Question

A block resting on the horizontal surface executes S.H.M. in horizontal plane with amplitude 'A'. The frequency of oscillation for which the block just starts to slip is ($\mu =$ coefficient of friction, g$=$ gravitational acceleration).

• A. $\frac{1}{2\pi }\sqrt{\frac{\mu g}{A}}$
• B. $\frac{1}{4\pi }\sqrt{\frac{\mu g}{A}}$
• C. $2\pi \sqrt{\frac{A}{\mu g}}$
• D. $4\pi \sqrt{\frac{A}{\mu g}}$

Answer 1

Answer: (A)

$\frac{1}{2\pi }\sqrt{\frac{\mu g}{A}}$

As the block just starts to slip, thus force acting on it is $\mu mg$.

$\therefore$ $\mu mg=kA$

where $A$ is the maximum displacement.

$⟹$ $\frac{k}{m}=\frac{\mu g}{A}$

Time period of oscillation $T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{A}{\mu g}}$

Thus frequency of oscillation $\nu =\frac{1}{T}=\frac{1}{2\pi }\sqrt{\frac{\mu g}{A}}$