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A particle is executing linear simple harmonic motion of amplitude A. At what displacement is the energy of the particle half potential and half kinetic?
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1720 Upvotes
Verified Answer
The correct answer is:
$\frac{\mathrm{A}}{\sqrt{2}}$
Hints : Total Energy $(\mathrm{E})=\frac{1}{2} m \omega^2 \mathrm{~A}^2$
P.E. $=\frac{1}{2} m \omega^2 x^2$
As P.E. $=\frac{E}{2}$
Then, $\frac{1}{2} m \omega^2 \mathrm{~A}^2 \times \frac{1}{2}=\frac{1}{2} m \omega^2 x^2$
$$
x^2=\frac{\mathrm{A}^2}{2} \Rightarrow x=\frac{\mathrm{A}}{\sqrt{2}}
$$
P.E. $=\frac{1}{2} m \omega^2 x^2$
As P.E. $=\frac{E}{2}$
Then, $\frac{1}{2} m \omega^2 \mathrm{~A}^2 \times \frac{1}{2}=\frac{1}{2} m \omega^2 x^2$
$$
x^2=\frac{\mathrm{A}^2}{2} \Rightarrow x=\frac{\mathrm{A}}{\sqrt{2}}
$$
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