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OSCILLATIONS 44/31, , irre] DISPLACEMENT, VELOCITY ye z, ACCELERATION TIME PERIOD OF SHM = s0x4x( F J cos|2 mxl5 + 4], 7, , I, , Formulae used. 2 ,, 1, Displacement inSHM, y=a sin (wrt 0) =50x4x(F) 0-707 = 139-56 ms~, when time is noted from the mean position of SHM 7, and X=a cos (cr + dq) Example PJ A particle executes SHM with, when time is noted from the extreme position of a time period of 3 s and amplitude 6 cm. Find, SHM (i) displacement (ii) velocity and (iii) acceleration,, seers . after 1/2 second ; starting from the mean position., BNelocity inSHM: = VE aya" = Solution. Here T=3s;A=6cm;t=1/2s, , 3. Acceleration in SHM, A =~? y and (i) As particle is starting from mean position,, , a= 2nv=a2 wT where ais the amplitude of the displacement of particle at time r is given by, simple harmonic motion and @ is its angular, 20, , frequency. $9 is the initial phase in SHM. x= Asinwe= Asin 2n, = 6sin anit, Units used. y and a are taken in m or cm, v in hertz E 3.2, , , , , , and T in seconds., #6 to io 586 em, example fl A body oscillates with SHM, 3 2, according to the equation, ‘ wy po 2A 2m, x = (5-0 m) cos [(2 m rad s~!) ¢ + n/4] (i Nelocity; B= dso Se, At t= 1:5 s, calculate the (a) displacement 2n 2n 1, =—x6cos—x—, , (6) speed and (c) acceleration of the body., NCERT Solved Example, , , , , , , , , , ee nm 4nxi1 22, Solution. Here, @ = 2 1 rad/s ; = SRC = 2 i 2s, 7a aR ats 1215s = 6-28 cm/s, o Tt . 7 2, (a) displacement, (dp Accelemtion; i= s ao s Asin 22, x= (5-0) cos [2 2 x 1-5 + 1/4] 2 rT, = 5-0 cos (3 m + 1/4) = — 5-0 cos n/4 4m x6 5201, =~ 5.0 x 0-707 =- 3-535 m 3 32, dx 82. 8x? V3, (b) speed, v = — =—-— sin = = -—— x, ceca) 3.) 3°)~US, =-5.0x2n sin (21+ 1/4) =~ 22:8 cm s?, =-5.0x 2 msin (2 2x 1-5 + 1/4) In magnitude, A = 22-8 cm s~, Here —ve sign, , % 22 indicates that acceleration is directed towards mean, , =50x2nxsin — =50x2x 7 x 0-707 position whereas displacement is directed away from, 4 a5 al mean position., , = Example) A particle moving ina straight, , (c) Acceleration, A = dv line has velocity v given by v? = a - B y?, where a, , dt and B are constants and y is its distance from a, fixed point in the line. Show that the motion of, =-5) 2 £ i ind its ti ; As, 5-0x4 n* cos} 2nt+ particle is SHM. Find its time Period and, 4 amplitude., , , , Scanned with CamScanner

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Ny, , , , 14/32 Pradeep's Fundamental Physics vor, Solution. Given, v? = 0.-B y* (i) _ 2n_2n_ Bit, Differentiating it wrt. time t, we have T 4 5, , dv dy dv Displacement of a particle in SHM in time 1,, ” a “pay a ia a “By when particle starts from mean position is, 2m, re x Aad = Asin—t, -. Acceleration, 4 = = =-By wii) y = Asin wf = Asin T, t, , As A & y and -ve sign shows that acceleration, is directed towards the mean position, so if the particle, is left free ; it will execute SHM., , Comparing (ii), with A =~ w* y, we have, w=B or w= vB, , 2, Time period, T= on ate, , o Bp, , We know that v = 0, when y=r. From (i),, , 0=a-B7 or r=Jop, -. Amplitude =r = Jup, Examplef[] The equations of simple, harmonic motion is given by, , y=3sin 5 nt +4 cos 5 nt, where y is in cm and tin, second. Determine the amplitude, period and, initial phase., Solution. Given,, y=3sin5 mt+4cos5 nt, , sui), Let, Acos$=3 and Asing=4, , Then, , A? (cos? 6 + sin? $) = 324. 