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

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

Chhattisgarh PMTChhattisgarh PMT 2005

Solution:

Potential energy of particle executing SHM is given by
$P E=\frac{1}{2} \,m \,\omega^{2} \,y^{2}$
maximum $P E\left(U_{\max }\right)=\frac{1}{2}\, m\, \omega^{2} \,a^{2}$
Given $P E(U)=\frac{1}{4} U_{\max }$
$\therefore \frac{1}{2}\, m \omega^{2}\, y^{2}=\frac{1}{4} \frac{1}{2} \,m\, \omega^{2}\, a^{2}$
or $y^{2}=a^{2} / 4$
$\Rightarrow y=\pm \frac{a}{2}$