A block resting on the horizontal surface executes S.H.M. in hori… - Vidyadip

MCQ

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)

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

Correct option: A.

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

(a) : Let $m$ be mass of the block. When the block is about to slip, then Force of friction $=$ Centrifugal force
$
\begin{aligned}
& \mu m g=m \omega^2 A \\
& \omega^2=\frac{\mu g}{A} \text { or } \omega=\sqrt{\frac{\mu g}{A}}
\end{aligned}
$
As $\omega=2 \pi \nu$
$
\therefore \quad v=\frac{\omega}{2 \pi}=\frac{1}{2 \pi} \sqrt{\frac{\mu g}{A}}
$

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