42=25 or A=Sem, Asing 4, , = > or tan >= 1-3333 = tan 53° 8”, Acosh 3, , 9 =53° 8’, From (i), y =A cos $ sin 5 mt+A sin 6 cos 5 mt, =A sin (5 mt+ o) +«(ii), It is a SHM with amplitude A (= 5 cm), Comparing (ii) with Y=A sin (wt +), we have, , @=S5n or oR asm or 7=2h, , on =0-45, , Initial phase, = 53° 8’, , example] A particle executes SHM of, , amplitude 20 cm and time Period 4s, What is the, , enn time required for the Particle to move, een two points 10 cm on either si, , sia T side of mean, , Solution. Here, 4 = 20cm, T=4 Sy, , , , = 10cm,, , . mt 10, 10 = 20sin 2+ or sin—=—=, , l uz, —=s5, - 200 a, , ala, , mt 1, or —=— or t==s, 2 6 3, , Time taken by particle to move between two, extreme positions 10-0 cm on either side of the mean, , position is given by, =2t=2x 13=0675, , (example (A Particle executing SHM, along a st. line has a velocity of 4 ms“, when ata, distance of. 3 m from its mean position and, 3ms“!, when at a distance of 4 m from it. Find the, time it takes to travel 2-5 m from the Positive, extremity of its oscillation., , Solution. Here (case I), V,=4ms"1, y,=3m;, case (II), V, = 3 ms7! ; y2=4m., , We know that, y= wa? -~y*, , Case (I) 4= ova2 —32 (i), Case (II) 3= ova2 ~42 (ii), ’ Dividing (i) by (ii), we get, 4_ wva?-9 16 a*-9, , 3 eve-16 9 a6, or 16 a*~ 256 =9 g2_ 8}, or 7 a?=256~81 = 175, , 175, 2, , =—=25, 7 or, , Substituting it in (d, we get, , 4 = wy5? —32 = w/25~9 =ox4, , oO O=4/4= 1} rad gl, , When the particle is at a distance 2-5 m from, the extre:, , ~ 1 /eMe position, then its distance from the meal, Position, X=5-2.5=2.5 m., , Me is to be noted from the extreme, therefore, we shall use the relation, *=acos wt., , 25=5 cos | xt=Scost, , , , or a, , a= 25=5m, , Since, the ti, Postion for SHM, , or, , Scanned with CamScanner

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| ae, , 14/33, , , , OSCILLATIONS, or cost = 251 yt, S usQ 3, nm 22, or f=—=— r, 3 Ix3 = 1-048 s, , Example ff A man stands on a weighing, machine placed on a horizontal platform. The, machine reads 50 kg. By means of a suitable, mechanism, the platform is made to execute, harmonic vibrations up and down with a, frequency of 2 vibrations per second. What will, be the effect on the reading of the weighing, machine ? The amplitude of vibration of platform, is 5m. Take g = 10 ms?, , Solution. Here, m = 50 kg, y= 2 s-),, a=5S5cm=0-05 m, Max. acceleration, A,,,, = @2a = (2 nv)2a, , * 22°, =4r?va=4x a x (2)? x 0-05 = 7-9 ms~2, , .. Max. force on the man = m (8 + Annax), = 50 (10 + 7-9) = 895-0 N = 89-5 kgf, Mini. force on the man = m (g — Ajax), = 50 (10 - 7:9) = 105-0 N = 10-5 kef., Hence, the reading of the weighing machine, varies between 10-5 kgf and 89:5 kgf, Example [] A particle executes simple, harmonic motion with a time period of 16 s. At, time ¢ = 2 s, the particle crosses the mean position, while at t = 4 5, its velocity is 4 ms-1, Find its, amplitute of motion., Solution. Here, T= 16s; Att=2s, y=0, andatt=4s;V=4ms!;a=?, For simple harmonic motion,, , ‘ a 21, y=a sin @t=a sin 7!, , When t=4 s, the time taken by particle to travel, from the mean position to a given position = 4 — 2, = 2s, The displacement,, , =asi 2K =asin FE \.8, (i), y=asin 16 = ayy, Velocity, V= wa? - y?, , 2 hel 28 Wat- 0272 28x%, 3 16 8 2, , N, , or a wo 2 344m, , a, , example f) The displacement of @ ae, executing periodic motion is given by, , y=4 cos? (5 } sin (100 f), Find the independent, , constituent simple harmonic motions., Solution. y = 4 cos? (1/2) sin (100 #), =2(1 +cos #) sin (100 ¢), [-: cos20= 2 cos? 6 - 1], =2 sin 100 ¢+ 2 sin 100 t cost, =2sin 100 r+ sin (100 + 1) ¢, +sin (100-1)¢, =2sin 100 f+ sin 101 ¢+ sin 99 ¢, [-: 2sin A sin B = sin (A + B) + sin (A -B)}, Thus, the given expression is the resultant of, three independent harmonics., jexamepte [[] Fig. 14.31 (a) and (b) depict, two circular motions. The radius of the circle, the, period of revolution, the initial position and sense, of revolution are indicated in the figures., Obtain the simple harmonic motions of the xprojection of the radius vector of the rotating, particle P in each case., , NCERT Solved Example, , , , y FIGURE 14.31 y, 1, 1, , , , , , , , , , Solution. Refer to Fig. 14.31(a), T=4s, initial, , phase, 9 = 45°= Trad with the positive direction of, , 2n, x-axis. After a time t, it covers an angle 7 in the, anticlockwise direction and makes an angle, (= + 4 with the x-axis. The projection of OP on, T, axis at time f is, 2nt 2nt, x= acos( 2+ 9) = aco( 74), , Refer to Fig. 14.31(b), T = 30 s, initial phase,, , the x, 6=90°=Zrad with the x-axis, After a time ¢, it, 2, , Scanned with CamScanner

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14/34, , covers an angle oe in the clockwise direction and, , nm 2nt, , makes ‘an angle 5.28) with the x-axis, Now,, , the projection of OP on the x-axis at time t is given, , by, 2, x= bcos m_2nt = bcos 2nt nt, 2 T T, , =bcos ont 2 =bcos RL, 30 2 15 2, , Example [§l Time period of a simple, pendulum is 2 s and it can go to and fro from, equilibrium position at a maximum distance of, 6 cm. If at the start of the motion the pendulum is, in the position of maximum displacement towards, the right of the equilibrium position, then write, the displacement equation of the pendulum., , Solution. Here, T=2s;A=6cm ;, , Let $y be the initial phase of the pendulum, then, displacement of particle executing SHM at time ¢ is, given by, , ade . (20, y=A sin (Wf + $9) = Asin (Fe), , or y=6sin [3+ }=osincnrs 69, , when 1t=0,y=6cm,so, 6 =6 sin (7x 0 + 4) =6 sin dy, , or sin $9 = 1 = sin n/2, , Or = 9 = (n/2) rad, , Hence displacement equation for the simple, pendulum is, , y =6 sin (nt + 1/2) =6 cos tt, , Example [—] Maximum velocity of a, particle in SHM is 16 cm s~!, What is the average, velocity during motion from one extreme position, to other extreme position ?, , Solution. Given, Max. velocity = aw = 16 cm s~!, , 20, , or a—=l16 o, 7 Ir, , s, Pradeef's Fundamental Physics (XD rere, , Average velocity,, , _ totaldisplacement _ 2a, , =44_4a, av total time Tl2, 16 32,, , v =4x— =—cms= 2n oO, , P.E., K.E., AND, T.£.1N SHM, , ii, , Formulae used, , ay Bast at, () PE..U =>mo? y =5%, KE, K = ma? (a? — y?)