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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, 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 simple harmonic motion 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

Oscillations

Solution:

As, $ \frac{U}{U_{\max }} =\frac{\frac{1}{2} m \omega^{2} y^{2}}{\frac{1}{2} m \omega^{2} a^{2}}=\frac{1}{4} $
$ \frac{y^{2}}{a^{2}} =\frac{1}{4} $
$ \Rightarrow y =\frac{a}{2} $