Question

A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its periodic time in second is.

Answer 1

formula for velocity: v=w√(A²-x²)

and for acceleraton: a=-w²x

where w=angular velocity

A=amplitude of the particle

x=position of the particle

and w=2pi/T

where T is time period

so given v=a (numerically)

w√(A²-x²)=-w²x

A²-x²=w²x²

A²=(w²+1)x²

25=(4pi^2/T²+1)*16

4(9.86)/T²=25/16-1

39.5/T²=9/16

T=√(39.5*16/9)

T=8.37sec

so the time period of oscillating particle is 8.37seconds.