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Excellent Notes, , By Prof. Chavan Sir, , Oscil lations, Q.3) Describe a spring and block oscillator with a neat, diagram., Ans., 1), Consider a block of mass m attached to an ideal spring, offeree constant K and free to move over a frictionless, horizontal surface., 2), In equilibrium postition, the spring is neither stretched, nor compressed, 3), If now the block is pulled towards right, the spring is, streched, Due to elastic properties, the spring tends to regain, its orignial shape and size. Therefore it exerts a restoring, force on block., 4), When the block returns to its mean position, it doesen’t, stop & continues to move beyond the mean position, towards left., , Q.1) Define linear S.H.M. Give its examples., Ans., Linear S.H.M. :, Linear S.HM. is defined as a linear, oscillatory, periodic, motion of like particle, in which;, i), The magnitude of restoring force (F) or acceleration, acting on a particle is directly proportional to the, displacement of particle measured from its mean, position and, ii), The force (or acceleration) is always directed towards, the mean position (i.e., opposite to linear displacement), i.e., F x, F = – kx, Negative sign indicate that force (or accleration) and, displacement are oppositely directed., Examples :, 1., Motion of the needle of sewing machine., 2., Motion of simple pendulum with small swing., 3., Motion of the body suspended from the spring., 4., Motion of pendulum of pendulum clock., Q.2) What is force constant? Give its units and, dimensions., Ans., 1), For a particle performing linear S.H.M.,, Force (displacement), F x, F = – Kx, where,, K = force constant or spring constant, Negative sign indicates that force and displacement, are oppositely directed., 2), In magnitude,, F = Kx, , 3), , 4), , 5), , During this motion, the spring is compressed and it, again exerts a restoring force on the block, 6., This process goes on repeating causing the block to, oscillate on the two sides of its mean position such, oscillatory or vibratory motion along a straight line is, called linear S.H.M., 7., In both the cases (compressed and relaxed state), restoring force is proportional to the displacement and, its direction is opposite to that of displacement, F x, x, F = – Kx, Where K is called force constant, For constant (K), Q.4), In oscillatory motion of a body, define., The magnitude of restoring force per unit, 1. Mean position, 2. Extreme position, displacement is called force constant., 3. Displacement, 4. Amplitude and, Units :, 5., Range, or, path, length., i. S.l. Unit : N/m, Ans :, ii. C.G. S. unit : dyne/cm, Consider a particle P performing linear S.H.M., Dimensions : [K] = [M1L0T–2], Note :, Restoring force in vector form :, , For-XII-NEET, Medical, Std. Board, , 5., , 2, , Oscillations (XII)

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Excellent Notes, , By Prof. Chavan Sir, , In the figure,, O = mean position, A and B = extreme positions, OP = x= displacement of the particle, OA = OB = a=amplitude of S.H.M., AB = range or path length, AB = 2 a, 1., Equilibrium or mean position :, The position about which a body oscillates equally on, two sides (or at which the net force acting on the boy, is zero )is called the mean equilibrium position of the, body., O is the mean position of the particle., 2., Extreme position :, The position in which the displacement of the particle, from its mean position is maximum is called its extreme, position., A and B are the -extreme positions., 3., Displacement (x) :, The distance covered by the particle performing, S.H.M. from its mean position is called displacement, The distance OP = x is a displacement of the particle, at any point P., S.I. unit : m, C.G.S. unit : cm, Dimensions : [x]= [M0 L1 T0], 4., Amplitude of S. H. M. (a) :, The magnitude of maximum displacement of a particle, performing S.H.M. from its mean position is called, amplitude of S.H.M, It is denoted by a., The distance OA = OB is a amplitude of S. H. M., OA = OB = a, S.I. unit : m, C.G.S. unit : cm, Dimensions : [M°L1T°], 5., Range or path length of S. H. M. :, The distance between two extreme positions of a, particle performing S.HM. is known as path length or, range of S.H.M., The distance AB = range or path length., The distance AB = 2 a., Questions Bank :, 1., Define apmlitude of S.H.M. (1 .M), Q.5) Define, 1. Period of S.H.M., 2. Frequency of SJLM. -2M*, Ans., 1. Period of S.H.M. (T) :, The time taken by a particle performing S.HM. to, complete one oscillation is called period of S.HM., For-XII-NEET, Medical, Std. Board, , 2., , It is denoted by T., S. L Unit : sec., Dimensions : [ T ] = [M0L0T1], Frequency (n) :, The number of oscillations of a particle performing, S.HM. in one second is called frequency of S.H.M., It is denoted by n. It is the reciprocal of time period., , S. L Unit : hertz (Hz), Dimensions : (n ) = [M0L0T–1], Q.6) The differential equation of linear S.H.M., Ans., i), A particle performing linear S.H.M. satisfies the, following two conditions,, ii), The magnitude of force is directly proportional to the, displacement from the mean position and, iii) The force acting on the particle is directed towards, the mean position, i.e., force and displacement are, appositely directed., F x, F = – Kx, .......(i), Where, K = the force constant (force per unit displacement ), velocity (v) of the particle is rate of change of, displacement w.r.t. time and acceleration is the rate of, change of velocity w.r.t. time., , a, , ........(ii), 3., , According to Newton’s II law of motion, Force = mass × acceleration, F=m×a, , 4., , Substituing this value in equation (i), wer get, dt, , 3, , Oscillations (XII)

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Excellent Notes, , Let,, , By Prof. Chavan Sir, , K, m, , This is the differential equation of lienar S.H.M., Q.7) From the differential equation of linear S.H.M., obtain an expression for acceleration, velocity, and displacement of a particle performing linear 6., S.H.M., Ans., 1., Differential equation of linear S.H.M: is given by,, , 2., , d2x, .......(i), dt 2, Where, x = displacement of the particle, = angular speed of the particle, From equation (i),, , Putting this value of C in equation (ii),, , .......(iii), This is the expression for velocity., , d2x, dt 2, But,, , 3., , But, v, , d2x, = acceleration, dt 2, , Acceleration is rate of change of velocity., i.e. acceleration, , But,, , 7., , Integrating the above equation on both sides, , dx, dt, where = constant of integration,, , 4., , 5., , This is the expression for displacement., Integrating the above equation on both sides,, Q.8) Write equation of displacement of a particle, performing linear S.H.M. Hence, find the, displacement of motion when particle starts from, (i) mean position (ii) extreme position., Ans., i), Displacement of a particle performing linear S.H.M., .......(ii), is, x = a sin ( t + ), ...... (i), Where C is the constant of integration., ii), If particle starts from mean position then x = 0, To find value of C :, when t = 0, At extreme position, x = a and v = 0, 0 = a sin [ (0) = ], Putting these values in equation (ii),, 0 = a sin [0 + ], , For-XII-NEET, Medical, Std. Board, , 4, , Oscillations (XII)

