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14.14 PHYSics—x1, , A body of mass, , I DL ky, according tothe eguatbor O.L ky is exeeuting SHM, , y =05 ws| In, {100 14 | metre, , Find (i) the freq, le frequency of oscillation, agente eae f oscillation (ii) initial phase (iii), , ) maximum ace, see ximtum acceleration and (v) total, , Solution. Gi 5 3), siven x =0.5 cos| 100 14 ) metre, \ t ‘ | metre, , For any SHM, x= A cos (wt + 4), , Comparing the above two equations, we get, , , , A=05m, o=100 rads!) Qy= rad, . » 100, ( Frequency, v= © = 1 ny, BNieheerek, 3n, (i) Initial phase, =~ rad, , (ii) =, , Dam =O A=1000.5 =50 ms, , =o? A=(100)? x 0.5 =5000 ms., , , , (e, , (@) Total energy, , == mo, =, , x 0.1 x (50)? = 125 J., , la, , SN PROBLEMS FOR PRACTICE, , 1. A simple harmonic oscillation is represented by the, 0.40 sin (4401 + 0.61), , Here yand f are in m and s respectively. What are, the values of (i) amplitude (ii) angular freq, , equation, y, , , , ney, (Gi frequency of oscillations (iv) time period of, oscillations and (p) initial phase ?, , [Ans. (i) 0.40 m (ii) 440 rad s~! (iii) 70 Hz, , (iv) 0.0143 s (v) 0.61 rad], , 2 The periodic time of a body executing SHM is 2 s., , After how much time interval from t =0, will its, displacement be half of its amplitude ?, , (Ans. 1/, , , , , , 3. A particle executes SHM represented by the, , equation : 10y = 0.1sin 50 xt, where the displacement, , y is in metre and time | in second. Find the, amplitude and frequency of the particle., , (Ans. A = 0.01 m, v = 25 Hz), , 4. The displacement of a particle executing periodic, , motion is given by y = 4.cos” (t /2)sin (10001). Find, , the independent constituent SHM's. (ur 93), , {Ans. sin (10011), sin (10001), sin (999t)), , , , 1200 ose, , through the mean, , , , 5, A particle executing SM compl, tions per minute and passes, , position with a velocity of 3) 4 ms ', Determine the, , Shot with my Galaxy M, , ¥, , , , 6., , 9,, , 10., , 1., , 13., , 14,, , , , maximum displacernent of the particle from the, mean Also leer, equation of the particle if its displieement be zor, at the instant f= 0, , [Ans. A, , position obtain the di, , , , 0.025 m, y = 0.025 sin (AOR) emetra}, , , , vration of a particle performing SHIM ig, , , , 12ems 7 ata d, , , , , , ance of 3 cm from the mean, position, Calculate its time-period. (Ans. 3.142 9), In a pendulum, the amplitude is 0.05 m and, , period of 2s. Compute the maximum velocity, (Ans. 0.1571 ms~), , In what time, , , , 1 its motion begins, will a particle, , , , oscillating according to the equation, y =7 sin 0.5 nf, move from the mean position to maximum, displacement ? [Himachal 08C] (Ans. 1s), , A particle executes SHM on a straight line path. The, amplitude of oscillation is 2. cm. When the displace, ment of the particle from the mean pos, the magnitude of its ac, velocity. Find the time period, maximum velocity, and maximum acceleration of SHM., , , , tion is Tem,, , , , , , eleration is equal to its, , (Ans. 3.635, 3.464 cms", 6cms-*), The velocity of a particle describing SHM is, ! at a distance of 8 cm from mean position, at a distance of 12cm from mean, , locms:, , , , and 8cms*, position. Calculate the amplitude of the motion., , (Ans. 13.06 cm), A particle is executing SHM. If u, and u are the, speeds of the particle at distances x, and x, from the, equilibrium position, show that the frequency of, oscillation,, , , , , a5, , . If a particle executes SHM of time period 4 s and, , amplitude 2.¢m, find its maximum velocity and that, at half its full displacement. “Also find the, acceleration at the turning points and when the, displacement is 0.75 cm. (Ans. 314m s*, , , , 272cms!, 493ems™, L8Sams~), , , , Show that if a particle is moving in SHM, its velocity at a distance V3 /2 of its amplitude from the cen=, tral position is half its velocity in central pesition., {Chandigarh 03 ; Central Schools 031, A particle executes SHM of period 12 s Two, seconds after it passes through the centre of, oscillation, the velocity is found to be 3.142 ems ,, Find the amplitude and the length of the path, , , , , , (Ans, 12 em, 4 em), , A block lying on a horizontal table executes SHM of, period [ second, horizontally, What is the maximum