Question

the spring. An object is undergoing SHM with period 0.900 s and amplitude 0.320 m. At t 0 the object is at x 0.320 m and is instantaneously at rest. Calculate the time it takes the object to go (a) from x 0.320 m to x 0.160 m and (b) from x = 0.160 m to x 0 14.10 A small block is attached to an ideal spring and is moving

Answer 1

General equation of simple harmonic motion showing position x of the object at any time t is given by, x = Asin(2X+++ )......(1) Here, Amplitude of oscillation is A, position of the particle is x at time t, time period of simple harmonic motion is T and initial phase is Differentiate with respect to time to get velocity Asin 2*11+0) v = 2 X7A x cos(2X+1 + p)..... (2) Since object is released from rest at t = Os hence velocity of object at t = Os is zero. At t = Os v=0 m/s Substitute 0 m/s for

In equation (2) 0 = 2 XpA x cos(2X1 x (0) ++) coso. = 0 $ = 90° 0. = 5 Now substitute 0.320m for 0.900s for and

in equation (1). Position of the object as a function of time t is, x = 0.320 x sin (2.900 * (t) +) x = 0.320 x sin (20X A+ x = 0.320 x cos (20 X )......(3) Now substitute 0.160m for in equation (3) 0.160 = 0.320 x cos(20 X 14,) Here is the time taken by the object to go from x = 0.320m x = 0.160m 1 = cos(20X24) cos - '(%) = (20,X24) (20,,) = 11 = 4x20X1 11 = X t1 = 0.150s

Therefore time taken by the object to go from x = 0.320m x = 0.160m 0.150s (b) Substitute От for in equation (3) 0 = 0.320 x cos(20X242) Here is the time taken by the object to go from x = 0.320m to x = Om

0 = cos(20,X712) 20,X112 = cos - (0) 20,X#t2 = t2 = 40 12 = 0.225s Since time taken by the object to go from 0.320m to 0.225s and time taken by the object to go from 0.320m to 0.160m 0.150s Therefore time taken by the object to go from 0.320m to 0.160m 1 - = 0.225 – 0.150 11-11 = 0.0750s