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Excellent Notes, , By Prof. Chavan Sir, , 0 = a sin, 0 = sin, sin = 0, But sin 0 = 0, =0, Substituting this value in equation (i), we get, iii), , 7., , In OPM sin( t, , )=, , OP, , a, x =a sin ( t + ), , ......(ii), , Velocity of the particle is the rate of change of, displacment w.r.t. time., , If particle starts from extreme position then x = a when, t=0, , 2, The acceleration is the rate of change of velocity w.r.t., time., , Q.9) Show that linear S.H.M. is a projection of uniform, circular motion on any diameter., Ans :, 1., Consider a particle P performing U.C.M. witr angular, velocity in an anticlockwise direct direction as shown, in figure., , acceleration =, , d, v, dt, , acceleration =, , d, dt, , ... from eqn (iii), , acceleration =, acceleration =, acceleration =, acceleration =, , 2., , 3., , 4., , 5., , acceleration = – 2x, , Let, O = centre of the circle, a = radius of the circle and, AB = CD = diameter of the circle., Suppose that particle starts from the point, P1 such that P1OD =, In time t, it moves from P1 to P such that P1OP= t,, Then, DOP = ( t + ), .......(i), Draw PM perpendicular to AB from point P., Therefore particle M is the projection of the particle, P on diameter AB., Expression for displacement (x) :, At time t, OP = a and the displacement of particle M, is OM = x, OPM = DOP, .......alternate angles, OPM = ( t, ), .......from equation (i), , For-XII-NEET, Medical, Std. Board, , ..... from eqn(ii), , Hence, acceleration of a particle (M) is, i), , directly proportional to its displacement from mean, position and, , ii), , the negative sign indicates that they are oppositely, directed., Hence, of the particle M performs linear S.H.M., , 10., , But particle Mis the projection of the particle P on, diameter AB. The particle P is performing UCM., Therefore, we can say that S.H.M. is a projection of, U.C.M. on any diameter., Question bank : 1., , 1., , 5, , Prove that the projection ofU.CM. on any diameter, of the circle is linear S.HM. (3.M), Oscillations (XII)

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Excellent Notes, , By Prof. Chavan Sir, , Q10) Define :, 1. Phase of S.H.M. and, 2. Epoch of S.H.M., 3., Ans. 1. Phase of SHM:, A quantity which describes the stale of oscillation, (position and direction) of the particle performing, S.HM. at any instant is called phase of S.HM., i), The general expression for displacement is, x = a sin ( t + a), In this expression, the angle ( t + a) is called phase, of S.H.M. or phase angle., 2., Epoch of S.H.M. :, A physicsl quantity which describes the state of, oscillations of a particles performing S.H.M. at time, t = 0 is called epoch of S.H.M., It is also known as initial phase or starting phase, S.H.M. or phase constant of S.H.M. In the equation, x = a sin ( t + ) the angle is called epoch or initial, phase of S.H.M., Q.11) Derive an expression for periodic time of a, particle performing S.H.M. in terms of, 1. Angular velocity, 2. Force constant and, 3. Acceleration., Ans., 1., Displacement (x) of a particle performing S.H.M. at, time (t) is given by, x = a sin ( t + a), .... (i), 4., 2., Let,, x1 be the displacement after time interval, , 2, , T in terms of force constant (K) :, m, m, Where, K = force constant, m= mass of particle, , m, , m, , T in terms of acceleration :, acceleration = – 2x, , Then, x, , But, sin (2 + ) = sin, , ......(ii), From equation (i) and (ii), x = x1, This shows that displacement is repeated after time, 2, 2, or particle completes one oscillation intime, ,, Hence,, , 2, , 2, acceleration, displacement, , is the period of S.H.M., ......(iii), , For-XII-NEET, Medical, Std. Board, , 6, , Oscillations (XII)

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Excellent Notes, , By Prof. Chavan Sir, , Q.12)If the displacement of a particle perfoming linear, S.H.M. is given by x = a sin( t + ) find an, expression for the displacement when the particle, starts from the, 1. mean position, 2. extreme position., Ans., 1., The general expression for displacement is, x = a sin ( t + ), ... (i), Where,, a = amplitude of S.H.M., = angular velocity ., = initial phase, t = time, 2., If the particle starts from mean position, = n, Hence the displacement is given by, 3., , 6., i), , If the particle starts from extreme position,, , ii), Putting this value in equation (i), we get, iii), , Characteristics of the graphs :, From graphs we can say displacement, velocity and, acceleration of S.H.M. are periodic functions of time., Displacement - time, curve and acceleration time curve, are sine curves and velocity - time curve is a cosine, curve., Phase difference between displacement and velocity, is, , iv), Q.13)Particle performing S.H.M. starting from mean, position. a. Plot a graph of displacement, velocity, and acceleration against time., v), Ans :, 1., If particle starts from mean position its displacement, is given by, vi), x = a sin t, vii), v=a, , cos, , acceleration = – a, Time(t), , rad., 2, Phase difference between displacement and, acceleration is rad., The displacement and acceleration are oppositely, directed,, All the curves are similar in nature. Thus displacement, velocity and acceleration vary harmonically with phase, t, viii) All curves represent same path after phase of (2 ), radian., Question bank:, 1., Represent graphically the displacement, velocity and, acceleration against time for a particle performing linear, SHM when it starts from the mean posim. (3M.), 2., Represent graphically displacement, velocity and, acceleration when the particle starts from mean, position. What conclusions can you draw from these, graphs?, (4 M.), 3., Represent graphically the displacement and, acceleration against time of a particle in SHM when it, starts from mean position. Write their corresponding, equations., (2 .M), , T, cos t, , 2, t, T, acceleration = – a, v=a, , 0, , T, 4, , 2, , sin t, , 2, , sin, , 2, t, T, , T, 2, , 3T, 4, , For-XII-NEET, Medical, Std. Board, , T, , rad., 2, Phase difference between velocity and acceleration is, , 5T, 4, , 7, , Oscillations (XII)

