Question

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 equation for the simple harmonic motion is
A. `y=-Asin[sqrt((k)/(m))t+sin^-1((h)/(A))]`
B. `y=-Acos[sqrt((k)/(m))t+sin^-1((h)/(A))]`
C. `y=Asin[sqrt((k)/(m))t+cos^-1((h)/(A))+(pi)/(2)]`
D. `y=Asin[sqrt((k)/(m))t+cos^-1((h)/(A))+(pi)/(4)]`

Answer 1

Correct Answer - A

To have the equilibrium of simple harmonic motion, it is best to represent simple harmonic motion as uniform circular motion.
At `t=0`, let particle is making an angle `delta` with `-ve` x-axis as shown then
`sindelta=(h)/(A)`
`impliesdetal=sin^-1((h)/(A))`
At time t, `y=-Asin(omegat+delta)`
So the equation of simple harmonic motion is
`y=-sqrt((mv_0^2)/(k)+h^2){sin[sqrt((k)/(m))t+sin^-1((h)/(A))]}`