Question:
In a simple harmonic oscillation, what fraction of total mechanical energy is in the form of kinetic energy, when the particle is midway between mean and extreme position.
Correct Option: , 2
Solution:
$\mathrm{K}=\frac{1}{2} \mathrm{m \omega}{ }^{2}\left(\mathrm{~A}^{2}-\mathrm{x}^{2}\right)$
$=\frac{1}{2} m \omega^{2}\left(\mathrm{~A}^{2}-\frac{\mathrm{A}^{2}}{4}\right)$
$=\frac{1}{2} \mathrm{~m} \omega^{2}\left(\frac{3 \mathrm{~A}^{2}}{4}\right)$
$\mathrm{K}=\frac{3}{4}\left(\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}\right)$
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