=4e(a? ~ 92), , * (iii) Total energy, E = mot a= 5 ta?, , Units used, Here, m in kg, w in Hz. or s“!, a or y in m, k in, Nm, U, K and E in J., , , , (Exampte[R] A body executes SHM of time, period 6 s. If its mass be 0-1 kg, its velocity, 1 s, after it passes through its mean position be, 3 ms“, find its (i) KE (ii) PE and (iii) total energy., , Solution. Here; T = 6 s, m= 0-1 kg, t=1s,, , v=3 ms,, v= Aacosat = Aa cos 2,, Tr T, 3= Ax 2B cos 2m = Ax2Byh, 6 6 6 2, Ante, Tt, 1 41, KB == mw = 5%01x@)? =045J, , 1 1 2 2, TE=5 mo? Az = sme xie, , T? x?, 1 4, = =x x yy 92 —, 2 XO x18 =18J, PE= TE - KE = 1.8~ 0.45 = 1.35 y, , ‘Exampie [{] An8k, = & body performs SHM, coos 30 cm. The restoring force is 60.N;, wien the displacement is 30cm. Find (a) time, , period (b) the accelerati PE, displacement is 12 ce ™ ee, , Scanned with CamScanner

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OSCILLATIONS, , Solution. Here, m = 8 kg;, , a= 30cm =0:30 m;, (a) F=60N;y=0.30m, , Fahy or k= _ 60 _ -1, y 030 200 Nm, , k 2, O= {; S a =5 rad/s“!, m 8, , Time period, T = 2% _ 2%22 _ 44, ® 7x5 35, (b) Here, y = 12cm = 0-12 m,., , “. Acceleration, A = @? y = (5)2 x 0.12, = 3-0 ms?, , 1 1, PE.= > by? = 5 % (200) x (0-12)? = 1-44 J, , KE.=5k(a?-y2), , 1, =5% 200 x [(0-30)? — (0-12)?] = 7-56 J, , Example [A] A point particle of mass, 0-1 kg is executing SHM with amplitude of 0-1 m., When the particle passes through the mean, Position, its kinetic energy is 8 x 10-3 joule. Obtain, the equation of motion of this particle if the initial, phase of oscillation is 45°., , Solution. Here, m = 0-1 kg, a= 0-1 m;, K=8x103J; 0, = 45°=“ rad., Kinetic energy in SHM is, K= smo? (a? - y*), when, y=0,K=8x 107), , 8x103= ; x Ol x w? (0-1? - 07), , or @=4 rad/s :, Equation of SHM is y =a sin [wt + $9], or y =0-1 sin [4 ¢ + 1/4], , {Exampte}{) A linear harmonic oscillator, of force constant 2 x 10°Nm~ and amplitude 2 cm, has a total mechanical energy 600 J. What is the, maximum K.E. and minimum P.E. ?, , Solution. Here, k=2x 10° Nm=!,, , a=2cm=0-:02m, , 14/35, , Total energy = 600 J, , 1, As maximum K.E, = max. P.E. = zh, , ; x (2. 109) x (0-02)? = 400 J, , In SHM PE. is minimum at the mean position, and K.E. is maximum at the mean position, whereas, total mechanical energy is constant throughout., , Therefore, at mean position total mechanical energy, = max K.E. + min. PE., , or 600 = 400 + min. PE., or min. PE. = 600 - 400 = 200 J, , Rea SPRINGS, Il, , Formulae used, , @ Spring constant k = F/y, , (ii) Spring constant of parallel combination of, , springs K=k +h, , (iii) Spring constant of series combination of. springs ~, Loti t, , —=—+—, Kk, (iv) Time period, T = anf, , Units used, k,, ky and K in Nm", m in kg and Tins, F in N., , , , ‘example[¥] Two identical springs of, spring constant k are attached to a block of mass, m and to fixed supports as shown in Fig. 14.32(a)., Show that when the mass is displaced from its, equilibrium position on either side, it executes a, simple harmonic motion. Find the period of, , oscillation. NCERT Solved Example, , , , , , FIGURE 14.32, rq m ‘a, k, , k, 8} TOIT = A DIOVO, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , 4 — Fo, Fy ‘tases © OOOO) TOOT, Dra, , , , , , , , Scanned with CamScanner