To the free end of spring hanging from a rigid support, a block of mass $m$ is hung and slowly allowed to…
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, $k$ )
$\frac{m g}{k}$
$2 d$
$\frac{m g}{3 k}$
$4 d$
Solution
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
$m g x=\frac{1}{2} k x^2$
$\therefore \quad x=2 \frac{m g}{k}=2 d \quad\left(\because d=\frac{m g}{k}\right)$