Question

A particle is performing S.H.M. along X-axis with amplitude 4 cm and time period 1.2sec. The minimum time taken by the particle to move from x = + 2 cm to x = + 4 cm

and back again is given by?​

Answer 1

Answer:

X = A sin w t

2 = 4 sin w t     so w t = 30 deg or pi/6 at 2 cm

1 = sin w t          and w t = 90 deg or pi/2 at 90 deg  (x = 4 cm)

pi/2 - pi/6 = pi/3    or 60 deg for the phase difference between 2 and 4 cm

w = 2 pi f = 2 pi / T = 2 pi / 1.2 = 5.24 for the angular frequency

w t = pi / 3     from 2 to 4 cm

t = (pi / 3) / w = .2 sec

So the time to move from 2 cm to 4 cm and back to 2 cm is .4 sec