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Excellent Notes, , By Prof. Chavan Sir, , Q.14)Particle perfoming S.H.M. starting from extreme iv., position. a. Plot a graph of displacement, velocity, and acceleration against time., Ans :, v., 1., If particle starts from extreme position its displacement, is given by, x = a cos t, vi., T, , T, acceleration = – a, , Time(t), , 0, , T, 4, , i., ii., , If the origin of one curve is shifted by, , viii., 2, , T, 2, , ix., 3T, 4, , x., , 5T, 4, , T, , rad., 2, Phase difference between velocity and acceleration is, , is, , rad., 2, Phase difference between displacement and, acceleration is rad., The displacement and acceleration are oppositely, directed., All the curves are similar in nature. Thus displacement, velocity and acceleration vary harmonically with phase, t., All curves represent same path after phase of (2 ), radian., The displacement and acceleration is maximum at, extreme position where as velocity is minimum at the, same path after phase of 2 radian., Question bank :, Draw graphs representing the variation of displacement, with phase for a paricle performing S.H.M. when it, starts from the i. Mean position ii. Extreme position, What conclusions can you draw from these graphs., Hint :, The nature of graph is same in both the cases., , vii., , cos t, 2, acceleration = –a 2 cos, T, , Phase difference between displacement and velocity, , 1), , 2, , radian, the, , other curve is obtained., From differential equation of linear S.H.M., obtain an, expression for velocity of a particle performing linear, S.H.M., 3), Represent graphically displacement, velocity and, acceleration when the particle starts from extreme, position. What conclusions can you draw from the, graphs?, 4), Give graphical representation of S.H.M. when particle, starts from the positive extreme position., 5), A particle performing S.H.M. starts from extreme, position. Plot the graphs of velocity and displacement, against time., 6), Define phase of S.H.M. Show variation of, displacement, velocity and acceleratin with phase for, a particle performing linear S.H.M. graphically, when, it starts from extreme position., [Definition of phase of SHM... 1M Graph of, displacement -1M... Graph of velocity -1M... Graph, of acceleration -1M], Q.15)Derive an expression for K. E. of a particle, performing S.H.M. Find its values at (a) mean, position (b) extreme position. Draw the graph, showing variation of K.E. with displacement., 2), , 6), i., ii., , iii., , Characteristics of graph :, From graphs we can say displacement, velocity and, acceleration of S.H.M. are periodic functions of time., Displacement- time curve, and acceleration time curve, are sine curves and velocity - time curve is a cosine, curve., When the particle starts from extreme position instead, of mean position, the curves are again of the same, nature, but origin is shifted by an angle of, , 2, , rad or, , 90o., For-XII-NEET, Medical, Std. Board, , 8, , Oscillations (XII)

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Excellent Notes, , By Prof. Chavan Sir, , Ans :, 1), , 2), , 3), 4), , Expression for K. E. :, Consider a particle performing linear S.H.M. along, a straight line AB., , 8), , Let,, O = the mean position of the particle, a = amplitude of S.H.M., m = mass of the particle, = angular velocity of the particle, Consider a point P at a distance x from O., OP = x, The velocity of particle performing S.H.M. at any, point P is given by,, , Q.16)Derive an expression for P.E. of a particle, performing S.H.M. Find its values at (a) mean, position (b) extreme position. Draw the graph, showing variation of P.E. with displacement., Ans :, Expression for P.E. :, 1), Consider a particle performing linear S.H.M. along a, straight line AB as shown in figure., , v=, 5), , The K.E. of particle is given by, K.E. =, =, , 1, mv2, 2, 1, m(, 2, , 2), , ), , 3), , 1, K.E. = m 2(a2 – x2), 2, Also,, 1, K(a2 – x2), 2, K.E. at mean position :, At mean position, x = 0, K.E. =, , 6), , Thus, K.E. is minimum at extreme position, Graph :, Graph showing variation of K.E. with displacement is, given below., , K.E. =, , 1, K(a2 – 0), 2, , K.E. =, , 1 2, Ka, 2, , 4), ...(, , m, , 2, , = K), , 5), , When particle performs linear S.H.M. it has P.E. due, to its position., The restoring force acting on the particle is given by, F=–Kx, where,, K = force constant, If the particle is given a small displacement dx against, the direction of force the workdone dw on the particle, is, workdone = force displacement, dw = – F dx, dw = – (–Kx) dx, dw = K dx, Total workdone when the particle is displaced from, O to x is given by integrating the above equation, between the limits 0 to x., =, , 7), , Therefore, at mean position K.E. is maximum., K.E. at extreme position :, At extreme position,, x=a, , =K, , K. E. =, , 1, K(a2 – a2), 2, , W=K, , K. E. =, , 1, K(0), 2, , W=, , 1, K[x2 – 02], 2, , W=, , 1, Kx2, 2, , For-XII-NEET, Medical, Std. Board, , 9, , Oscillations (XII)

