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)

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) (1/2π)√(μ g/A) (B) (1/4π)√(μ g/A) (C) 2π√(A/μ g) (D) 4π√(A/μ g)

Tardigrade Admin · Accepted Answer

When restoring force will become equal to the frictional force, block will start to slip.

∴

Restoring force

=

Friction force

\Rightarrow k A = μ m g ... (i)

Now, frequency

f = \frac{1}{2 π} \frac{k}{m}

From Eq. (i),

f = \frac{1}{2 π} \frac{μg}{A}

Q. 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)

$\frac{1}{2 π} \frac{μg}{A}$

$\frac{1}{4 π} \frac{μg}{A}$

$2 π \frac{A}{μg}$

$4 π \frac{A}{μg}$

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

Q. 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)

2π1​Aμg​​

4π1​Aμg​​

2πμgA​​

4πμgA​​