1500 N /m, and, , K 2 = 3000 N /m, , are stretched by the same, , force. The ratio of potential energy stored in, spring will be, [RPET 2001], (a) 2 : 1, , (b) 1 : 2, , (c) 4 : 1, , (d) 1 : 4

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45., , If a spring extends by x on loading, then energy, stored by the spring is (if T is the tension in the, spring and K is the spring constant), (a), , (c), 46., , T2, 2x, , (b), , 2K, T, , (d), , 2, , 50., , T2, 2K, 2T, K, , 51., , constant 200 N/m is kept straight and, unstretched on a smooth horizontal table and its, ends are rigidly fixed. A mass of 0.25 kg is, attached at the middle of the spring and is slightly, displaced along the length. The time period of the, oscillation of the mass is, [MP PET 2003], , (c), 47., , , 20, , , s, 5, , s, , (b), , , 10, , 52., , s, , , , (d), , (a), , 1, sec, 7, , (b) 1 sec, , (c), , 2, sec, 7, , (d), , 2, , A weightless spring of length 60 cm and force, , (a), , When a mass m is attached to a spring, it, normally extends by 0.2 m. The mass m is given a, slight[AFMC, addition, 2000], extension and released, then its, time period will be, [MH CET 2001], , 2, sec, 3, , If a body of mass 0.98 kg is made to oscillate on a, spring of force constant 4.84 N/m, the angular, frequency of the body is, (a) 1.22 rad/s, , (b) 2.22 rad/s, , (c) 3.22 rad/s, , (d) 4.22 rad/s, , A mass m is suspended from a spring of length l, and force constant K. The frequency of vibration, of the mass is f1 . The spring is cut into two equal, parts and the same mass is suspended from one of, the parts. The new frequency of vibration of mass, is f2 . Which of the following relations between, , s, , 200, , the frequencies is correct, [NCERT 1983; CPMT 1986; MP PMT 1991; DCE 2002], , The time period of a mass suspended from a, spring is T. If the spring is cut into four equal, , (a) f1 = 2 f2, (b) f1 = f2, parts and the same mass is suspended from one of, (c) f1 = 2 f2, (d) f2 = 2 f1, the parts, then the new time period will be[MP PMT 2002; CBSE PMT, 2003], 53. A mass m oscillates with simple harmonic motion, T, , (a) T, (b), with frequency f =, and amplitude A on a, 2, 2, (c) 2 T, 48., , 49., , (d), , spring with constant K , therefore, , T, 4, , (a) The total energy of the system is, , 1, KA 2, 2, , A mass M is suspended from a spring of negligible, , 1, 2, , K, M, , mass. The spring is pulled a little and then, released so that the mass executes S.H.M. of time, , (b) The frequency is, , period T. If the mass is increased by m, the time, period becomes 5T/3. Then the ratio of m/M is, , (c) The maximum velocity occurs, when x = 0, , (a), , 5, 3, , (b), , 3, 5, , (c), , 25, 9, , (d), , 16, 9, , (d) All, the above, [AIEEE, 2003]are correct, 54., , Two masses m1 and m2 are suspended together by, a massless spring of constant K. When the masses, are in equilibrium, m1 is removed without, disturbing the, oscillations is, , An object is attached to the bottom of a light, vertical spring and set vibrating. The maximum, speed of the object is 15 cm/sec and the period is, , (a), , m1 g, K, , 628 milli-seconds. The amplitude of the motion in, centimeters is, [EAMCET 2003], , (b), , m2g, K, , (c), , (m 1 + m 2 )g, K, , (a) 3.0, (c) 1.5, , (b) 2.0, (d) 1.0, , system., , The, , amplitude, , [J & K CET 2005], , m1, m2, , of, , [

