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
$\frac{KE}{PE}=\frac{\frac{1}{2}m{\omega }^2\left[a^2-{\left(\frac{a}{N}\right)}^2\right]}{\frac{1}{2}m{\omega }^2\frac{a^2}{N^2}}$
$=\frac{a^2-\frac{a^2}{N^2}}{\frac{a^2}{N^2}}=\frac{\frac{a^2}{N^2}\left(N^2-1\right)}{\frac{a^2}{N^2}}=N^2-1$