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Excellent Notes, , But,, , W=m, , By Prof. Chavan Sir, 2, , x –1M..., , 1, m 2x2, 2, This workdone stores in the form of P.E., P. E. = W, W=, , 6), , 7), , 8), , a, 2, , –1/2M...], Q.17. Assuming the expressions for P.E. and K.E. for, a particle performing simple harmonic motion,, obtain the expression for its total energy. What, 1, P. E. = m 2x2, conclusions can be drawn from the expression, 2, for the total energy?, OR, Ans :, 1, Expression for T. E. :, ( m 2 = K), P. E. = Kx2, 2, 1), The general expressions for the P.E. and K.E. of a, This is the expression for potential energy of a particle, particle performing S.H.M. are given by, performing linear S.H.M. at a distance x., 1, P. E. at mean position :, P. E. = m 2x2, 2, At mean position, x = 0, 1, 1, P. E. = m 2, ...(i), P. E. = Kx2, 2, 2, 1, 1, P. E. = K(0)2, K. E. = m 2(a2 – x2), 2, 2, P. E. at extreme position :, At extreme position, x = 0, 1, P. E. = Kx2, 2, 1, (P. E.)max = Ka2, 2, OR, , 2), , Thus, P. E. is minimum at mean position and maximum, at extreme position., Graph :, Graph showing variation of P.E. with displacement is, given below., , Question bank :, State an expression for K.E. (kinetic energy) and P.E., (potential energy) at displacement ‘x’ for a particle, performing linear S.H.M. Represent .... graphically., Find the displacement at which K.E. is equal to P.E., [Formula for K.E. –1/2M... formula for P.E. –1/2M..., Graphical representation of K.E. and P.E. with, distance, , For-XII-NEET, Medical, Std. Board, , 1, K(a2 – x2), 2, where,, m = mass of the particle, a = amplitude of S.H.M., x = displacement of the particle and, = angular velocity, The total energy of the particle at any point is the sum, of its K.E. and P.E. at any point P., T.E. = K.E. + P.E., K. E. =, , T.E. =, , 1, 1, K(a2 – x2) + Kx2, 2, 2, , T.E. =, , 1, K(a2 – x2 + x2), 2, , 1, T.E.= Ka 2, 2, , i., , 1), , 1 2 1, Kx = K(a2 – x2) –1/2M... x =, 2, 2, , ii., iii., iv., , v., , 10, , This is the expression for total energy of particle, performing linear S.H.M., Conclusions :, As m, and a are constant, the T. E. of a particle, always remains constant i.e., conserved., T. E. is directly proportional to the mass of the particle, (i. e., T. E. m), T. F. is directly proportional to the square of the, 2, angular velocity (i.e., T. E., ), T. F. is directly proportional to the square of the, amplitude (i.e., T. E. a2)., T. E. =, , 1, m 2a2, 2, Oscillations (XII)

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Excellent Notes, , By Prof. Chavan Sir, , 1 2, Ka, 2, T. E. K, Hence, T. E. is directly proportional to the force, constant., T. E. =, , vi., , T. E. =, , K. E. =, , 1, K(a2 – 0), 2, , 1 2, 1, Ka and, P. E. = K(0), 2, 2, P. E. = 0, (T.E.)mp = K. E. + P. E., , K. E. =, , 1, m 2a2, 2, , 1, (4 2 n2)a2, 2, T. E. n2, T. E. is directly proportional to the square of the, frequency., , =, , 1 2, Ka + 0, 2, , =, , 1 2, Ka, 2, , T. E. =, , vii., , T. E. =, , 1, m 2a2, ...(iv), 2, At mean position, T. E. is equal to the maximum K.E., Case II : T. E. at extreme position :, At extreme position, x = a, (T. E.)mp =, , 1, m 2a2, 2, , T. E. =, , 1, m, 2, , a2, , K. E. =, , 1, T2, T. E. is inversely proportional to the square ot the, period., Q.18)Show that T. E. of a particle performing linear, S.H.M. is conserved., Ans :, 1), The K.E. of a particle performing linear S.H.M. at, any instant is, , 1, K(a2 – a2), 2, , T. E., , 2), , 3), , 1, K(0), 2, K. E. = 0 and,, 1, P. E. = Kx2, 2, 1, P. E. = Ka2, 2, (T. E)ep = K.E. + P.E., 1, 1 2, K. E. = K(a2 – x2), ...(i), Ka, =, 0, +, 2, 2, The P. E. of a particle performing linear S.H.M. at, 1, any instant is, = Ka2, 2, 1, 1, P. E.= Kx2, ...(ii), (T. E.)ep = Ka2, 2, 2, The T. E. of a particle performing linear S.H.M. at, 1, ., any position is given by,, (T. E.)ep = m 2a2, .........(v), 2, T. E. = K. E. + P. E., At extreme position, T. E. is equal to the maximum, 1, 1, P.E., 2, 2, 2, T. E. = K(a – x ) + Kx, 2, 2, Thus, from equation (iii), (iv) and (v),, (T.E.) any position = (T. E.)mp =(T. E.)ep, 1, 2, 2, 2, T. E. = K(a – x + x ), Thus, total energy of a particle performing S.H.M. is, 2, always constant i.e., conserved., 1 2, Q.19)Discuss analytically the composition of two, (T. E.)any = Ka, 2, S.H.M.’s of the same period and parallel to each, other. Obtain the expression for their resultant, 1, 2 2, 2, amplitude and resultant initial phase. Hence, find, (T. E.)any = m a, ....( K = m )...(iii), 2, the resultant amplitude when phase difference, Case I : T. E. at mean position :, of two S.H.M.’s is, At mean position, x = 0, (i) 0o, (ii)180o, (iii) 90o., , For-XII-NEET, Medical, Std. Board, , =, , 11, , Oscillations (XII)

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Excellent Notes, , By Prof. Chavan Sir, , Ans :, 1), Two S.H.M.’s having same period and parallel to each, other (same path) but different amplitudes and different, initial phases are represented by,, x1 = a1 sin( t + 1) and, x2 = a2 sin( t + 2), 2), The resultant displacement (x) of their composition is, equal to the vector sum of their displacement by,, x = x1 + x2, = a1 sin( t + 1) + a2 sin( t + 2), Using the formula,, sin (A + B) = sinA cos B + cosA sin B, x = a1 (sin t cos 1 + cos t sin 1), + a2 (sin t cos 2 + cos t sin 2), x = a1 sin t cos 1 + a1 cos t sin 1, + a2 sin t cos 2 + a2 cos t sin 2, x = sin t (a1 cos 1 + a2 cos 2), + cos t (a1 sin 1 + a2 sin 2), 3), Let, a1cos 1 + a2cos 2 = Rcos, ...(i), a1 sin 1 + a2sin 2 = Rsin, ...(ii), x = sin t (R cos ) + cos t(R sin ), = R (sin t cos + cos t sin ), x = R sin ( t + ), ...(iii), This equation shows that, the resultant motion after, composition of two S.H.M. with same period having, resultant amplitude R and resultant initial phase ., 4), Expression for resultant amplitude (R) :, The resultant amplitude can be found by squaring and, adding equation (i) and (ii), thus,, R2 sin2 + R2 cos2 =, (a1 sin 1 + a2 sin 2)2 + (a1 cos 1 + a2 cos 2)2, R2 (sin2 + cos2 ) =, 2, a 1 sin2 1 + a22 sin2 2 + 2a1a2 sin 1 sin 2, + a21 sin2 1 + a22 cos2 2 + 2a1a2 cos 1 cos 2, R2 = a21(sin2 1 + cos2 1), + a22 (sin2 1 + cos2 2), + 2a1a2(sin 1sin 2+cos 1cos 2), But sinA sin B + cosA cos B = cos (A – B), R2 = a12 + a22 + 2a1a2 cos( 1 – 2), , tan =, , Case I :, When two S.H.M.’s are in same phase, i.e., phase, difference is zero i.e., 1 – 2 = 0 then, cos( 1 – 2), = (0) = 1, Putting this value in equation (iv), we get, R=, =, , Case II :, When two S.H.M.’s are out of phase,, i.e. 1– 2 = rad, then, cos ( 1 – 2) = cos ( ) = –1, Putting this value in equation (iv), we get, R=, =, , R=, , Note :, When two S.H.M.’s have equal amplitude,, i.e., a1 = a2 = a, from case (I),, R = a + a = 2a, From case (II),, R=a–a=0, From case (III),, , where, ( 1 – 2) is the phase difference of two S.H.M., Expression for resultant initial phase (Epoch) ( ) :, To find the resultant initial phase, dividing equation (ii), by equation (i), we get, R sin, =, R cos, , R=, =, , For-XII-NEET, Medical, Std. Board, , =, , Case III :, When phase difference of two S.H.M.’s is90o or /2, rad. Then, cos( 1 – 2) = cos( /2) = 0, Putting this value in equation (iv), we get, , ...(iv), , 5), , 2, , =, , 12, , 2a 2, Oscillations (XII)

