Q143 / 160 · AP EAMCET 2014 · Physics
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$N^2$
- Correct answerD$N^2-1$
Solution
The kinetic energy
$$
\mathrm{KE}=\frac{1}{2} m \omega^2\left[\mathrm{a}^2-\left(\frac{\mathrm{a}}{\mathrm{N}}\right)^2\right]
$$
The potential energy
$$
\mathrm{PE}=\frac{1}{2} m \omega^2 \frac{\mathrm{a}^2}{\mathrm{~N}^2}
$$
From the Eqs. (i) and (ii), we get
$$
\begin{aligned}
\frac{\mathrm{KE}}{\mathrm{PE}} & =\frac{\frac{1}{2} m \omega^2\left[a^2-\left(\frac{a}{N}\right)^2\right]}{\frac{1}{2} m \omega^2 \frac{a^2}{N^2}} \\
& =\frac{a^2-\frac{a^2}{N^2}}{\frac{a^2}{N^2}}=\frac{\frac{a^2}{N^2}\left(N^2-1\right)}{\frac{a^2}{N^2}}=N^2-1
\end{aligned}
$$
Hence, the correct answer is option: 'D'.