A body executes SHM with an amplitude A. At what displacement from the mean position is the potential energy of the body one-fourth of its total energy

# A body executes SHM with an amplitude A. At what displacement from the mean position is the potential energy of the body one-fourth of its total energy

1. A

A/4

2. B

A/2

3. C

3A/4

4. D

Some other fraction of A

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### Solution:

$\frac{1}{2}{\mathrm{m\omega }}^{2}{\mathrm{y}}^{2}=\frac{1}{4}\left(\frac{1}{2}{\mathrm{m\omega }}^{2}{\mathrm{A}}^{2}\right)⇒\mathrm{y}=\frac{\mathrm{A}}{2}$

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