Question

A mass m is undergoing SHM in vertical direction about the mean position with amplitude A and angular velocity $\omega$ . 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 $y_0$ (measured from the mean position) such that the height 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. $\frac{2\sqrt{2}g}{{\omega }^2}$

Answer 1

The correct option is A

$\frac{g}{{\omega }^2}$

Total height attained by block after detaching from spring above the mean position,

$h=\frac{y+v^2}{2g}$

where v is the velocity of block at a distance y above the mean position.

$v=\omega \sqrt{(A^2-y^2)}$

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

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

or $y_0=\frac{g}{{\omega }^2}$