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Excellent Notes, , =a 2, , Question bank :, 1), Two S.H.M.s are represented by, x1 = a1 sin ( t + 1) and x2 = 2 sin( t + 2); obtain, the expression for the displacement, amplitude and, initial phase of the resultant motion., 2), Discuss analytically the composition of two S.H.M.s, of the same period and parallel to each other. Find, the resultant amplitude when phase difference is, (i) 0, (ii) rad (iii) /2 rad (iv) /2 rad., Q.20)What is an ideal simple pendulum? Can it be, realised in practice? Why?, Ans :, 1), Ideal simple pendulum :, An ideal simple pendulum is defined as a heavy point, mass suspended by weightless and inextensible string, from rigid support., 2), Ideal simple pendulum can not be realised in practice, because, The requirements of heavy point mass and weightless, inexlensible string can not be fulfilled in practice., Q.21)Define practical simple pendulum show that, motion of simple pendulum in linear S.H.M. and, hance obtain an expression for its period., Ans :, Practical simple pendulum :, A small heavy point sphere (called bob) suspended, by a light and inextensible string from a rigid support., To show motion of simple pendulum is linear S.H.M.:, 1), A simple pendulum consisting of small bob of mass m, suspended from a rigid support is shown in the, following figure., , By Prof. Chavan Sir, , 4), , 5), , 6), i), ii), 7), i), ii), 8), 9), , 10), 11), , The distance between the point of suspension (S) and, centre of gravity of bob is called the length of simple, pendulum., Let,, L = length of the simple pendulum, m = mass of the bob, Suppose that the bob is displaced through small angle, from the mean position O to the position B, such, that OB = x., In the displaced position, the force acting on the bob, are, its weight (mg) acting vertically downwards., the tension (T) in the string., The weigh ‘mg’ can be resolved into two components., mg cos along the string and, mg sin perpendicular to the string, The component mgcos is balanced by tension T in, the string., The component mgsin acts as a restoring force (F), and it tends to bring the bob to its mean position, F = – mg sin, Negative sign shows that the force and angular, displacement ( ) are oppositely directed., If is small then sin, F = – mg, We know,, angle =, =, , arc, radius, , x, L, , F = – mg, , x, L, , F, g, = x, m, L, But,, , F, = acceleration, m, , accn = –, , g, x, L, , ...(i), , For a simple pendulum at a place, 2), , 3), , Let O be the mean position of the bob., The pendulum remains in equilibrium in the position i), SO, with the C.G. of the bob vertically below the point ii), of suspension S., If the pendulum is displaced through smaller angle and, released, it begins to oscillate on the two sides of its 12), mean or equilibrium position., , For-XII-NEET, Medical, Std. Board, , 13, , g, is constant., L, , acceleration – x, Thus, acceleration of the bob of simple pendulum is, Directly proportional to its displacement (x) and, Acceleration and displacement are oppositely, directed. Hence the motion of simple pendulum is linear, S.H.M., Expression for period (T) :, From equation (i), we get, Oscillations (XII)

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Excellent Notes, , By Prof. Chavan Sir, , Q.22)State the laws of simple pendulum., Ans. Laws of simple pendulum :, The period of a simple pendulum at given place is, g, , 1., But,, , acc, x, , n, , 2., , where, L = length of die simple pendulum, g = acceleration due to gravity ., Law of length : The period of simple pendulum at a, given place is directly proportional to the square root, of its length., (if g is constant), Law of acceleration due to gravity : The period, of simple pendulum at a given place is inversely, proportional to the square root of the acceleration, to gravity., T, , g (if L is constant), , Law of mass : The period of simple pendulum does, not depend upon the mass (m) or material'of the, bob of the pendulum., 4., Law of isochronism : The period of simple, pendulum does not depend upon the amplitude of, oscillation, provide that the amplitude is small., This law is called the law of isochronism., Q.23) Define Second's pendulum. Obtain an, expression for its length. Show that the length of, Second's pendulum is constant at a place, Ans. Second's pendulum :, A simple pendulum whose period is two second is, caleld Second's pendulum., Expression for length of Second's pendulum :, 1., The time period of simple pendulum is given by,, 3., , g, , 13., , This is the expression for time period of simple, pendulum. Time period does not depend upon mass, of the bob., time period of simple pendulum depends on length, (L) of simple pendulum and acceleration due to gravity, (g), Expression for frequency of simple pendulum:, We know, n =, , 1), 2), 3), , 4), , 1, T, , n, , Question bank :, Show that for a small amplitude, the motion of a simple, pendulum is linear S.H.M. Hence find its period., Show that under certain conditions, a simple pendulum, 2., performs linear S.H.M., Define an ideal simple pendulum. Show that, under, certain conditions, simple pendulum perfroms linear, simple harmonic motion., [Definition -1M... Diagram and forces acting on the, bob –1M... F = –mg –1/2M... q = x/l –1/2M... accn, = –g x/l –1/2M... Hence motion of the simple pendulum, is linear SHM –1/2M], Derive an expression for the period of motion of a, simple pendulum. On which factors does it depend?, [Explanation –1/2M... Forces acting on the bob, –1/2M... F = – mg –1/2M... = x/l –1/2M... accn = –, g x/l –1/2M... Formula for T –1/2M... Factors on which, T depends, , For-XII-NEET, Medical, Std. Board, , 14, , g, , ........(i), , For Second’s pendulum,, T = 2 sec and L = Ls, where, Ls = length of Second’s pendulum., Putting the values in equation (i), we get, g, , On squaring both sides, we get, 1, , g, Oscillations (XII)

