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  4. When the potential energy of a particle executing SHM is one-fourth of its maximum value during the oscillation, the displacement of the particle from the equilibrium position in terms of its amplitude a is

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Q. When the potential energy of a particle executing SHM is one-fourth of its maximum value during the oscillation, the displacement of the particle from the equilibrium position in terms of its amplitude a is

Chhattisgarh PMTChhattisgarh PMT 2011

Solution:

Maximum potential energy of particle executing SHM
$U_{\max }=\frac{1}{2} m \omega^{2} a^{2}$
Potential energy $U =\frac{1}{4} U_{\max }$ (given)
$\frac{1}{2} m \omega^{2} y^{2}=\frac{1}{4}\left(\frac{1}{2} m \omega^{2} a^{2}\right)$
$y^{2}=\frac{a^{2}}{4}$
$\Rightarrow y=\frac{a}{2}$