A block rests on a horizontal table which is executing SHM in the horizontal plane with an amplitude ' a '. If the coefficient of friction is ' μ ', then the block just starts to slip when the frequency of oscillation is

Question

A block rests on a horizontal table which is executing SHM in the horizontal plane with an amplitude ' a '. If the coefficient of friction is ' μ ', then the block just starts to slip when the frequency of oscillation is (A) (1/2 π) √(μ g / a ) (B) √(μ g / a ) (C) 2 π √(a/μ g) (D) √(a/μ g)

Tardigrade Admin · Accepted Answer

For the block is about to slip, μ g=ω2 a ω=√(μ g/a) ⇒ v=(1/2 π) √(μ g/a)

Q. A block rests on a horizontal table which is executing SHM in the horizontal plane with an amplitude ' $a$ '. If the coefficient of friction is ' $μ$ ', then the block just starts to slip when the frequency of oscillation is

$\frac{1}{2 π} \frac{μg}{a}$

$\frac{μg}{a}$

$2 π \frac{a}{μg}$

$\frac{a}{μg}$

For the block is about to slip, $μg = ω^{2}$ a
$ω = \frac{μg}{a} \Rightarrow v = \frac{1}{2 π} \frac{μg}{a}$

Q. A block rests on a horizontal table which is executing SHM in the horizontal plane with an amplitude ' a '. If the coefficient of friction is ' μ ', then the block just starts to slip when the frequency of oscillation is

2π1​aμg​​

aμg​​

2πμga​​

μga​​