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Excellent Notes, , By Prof. Chavan Sir, , Fd = –bv, Where b is a damping constant and negative sing, g, 7., 8., This is the expression for length of Second’s pendulum., At a place, g and are constants. Ls=constant., Q.24)Define and explain damped oscillations. Derive, an expression for displacement and period of, damped oscillations., Ans :, Damped oscillations :, Periodic oscillations of gradually decreasing amplitude, are called damped harmonic oscillations and the, oscillator is called a damped harmonic oscillator., Explanation :, 1), We know that the motion of a simple pendulum 9., swinging in air, dies eventually, because the air exerts, a drag force on the pendulum and friction at the support, opposes he motion of the pendulum and dissipate its, energy gradually., 2), When the motion of an oscillator is redeced by an, external force, the oscillator and its motion are said to, be damped., 3), Example :, An idealized example of a damped oscillator is shown, in following figure., , 10., , 1., 2., 3., , 4., , 5., 6., , 11., Block of mass m oscillates vertically on a spring with, spring constant k., From the block, a rod extends to vane that is, submerged in a liquid., 12., As the vane moves up and down, the liquid exerts, drag force on it and thus on the complete oscillating, system., With time, the mechanical energy of the block-spring, system decreases, as energy is transferred to thermal, energy of the liquid and vane., The damping force depends on the nature of the, surrounding medium., The damping force F is directly proportional ( v ) of 1., d, , indicates that Fd opposes the motion. SI unit of, damping constant is kilogram per second (kg/s)., The force on the block from the spring is Fs = –kx., Let us assume that the gravitational force on the block, is negligible compared to Fd and Fs. Thus the total, force acting on the mass at any time t is, F = Fd + Fs, But F = ma, ma = Fd + Fs, ma = –bv – kx, ma + bv + kx = 0, dx, d2 x, + kx = 0, ...(i), 2 +b, dt, dt, The solution of equation (i) describes the motion of, the block under the influence of a damping force which, is proportional to velocity., The solution is found to be of the form, x = Ae–bt/2mcos( ’t + ), –be/2m, Ae, is the amplitude of the damped harmonic, oscillations. Amplitude decreases with time –, exponentially as shown in displacement against time, graph., m, , The expression cos ( ’t + ) show that the motion is, still periodic and S.H.M, The angular frequency is, ’ =, 2, Period of oscillation is, T =, The period of oscillation increases due to the presence, of the term, , in the denominator.., , Thus damping increases the period and decreases the, amplitude., Question bank :, Obtain the differential equation of damped S.H.M., , the vane and block., For-XII-NEET, Medical, Std. Board, , 15, , Oscillations (XII)

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Excellent Notes, , By Prof. Chavan Sir, , Hence, from Eq. (1),, , a g, The acceleration per unit displacement, =, x R, 2, The period of the motion, T = | a / x | = 2, Data : R = 2.5 m, g = 9.8 m/s2, The period of oscillation is, , R, g, , 2.5, = 6.284 0.5051 = 3.174 s., 9.8, Q.25)Explain angular or torsional oscillations., Hence obtain the differential equation of the, motion., Ans : Suppose a disc is suspended from its centre by a wire, or a twistless thread such that the disc remains, horizontal, The rest position of the disc is marked by, T = 2 3.142, , d2, I 2 =–c, dt, d2, I 2 +c =0, dt, This is the differential equation of angular S.H.M., [Note : Angular displacement being a dimensionless, quantity, the SI unit of torsion constant is the same as, that of torque = the newton-metre (N.m)], Q.26)Define angular SHM. State the differential, equation of angular SHM. Hence derive an, expression for the period of angular SHM in, terms of (i) the torsion constant (ii) the angular, acceleration., Ans : Definition : Angular SHM is defined as the oscillatory, motion of a body in which the restoring torque, responsible for angular acceleration is directly, proportional to the angular displacement and its, direction is opposite to that of angular displacement., The differential equation of angular SHM is, d2, +c =0, dt 2, where I = moment of inertia of the oscillating body,, I, , a reference line. When the disc is rotated in the, horizontal plane by a small angular displacement =, from its rest position ( = ), the suspension wire, m, is twisted. When the disc is released, it oscillates about, the rest position in angular or torsional oscillation with, angular amplitude m., The device is called a torsional pendulum and the, springiness or elasticity of the motion is associated, with the twisting of the suspension wire. The twist in, either direction stores potential energy in the wire and, provides stores potential energy in the wire and, provides an alternating restoring torque, opposite in, direction to the angular displacement., The motion is governed by this torque., If the magnitude of the restoring torque ( ) is, proportional to the angular displacement ( )., (– ) or = – c, where the constant of proportionality c is called the, torsion constant, that depends on the length, dis-meter, and material of the suspension wire. in this case, the, oscillations will be simple harmonic., Let I be the moment of inertia (MI) of the oscillating, disc., Torque = MI angular acceleration, =I =I, , d2, = angular acceleration of the body when its, dt 2, angular displacement is , and c = torsion constant of, the suspension wire,, , c, I, frequency,, Let,, , a constant. Therefore, the angular, and the angular acceleration,, , 2, , a, , ........(2), , The minus sign shows that the and have opposite, directions. The period T of angular SHM is, .......(3), This is the expression for the period in terms of torque, constant. Also, from Eq. (2),, , ement, , d2, dt 2, , 2, em ent, , For-XII-NEET, Medical, Std. Board, , 16, , Oscillations (XII)

