Question

A ball of mass $2kg$ hanging from a spring oscillates with a time period $2s$. Ball is removed when it is in equilibrium position, then spring shortens by:

• A. $g/{\pi }^2$ meter
• B. $g/2$ meter
• C. $2g$ meter
• D. $2$ meter

Answer 1

Answer: (A)

$g/{\pi }^2$ meter

The block executes $SHM$ about mean position O. When the block is detached from the spring at point O, the spring again reaches upto its natural length.

Let the spring shortens by $x$ meter.

Thus, $F=Mg=kx$

$⟹x=\frac{Mg}{k}$

Also time period of oscillation, $T=2\pi \sqrt{\frac{M}{k}}$

$2=2\pi \sqrt{\frac{M}{k}}⟹k=M{\pi }^2$

$\therefore x=\frac{Mg}{M{\pi }^2}⟹x=\frac{g}{{\pi }^2}$ meter