Question

a block of mass m rests on a horizontal table executing shm along horizontal with amplitude a when just starts to slip then the period is? (meu is coefficient of friction)

Answer 1

Answer:

Explanation:

F restoring =  F friction

F friction  = Ц mg

F restoring =  mw²A

Restoring force on the block = mω2A = μmg.

∴ Acceleration in the block = μg

Frequency of oscillation,

v  =   1  / 2 √acceleration / Displacement

=  1/2 π √μg / A

Answer 2

When the block just starts to slip the time period will be $2\pi\sqrt{\dfrac{A}{\mu g}}$.

Explanation:

The displacement of particle from its mean position is given by :

$y=A\ cos(kx-\omega t)$

Where

k is the propagation constant

$\omega$ is the angular speed

A is the amplitude of wave

When the mass m just starts to slip, then the condition is given by :

maximum restoring force = limiting friction force

$m\omega^2A=\mu mg$

$\omega^2 A=\mu g$

$\omega=\sqrt{\dfrac{\mu g}{A}}$

`Since, $\omega=\dfrac{2\pi }{T}$

T is the period

$T=2\pi\sqrt{\dfrac{A}{\mu g}}$

Hence, when the block just starts to slip the time period will be $2\pi\sqrt{\dfrac{A}{\mu g}}$. Hence, this is the required solution.

Learn more :

Simple Harmonic Motion

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