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. The amplitude of oscillation is

Options

A. h
B. `sqrt((mv_0^2)/(k)+h^2)`
C. `sqrt((m)/(k)v_0)+h`
D. none of these

Answer

B

Explanation

The angular frequency of simple harmonic motion is given by
`omega=sqrt((k)/(m))`
The velocity of block when it is a displacement of y from mean position is given by `v=omegasqrt(A^2-y^2)`,
From given initial condition `v_0=sqrt((k)/(m))sqrt(A^2-h^2)`
`impliesA^2=(mv_0^2)/(k)+h^2impliesA=sqrt((mv_0^2)/(k)+h^2)`