Question

The ratio between kinetic and potential energies of a body executing simple harmonic motion, when it is at a distance of $ \frac{1}{N} $ of its amplitude from the mean position is

• A. $ N^{2}+1 $
• B. $ \frac{1}{N^{2}} $
• C. $ N^{2} $
• D. $ N^{2}-1 $

Answer 1

Answer: (D)

The kinetic energy
$KE =\frac{1}{2}\, m \omega^{2}\left[ a ^{2}-\left(\frac{ a }{N}\right)^{2}\right]\,\,\,...(i)$
The potential energy
$PE =\frac{1}{2}\, m \omega^{2} \frac{ a ^{2}}{N^{2}}\,\,\,...(ii)$
From the Eqs. (i) and (ii), we get
$\frac{ KE }{ PE }=\frac{\frac{1}{2} \,m \omega^{2}\left[ a ^{2}-\left(\frac{ a }{N}\right)^{2}\right]}{\frac{1}{2} \,m \omega^{2} \frac{ a ^{2}}{N^{2}}}$
$=\frac{a^{2}-\frac{a^{2}}{N^{2}}}{\frac{a^{2}}{N^{2}}}=\frac{\frac{a^{2}}{N^{2}}\left(N^{2}-1\right)}{\frac{a^{2}}{N^{2}}}=N^{2}-1$