A block of mass m is connected to a spring constant k and is at rest in equilibrium as shown. Now, the block is Displacement by h below its equilibrium position and imparted a speed `v_0` towards down as shown in the Fig. As a result of the jerk, the block executes simple harmonic motion about its equilibrium position. Based on this information, answer the following question.
Q. Find the amplitude.

Options

A. `(2pi-cos^(-1)((h)/(A)))/sqrt((k)/(m))`
B. `((pi)/(2)-cos^(-1)((h)/(A)))/sqrt((k)/(m))`
C. `(pi-sin^(-1)((h)/(A)))/sqrt((k)/(m))`
D. `(pi-sin^(-1)((h)/(A)))/(2sqrt((k)/(m)))`

Answer

C

Explanation

To compute the time taken by the block to cross mean position for the first time, we can make use of circular motion representation.
`t=(pi-Sigma)/(omega)=(pi-sin^(-1)((h)/(A)))/(sqrt(k)/(m))`