Question

To the free end of spring hanging from a rigid support, a block of mass $m$ is hung and slowly allowed to come to its equilibrium position. Then stretching in the spring is $d$. If the same block is attached to the same spring and allowed to fall suddenly, the amount of stretching is : (force constant, $k$ )

• A. $\frac{m g}{k}$
• B. $2\, d$
• C. $\frac{m g}{3 k}$
• D. $4 \,d$

Answer 1

Answer: (B)

To leave the block, it oscillates in vertical plane. If maximum extension in spring in extreme position of block is $x_{1}$, then
Work done by weight of the block $=$ potential energy stored in spring
$m g\, x=\frac{1}{2} k x^{2}$
$\therefore x=2 \frac{m g}{k}=2 d$
$\left(\because d=\frac{m g}{k}\right)$