Question

potential energy of a particle performing liner S.H.M is 0.1π^2 x^2 joule .if mass of the the particle is 20g. find the frequency of S.H.M​

Answer 1

The potential energy of a particle executing SHM varies with displacement x as,

$\longrightarrow\sf{U=\dfrac{1}{2}\,m\omega^2x^2}$

where,

• m = mass of the particle

• ω = angular frequency

Here,

$\longrightarrow\sf{U=0.1\pi^2x^2}$

$\longrightarrow\sf{\dfrac{1}{2}m\omega^2x^2=0.1\pi^2x^2}$

Since $\sf{m=20\ g=0.02\ kg,}$

$\longrightarrow\sf{\dfrac{1}{2}\times0.02\omega^2=0.1\pi^2}$

$\longrightarrow\sf{0.01\omega^2=0.1\pi^2}$

From this we get,

$\longrightarrow\sf{\dfrac{\omega^2}{\pi^2}=\dfrac{0.1}{0.01}}$

$\longrightarrow\sf{\dfrac{\omega^2}{\pi^2}=10}$

Taking square roots,

$\longrightarrow\sf{\dfrac{\omega}{\pi}=\sqrt{10}}$

And dividing both sides by 2,

$\longrightarrow\sf{\dfrac{\omega}{2\pi}=\dfrac{\sqrt{10}}{2}}$

$\longrightarrow\sf{\dfrac{\omega}{2\pi}=\dfrac{\sqrt{5}\times\sqrt2}{2}}$

$\longrightarrow\sf{\dfrac{\omega}{2\pi}=\sqrt{\dfrac{5}{2}}}$

But, $\sf{\dfrac{\omega}{2\pi}=\nu,}$ frequency. Therefore, frequency of the SHM is,

$\longrightarrow\underline{\underline{\sf{\nu=\sqrt{\dfrac{5}{2}}\ s^{-1}}}}$