Question

A mass-spring system oscillates such that the mass moves on a rough surface having coefficient of friction $\mu$. It is compressed by a distance a from its normal length and, on being released, it moves to a distance $b$ from its equilibrium position. The decrease in amplitude for one half-cycle $(-atob)$ is

• A. $\frac{\mu mg}{K}$
• B. $\frac{2\mu mg}{K}$
• C. $\frac{\mu g}{K}$
• D. $\frac{K}{\mu mg}$

Answer 1

The correct option is B $\frac{2\mu mg}{K}$

From $-a$ to $b(\Delta KE=0$ as mass will stop for a moment at extreme position) decrease in elastic potential energy = work done against friction.
$\therefore \frac{1}{2}K{a}^2-\frac{1}{2}K{b}^2=\mu mg(a+b)$
or $(a-b)=\frac{2\mu mg}{K}$
$\therefore \text{Decrease in amplitude}=\frac{2\mu mg}{K}$