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)+k`
D. None of theses

Answer

B

Explanation

The angular frequency of simple harmonic motion is given by, `omega = sqrt((k)/(m))`
The velocity of block, when it at a displacement of `y`from mean position is given by, `v = omega sqrt(A^(2) -y^(2))` , where `A` is the amplitude of oscillation. For given initial condition,
`v_(0) = sqrt((k)/(m)) sqrt(A^(2) - h^(2)) rArr A^(2) = (mv_(0)^(2))/(k) +h^(2)`
`rArr A = sqrt((mv_(0)^(2))/(k)+h)`