Oscillations

JEE Main level, single correct, medium

77 of 194

Question 77 of 194

A particle executes SHM of amplitudeAA. The distance from the mean position when its kinetic energy becomes equal to its potential energy is:

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Answer: (A) 12A\dfrac{1}{\sqrt{2}}A

The potential energy is given by
U=mω2x22U=\dfrac{m{\omega }^{2}{x}^{2}}{2}
The kinetic energy is given by K=mω2(A2−x2)2K=\dfrac{m{\omega }^{2}({A}^{2}-{x}^{2})}{2}.
Using the condition that K=UK=U,
12mω2(A2−x2)=12mω2x2\dfrac{1}{2}m{\omega }^{2}\left({A}^{2}-{x}^{2}\right)=\dfrac{1}{2}m{\omega }^{2}{x}^{2}
⇒x=A2\Rightarrow x=\dfrac{A}{\sqrt{2}}

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