Test Index

Oscillations Part 1

Section: Physics

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Question : 20 of 100

Marks: +1, -0

For what value of displacement the kinetic energy and potential energy of a simple harmonic oscillation become equal?

[17 Mar 2021 Shift 1]

$x = 0$
$x = \pm A$
$x = \pm \;\frac{A}{\sqrt{2}}$
$x = \;\frac{A}{2}$

Solution:

Potential energy of

\text{SHM} = \;\frac{1}{2} k x^{2} = \;\frac{1}{2} m \omega^{2} x^{2}

where

m=

mass of particle,

\omega=

angular velocity

and

x=

displacement.

Kinetic energy of

\text{SHM} = \;\frac{1}{2} m \left( \omega \sqrt{A^{2} - x^{2}} \right)^{2}

Here,

A

is the amplitude of SHM.

According to question,

Potential energy of

\text{SHM}=

Kinetic energy of

\text{SHM}

\;\frac{1}{2} m \omega^{2} x^{2} = \;\frac{1}{2} m \left( \omega \sqrt{A^{2} - x^{2}} \right)^{2}

\Rightarrow \;\; \omega^{2} x^{2} = \omega^{2} \left( \sqrt{A^{2} - x^{2}} \right)^{2} \Rightarrow \omega^{2} x^{2} = \omega^{2} \left( A^{2} - x^{2} \right)

\Rightarrow \;\; \omega^{2} x^{2} = \omega^{2} A^{2} - \omega^{2} x^{2} \Rightarrow 2 \omega^{2} x^{2} = \omega^{2} A^{2}

\Rightarrow \;\; 2 x^{2} = A^{2} \Rightarrow x^{2} = \;\frac{A^{2}}{2}

\Rightarrow \;\; x = \pm \sqrt{\;\frac{A^{2}}{2}} \Rightarrow x = \pm \;\frac{A}{\sqrt{2}}

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