Question

A particle executes s.h.m with a period of 10 seconds.Find a time in which its potential energy will be half of its total energy ​

Answer 1

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▶️ ${{ \underline { \bf{t=1.25sec}}}}$

Given :-

T = 10 sec.

We have to find out

t = ?

Explanation :-

$p.e = \frac{1}{2} \: t.e$

$p.e = \frac{1}{2} \: mw {}^{2} x {}^{2}$

$t.e = \frac{1}{2} \: mw {}^{2} a {}^{2}$

$p.e = \frac{1}{2} \: t.e.$

$\frac{1}{2} \: mw {}^{2} x {}^{2} = \frac{1}{2} ( \frac{1}{2} \: mw {}^{2} a {}^{2} )$

$x {}^{2} = \frac{a {}^{2} }{2}$

$x = \frac{a}{ \sqrt{2} }$

$x = a \sin \: wt$

$\frac{a}{ \sqrt{2 } } = a \sin( \frac{2\pi}{t} )t$

$\frac{a}{ \sqrt{2} } = \: \sin( \frac{2\pi}{10} ) t$

$\frac{1}{ \sqrt{2} } = \sin( \frac{\pi}{5}t )$

$\frac{\pi}{5} t = sin { - }^{1} ( \frac{1}{ \sqrt{2} } )$

$\frac{\pi}{5} t = \frac{\pi}{4}$

$t = \frac{5}{4}$

$t = 1.25 \: sec.$

Hence,

t = 1.25 sec Is the right answer.

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Answer 2

The time in which potential energy of a particle will be half of its total energy ​is 1.25 s.

Explanation:

The time period of a particle executes S.H.M, T = 10 s.

The equation for S.H.M is given as

$x =A sin\omega t$

Here, x is the displacement of the particle and A is the amplitude of s.h.m and $\omega$ is the angular frequency.

Now potential energy in S.H.M is

$P=\frac{1}{2}m\omega^2x^2$

Total energy in S.H.M is

$T=\frac{1}{2}m\omega^2A^2$

As per question,

$P = \frac{1}{2} T$

Therefore,

$\frac{1}{2}m\omega^2x^2=\frac{1}{2} (\frac{1}{2}m\omega^2A^2)$

$x=\frac{A}{\sqrt{2} }$

So, from equation of S.H.M,

$\frac{A}{\sqrt{2} }=A sin\omega t$

$\omega t=sin^{-1} (\frac{1}{\sqrt{2} } )$

or

$\frac{2\pi}{T} t=\frac{\pi}{4}$                         ( as   $\omega =\frac{2\pi}{T}$ )

$t=\frac{10}{8}$

t = 1.25 s.

Thus, the time in which potential energy of a particle will be half of its total energy ​is 1.25 s.

#Learn More: S.H.M

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