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  4. When the displacement of a particle executing simple harmonic motion is half its amplitude, the ratio of its kinetic energy to potential energy is

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Q. When the displacement of a particle executing simple harmonic motion is half its amplitude, the ratio of its kinetic energy to potential energy is

Oscillations

Solution:

$x=A \cos \omega t$
$x=A / 2 $
$\text { we get } \omega t=60^{\circ} $
$\frac{K E}{P E}=\frac{\frac{1}{2} m \omega^{2} A^{2} \sin ^{2} \omega t}{\frac{1}{2} m \omega^{2} A^{2} \cos \omega t}=\tan ^{2} \omega t $
$=\tan ^{2} 60^{\circ}=3$