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 ($μ =$ 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)

When restoring force will become equal to the frictional force, block will start to slip.
$\therefore $ Restoring force $=$ Friction force
$\Rightarrow \,\,\, k A=\mu m g \,\,\,\,\,\, ...(i)$
Now, frequency $f=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}$
From Eq. (i),
$f=\frac{1}{2 \pi} \sqrt{\frac{\mu g}{A}}$