Q117 / 200 · NEET 2009 (Screening) · Physics
A simple pendulum performs simple harmonic motion about $\mathrm{x}=0$ with an amplitude a and time period $T$. The speed of the pendulum at $\mathrm{x}=\frac{\mathrm{a}}{2}$ will be
- A$\frac{\pi \mathrm{a} \sqrt{3}}{2 \mathrm{~T}}$
- B$\frac{\pi \mathrm{a}}{\mathrm{T}}$
- C$\frac{3 \pi^2 a}{T}$
- Correct answerD$\frac{\pi \mathrm{a} \sqrt{3}}{\mathrm{~T}}$
Solution
$$
\begin{aligned}
v=\frac{d y}{d t}=A \omega \cos \omega t & =A \omega \sqrt{1-\sin ^2 \omega t} \\
& =\omega \sqrt{A^2-y^2}
\end{aligned}
$$
Here, $\mathrm{y}=\frac{\mathrm{a}}{2}$
$$
\therefore \mathrm{v}=\omega \sqrt{\mathrm{a}^2-\frac{\mathrm{a}^2}{4}}=\omega \sqrt{\frac{3 \mathrm{a}^2}{4}}=\frac{2 \pi}{\mathrm{T}} \frac{\mathrm{a} \sqrt{3}}{2}=\frac{\pi \mathrm{a} \sqrt{3}}{\mathrm{~T}}
$$
Hence, the correct answer is option: 'D'.