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Excellent Notes, , By Prof. Chavan Sir, , Q.27)Prove that a magnet vibrating in uniform Q.28)Obtain the expression for the period of a magnet, in angular simple harmonic oscillations in a, magnetic field performs angular SHM., uniform magnetic field., Ans : Consider a bar magnet of magnetic mom suspended, 2, horizontally by a light twistless fibre in a region where, 2, the horizontal component of the Earth's magnetic field Ans :, is Bh. The bar magnet is free to rotate in a horizontal, 2, plane. It comes to rest in approximately the Northwhere, = the angular acceleration and I = the, South direction, along Bh. If it is rotated in the, horizontal plane by a small, moment of inertia of the magnet about the axis of, oscillation which is the transverse symmetry axis of, the bar magnet., The equation of motion of the bar magnet suspended, horizontally in the Earth's magnetic field is, , For this axis, the MI of a rectangular bar magnet of, geometric length L and width b is, I, From Eq. (2), the angular acceleration per unit angular, h, , displacement,, , The period of the oscillations,, , ., , displacement from its rest position ( = 0), the, suspension fibre is twisted. When the magnet is, released, it oscillates about the rest position in angular, or torsional oscillation., , h, , ****, , The bar magnet experiences a torque due to the, field Bh. Which tends to restore it to its original r, orientation parallel to Bh. For small , this restoring, torque is, , ....... (1), where the minus sign indicates that the torque is, opposite in direction to the angular displacement ., Equation (1) shows that the torque (and hence the Mt, angular acceleration) is directly proportional in, magnitude of the angular displacement but opposite, in direction. Hence, for small angular displacement,, the oscillations of the bar magnet in a uniform magnetic, field is simple harmonic., For-XII-NEET, Medical, Std. Board, , 17, , Oscillations (XII)

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Excellent Notes, , By Prof. Chavan Sir, , Examples, 1., , 3., , A body of mass 0.1 kg performs linear S.H.M. It, experiences a restoring force of 1N when its, displacement.is 5 cm., Find (i) Force constant, (ii) Period of S.H.M., (iii) Acceleration of the body when Its, displacement from the mean position, is 1 cm., The differential equation of S.H.M. is given by, , 12., , A particle is executing S.H.M. according to the, equation x = 5 sin t where x is to cm. How long will, the particle take to move from the position of, equilibrium to the position 01 maximum displacement, , Ans : T = 0.5 second, 16., , A particle executes SHM with amplitude Jem and, period 2 sec. Find the speed of the particle at a point, where its acceleration is half the maximum value., , Ans :, , 2, , d x, ., dt 2, Find -(i) Period and, Ans : (i), 4., , 17., (ii) frequency of S.H.M., , , (ii), , A body of mass 2 kg performs linear S.H.M. The, restoring force acting on it when its displacement is, 0.6 m from the mean position is 3N. Write down the, differential equation of its motion, , Ans :, 18., , 6., , The periodic time of vibration of mass m1 suspended, from a light spring is T second. When a mass m2 is, added to first mass and the system is made to vibrate,, the periodic time is 2T. Compare m1 and m2., A vertical light spring is stretched by 1cm when a body, of mass 25 gram is tied to its free end. The body is, further pulled down by 0.25cm and released so that, it performs S.H.M. Find the (i) Period and (ii), amplitude of S.H.M. (g = 980 cm/s2), , Ans :, , Ans :, 11., , If the quantities are expressed in S.I. units, determine, the (i) amplitude (ii) angular velocity (ill) period (Iv), Frequency and M epoch of the S.H.M., Ans :, 21., , The shortest distance travelled by a para performing, S.H.M. from mean position in 2 seconds is equal to, 3 / 2 of its amplitude. Find its period., , Ans :, 22., , Show that the velocity of a particle performing S.H.M., is half the maximum velocity at a displacement, , A body describes S.H.M. in a path 0.12 m long. Its, velocity at the centre of the line 1s 0.12 m/s. Find the, period and magnitude of velocity at a, distance, , 10., , ., , ,, , Ans :, 8., , The displacement of a particle in near S.H.M. is given, by,, , d2x, Ans : dy2, 5., , A particle performs S.H.M. of period 12 second and, amplitude 8 cm. If initially the particle is at the positive, extremity, how much time will it take to cover a, distance of 6cm from the extreme position., , times its amplitude., , from the central position., , Ans :, , ,, , 23., , A body executing S.H.M. completes 120 osc/ min, having amplitude of oscillation 10 cm. Find the velocity, and acceleration of the particle, when it is at distance, of 5 cm from the mean position., 2, , A particle performing linear S.H.M. has maximum, velocity of 20 cm/s and maximum acceleration of 80, cm/s2. Find the amplitude and period of oscillation., , Ans :, For-XII-NEET, Medical, Std. Board, , 3, 2, , Show that the average speed over one oscillation of a, particle performing linear S.H.M. of amplitude a and, angular speed, , is, , Ans :, 24., , A particle performing S.H.M. has velocities of 8 cm/, s and 6 cm/s, when its displacements are 3cm and, 4 cm respectively. Find the amplitude and period., , Ans :, , 18, , Oscillations (XII)

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Excellent Notes, , 25., , A particle executes S.H.M. of period, , By Prof. Chavan Sir, , 2, , second, , 3, along a straight line 4cm long. Find the displacement, of the particle at which the velocity is numerically equal, to the acceleration., , The displacement of the particle performing S.H.M., 36., is given by,, x = 10 sin (5 t + /6) cm., If t is in seconds. Find the velocity at t = 0.4 s., , Ans :, 29., , 20 2 cm. Find at what distance from the mean, position its K.E. is equal to its P.E, , A particle performs linear S.H.M. along the path of 38. The maximum velocity of a particle of mass. 4 × 10–3, 10 cm. The particle starts at a distance of 2 cm from, kg performing a linear S.H.M. is 8 ×10–2 m/s. Find its, the mean position and moves towards the positive, total energy., extremity. Find its initial phase., Ans :, A particle performing SHM has velocities v1 and v2, when it is at distances x1 and x2 from mean position., Show that the period is given by,, , Ans :, , 40., , A particle performing S.H.M. has a K.E. of 480 erg, when it is at 2 cm from mean position, and 420 erg, when it is at a distance of 4cm from mean position., Determine total energy and amplitude of vibration., , Ans :, 41., , 31., , A particle performs a linear S.H.M. along a line, , Ans :, , Ans :, 27., , 1, rd of lhe, 3, amplitude. What fraction of total energy , is kinetic, energy and what fraction is potential, When the displacement in S.H.M. is, , Ans :, , Ans :, 26., , 35., , A particle executes S.H.M. with a period 8s. Find the, time in which half the total energy is potential., , Ans :, , A particle of mass 10 g executes linear S.H.M. of, amplitude 2 cm with a period of 2 seconds. Find its, RE. and K.E. (1/6)th second after it has passed, through mean position,, , Calculate the ratio of displacement to amplitude when Ans :, kinetic energy of a body is thrice its potential energy. 42. A particle is'performing linear S.H.M. of amplitude a., What fraction of the total energy Is kinetic, when the, displacement is half the amplitude ?, Ans :, 32., , 33., , Ans :, A particle executes S.H.M. of amplitude a, i) At what distance from mean position is its kinetic 43. For a particle performing S.H.M. in a given position,, energy equal to its potential energy?, the ratio of its K.E. to RE. is 3. If the amplitude of its, ii) At what points is its speed half the maximum speed?, motion is 20 cm, find the distance of the point from, mean, , Ans :, 34., , Ans :, 44., Particle is executing SMM. of amplitude a, i) What fraction of total energy is kinetic energy when, the displacement is half the amplitude?, ii) At what displacement is the kinetic energy three, times that of potential energy?, , Ans : Displacement, , For-XII-NEET, Medical, Std. Board, , The displacement of particle performing SUM. is given, by, , Determine the amplitude, period and initial phase of, the motion., , Ans :, , 19, , Oscillations (XII)

