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A body performs linear simple harmonic motion of amplitude 'A". At what
displacement from the mean position, the potential energy of the body is one
fourth of its total energy?
Options:
displacement from the mean position, the potential energy of the body is one
fourth of its total energy?
Solution:
1642 Upvotes
Verified Answer
The correct answer is:
$\frac{\mathrm{A}}{2}$
Total energy $=\frac{1}{2} \mathrm{KA}^{2}$
Potential energy $=\frac{1}{2} \mathrm{kx}^{2}$
If P.E. $=\frac{1}{4}(\mathrm{~K} . \mathrm{E} .)$ then
$\frac{1}{2} \mathrm{kx}^{2}=\frac{1}{4}\left(\frac{1}{2} \mathrm{kA}^{2}\right)$
$\therefore x^{2}=\frac{A^{2}}{4}$ or $x=\frac{A}{2}$
Potential energy $=\frac{1}{2} \mathrm{kx}^{2}$
If P.E. $=\frac{1}{4}(\mathrm{~K} . \mathrm{E} .)$ then
$\frac{1}{2} \mathrm{kx}^{2}=\frac{1}{4}\left(\frac{1}{2} \mathrm{kA}^{2}\right)$
$\therefore x^{2}=\frac{A^{2}}{4}$ or $x=\frac{A}{2}$
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