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 time taken by the block to cross the mean position for the first time.

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))`
`Rightarrow A^(2)=(mv_(0)^(2))/(k)+h^(2) RightarrowA=sqrt((mv_(0)^(2))/(k))+h^(2)`