Question

A mass $m$ is undergoing SHM in the vertical direction about the mean position $y_0$ with amplitude A and angular frequency $\omega$. At a distance $y$ from the mean position, the mass detaches from the spring. Assume that the spring contracts and does not obstruct the motion of $m$. Find the distance $y^{\ast }$ (measured from the mean position) such that the height $h$ attained by the block is maximum. $(A\omega {)}^2>g$

• A. $\frac{g}{{\omega }^2}$
• B. $\frac{2g}{{\omega }^2}$
• C. $\frac{g}{2{\omega }^2}$
• D. None of these

Answer 1

The correct option is A $\frac{g}{{\omega }^2}$

Let $v$ be the upward velocity of the block when it detaches from the spring at a distance $y$ above the mean position.

$v=w\sqrt{A^2-y^2}⟹v^2=w^2(A^2-y^2)$ ..........(1)

At the detaching point, the block will continue to move upwards due to inertia and reach the height $h$ above the mean position. At that height the velocity of the block will be zero.

Thus $0-v^2=2(-g)s⟹s=\frac{v^2}{2g}$

Now the total height $h=s+y=\frac{v^2}{2g}+y$

$\therefore h=\frac{w^2(A^2-y^2)}{2g}+y$

For $h$ to be maximum, $\frac{dh}{dy}=0$

Thus $\frac{w^2}{2g}(-2y)+1=0$

$⟹y=\frac{g}{w^2}$