Question

An object of mass $M$ is attached to a vertical spring which slowly lowers to equilibrium position. This extends the spring by $x$. If the same object is attached to the same vertical spring but permitted to fall suddenly instead, then the extension of the spring is

• A. $x/2$
• B. $2x$
• C. $3x$
• D. $x$

Answer 1

Answer: (B)

$2x$

When the loaded spring is stretched slowly to its equilibrium position through $x$, then the equilibrium
$Mg=kx$
or $x=\frac{Mg}{k}...\left(i\right)$
when object is allowed to fall suddenly, then the decrease in the potential energy of the body is stored as the potential energy of the extended spring. If extension is $x_0$, then
$Mg{x}_0=\frac{1}{2}k.x_0$
$\Rightarrow x_0=\frac{2Mg}{k}...\left(ii\right)$
From Eqs. (i) and (ii) we get $x_0=2x$