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Excellent Notes, , 45., , The displacement of a S.H.M. is given by, x = 12 sin 56., (0.8 t) cm. Find the amplitude, period, frequency, and initial phase of S.H.M., , Calculate the maximum velocity, with which an, oscillating pendulum of length 90 r.m will attain, if its, amplitude is 10 cm, , Ans :, , Ans :, 46., , By Prof. Chavan Sir, , Two parallel S.H.M.’s are given by,, 57. The period of oscillation of a simple pendulum is, doubled when its length is increased by 1.2 m., x1 = 20 sin 8 t cm and, Calculate its original length and period, (g = 9.8 m/s2), x2 = 10 sin (8 t + /6) cm., Find the amplitude and initial phase of resultant S.H.M. Ans :, 58., , Ans :, 47., , The period of simple pendulum is found to increase, by 50% when the length of the pendulum is increased, by 0.6 m. Calculate the initial length and initial period, of oscillation at a place where g = 9.8 m/s2., , The displacement of a particle in motion is given by,, x = a sin t + b cos tot meter where a and b are, constants. Show that the motion is simple harmonic., If a = 3 cm, b = 4cm, = 2 rad/s. Find the period, Ans :, maximum velocity and-maximum acceleration., 62. Calculate the length of Second's pendulum on a planet, where the acceleration due to gravity has a value of, Ans :, (1/8)th that on earth. Length of second's pendulum on, 48. Simple pendulum of length 1 m has mass 10 g and, earth = 80 cm., oscillates freely with amplitude of 5 cm. Find its, potential energy at extreme position, (g = 9.8 m/s2), Ans : L 2 p 10cm, Ans :, 49., , 63., , A simple pendulum of length 1m has a mass of 10 g, and oscillates freely with amplitude 2 cm. Find its, potential energy at extreme point. (g.= 9.8 m/s2), , If period of simple pendulum on earth is 4 s. What is, the period of simple pendulum on surface of moon, having acceleration due to gravity 1.6 m/s 2? Ans :, (acceleration due to gravity on earth, g = 9.8 m/s2), 69. The equation of S.H.M. is given by,, , Ans :, , d2x, dt 2, , . Find period and frequency.., The period of oscillation of simple pendulum, increases by 20% when its length is increased by 44, Ans. (i) n = 1.2738 Hz, (ii) T = 0.785 sec, cm. Find its (i) initial length (ii) initial period, 70. A light vertical spring is stretched by 2cm when weight, Ans :, of 10 g is attached to its free end. the weight is further, pulled down by 0.1 cm and released. Calculate the, 52. A clock regulates by a second's pendulum keeps, frequency of motion and amplitude of S.H.M., correct time, during summer the length of the pendulum, (g = 980 cm/s2)., increases to 1.01 m. How much will the clock gain or, Ans. (i) n = 3.524 Hz (ii) a = 0.1 cm, lose in one day?, 71. A body of mass 100 g is suspended from a rigid support, Ans :, by a light spring and performs linear S.H.M. in the, vertical direction. If the force constant of spring is, 54. Calculate length of seconds pendulum at a plane where, 2, 4.9 × 103 dyne/cm, find the frequency of S.H.M., g = 9.78 m/s ., Ans : n=1.1146 Hz, Ans : 0.9907 m, 73. The displacement of a particle in linear S.H.M. is given, 55. When length of simple pendulum is increased by 22, cm, the period changes by 20% the original length of, by, x = 0.2 sin, . If the quantities are, simple pendulum., expressed in S.I. units. Determine (i) displacement, Ans :, (ii) velocity and (iii) Phase all at time t = 2 second., 51., , For-XII-NEET, Medical, Std. Board, , 20, , Oscillations (XII)

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Excellent Notes, , By Prof. Chavan Sir, , Ans. (i) x = 0.2 m (ii) v = 0 (iii) ( t + ) = 2.5 rad, 74. A particle in S.H.M. has a period of 2 seconds and, amplitude of .10 cm. Calculate the acceleration when, it is at 4 cm from its positive extreme position., Ans. Accn = 2.366 m/s2, 78. A particle performing S.H.M. starting from position, has period 2 seconds and of amplitude 5cm., Determine its acceleration at the end of half sec, Ans : acceleration = 49.298 cm/s2, 80., , A particle performs S.H.MI of period, sec, along a path 4 cm long. Calculate the displacement of, the particle at which its velocity is numerically equal, to acceleration., Ans: x = ± 1.732 cm, 1, 5, cm. Find at what distance kinetic energy is 4 times its, P.E., Ans. x = +0.2 m, 84. An object of mass 0.04 kg executes S.H.M. of, amplitude 10 cm. If its P.E. in the extreme position is, 1.97 × 10–3 , calculate the period of S.H.M., Ans. T = 2.03 sec., 85. A particle of mass 10 g executes linear S.H.M. of, amplitude 5cm with a period of 2 second. Find its, P.E. and K.E., (1/6)th second after it has,passed, through the mean position., Ans : K.E. = 924.3 erg, *****, , 82., , A particle performs a linear S.H.M. of amplitude, , For-XII-NEET, Medical, Std. Board, , 21, , Oscillations (XII)