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(d), 55., , (m 1 − m 2 )g, K, , (c) Ellipse, 6., , A spring executes SHM with mass of 10kg, attached to it. The force constant of spring is, 10N/m.If at any instant its velocity is 40cm/sec,, the displacement will be (where amplitude is, 0.5m), [RPMT 2004], , 2., , Two mutually perpendicular simple harmonic, vibrations have same amplitude, frequency and, phase. When they superimpose, the resultant, form of vibration will be, [MP PMT 1992], , (a) 0.09 m, , (b) 0.3 m, , (a) A circle, , (b) An ellipse, , (c) 0.03 m, , (d) 0.9 m, , (c) A straight line, , (d) A parabola, , Superposition of S.H.M’s and Resonance, 1., , (d) Figure of eight, , 7., , the relation x = 4(cost + sint). The amplitude of, the particle is, , The S.H.M. of a particle is given by the equation, y = 3 sin  t + 4 cos  t . The amplitude is[MP PET 1993], (a) 7, , (b) 1, , (c) 5, , (d) 12, , If the displacement equation of a particle be, represented by y = A sin PT + B cos PT , the particle, , The displacement of a particle varies according to, , [AIEEE 2003], , 8., , (a) 8, , (b) – 4, , (c) 4, , (d) 4 2, , A, , S.H.M., , is, , represented, , by, , x = 5 2 (sin 2t + cos 2t). The amplitude of the S.H.M., , executes, , is, [MP PET 1986], , [MH CET 2004], , (a) 10 cm, , (b) 20 cm, , (c) 5 2 cm, , (d) 50 cm, , (a) A uniform circular motion, (b) A uniform elliptical motion, 9., , (c) A S.H.M., , Resonance is an example of, [CBSE PMT 1999; BHU 1999; 2005], , (d) A rectilinear motion, 3., , The motion of a particle varies with time, according to the relation y = a(sin  t + cos  t) , then, (a) The motion is oscillatory but not S.H.M., , 4., , 10., , (b) Forced vibration, , (c) Free vibration, , (d) Damped vibration, , (b) The motion is S.H.M. with amplitude a, , In case of a forced vibration, the resonance wave, becomes very sharp when the, , (c) The motion is S.H.M. with amplitude a 2, , (a) Restoring force is small, , (d) The motion is S.H.M. with amplitude 2a, , (b) Applied periodic force is small, , The resultant of two rectangular simple harmonic, , (c) Quality factor is small, , motions of the same frequency and unequal, amplitudes but differing in phase by, , , is[BHU 2003;, 11., 2, , CPMT 2004; MP PMT 1989, 2005; BCECE 2005], , 5., , (a) Tuning fork, , (a) Simple harmonic, , (b) Circular, , (c) Elliptical, , (d) Parabolic, , The composition of two simple harmonic motions, of equal periods at right angle to each other and, with a phase difference of  results in the, , (d) Damping force is small, Amplitude of a wave is represented by, A=, , c, a+b −c, , Then resonance will occur when, (a) b = −c / 2, (b) b = 0 and a = – c, (c) b = −a / 2, , (d) None of these, , displacement of the particle along, , A particle with restoring force proportional to, displacement and resisting force proportional to, [CBSE, 1990] to a force F sin t . If the, velocity, is PMT, subjected, , (a) Straight line, , amplitude of the particle is maximum for  = 1, , (b) Circle, , 12., , [, , [

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and the energy of the particle is maximum for,  =  2 , then (where 0 natural frequency of, oscillation of particle), , [CBSE PMT 1998], , (a) 1 =  0 and  2   o (b) 1 =  0 and  2 =  o, (c) 1   0 and 2 = o, 13., , (d) 1   0 and 2  o, , A simple pendulum is set into vibrations. The bob, of the pendulum comes to rest after some time, due to, [AFMC 2003; JIPMER 1999], , (a) Air friction, (b) Moment of inertia, (c) Weight of the bob, (d) Combination of all the above, 14., , A simple pendulum oscillates in air with time, period T and amplitude A. As the time passes, (a) T and A both decrease, (b) T increases and A is constant, (c) T increases and A decreases, (d) T decreases and A is constant, , [CPMT 2005]