Oscillations - Result Question 41
45. A simple pendulum performs simple harmonic motion about $x=0$ with an amplitude a and time period T. The speed of the pendulum at $x=\frac{a}{2}$ will be:
[2009]
(a) $\frac{\pi a}{T}$
(b) $\frac{3 \pi^{2} a}{T}$
(c) $\frac{\pi a \sqrt{3}}{T}$
(d) $\frac{\pi a \sqrt{3}}{2 T}$
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Answer:
Correct Answer: 45. (c)
Solution:
(c) Speed $v=\omega \sqrt{a^{2}-x^{2}}, x=\frac{a}{2}$
$\therefore \quad v=\omega \sqrt{a^{2}-\frac{a^{2}}{4}}=\omega \sqrt{\frac{3 a^{2}}{4}}=\frac{2 \pi}{T} \frac{a \sqrt{3}}{2}=\frac{\pi a \sqrt{3}}{T}$
