MHT CET · Physics · Oscillations
A simple pendulum starts oscillating simple harmonically from its mean position ( \(\mathrm{x}=0\) ) with amplitude ' a ' and periodic time ' T '. The magnitude of velocity of pendulum at \(X=\frac{a}{2}\) is
- A \(\frac{3 \pi^2 a}{T}\)
- B \(\frac{\sqrt{3} \pi a}{2 \mathrm{~T}}\)
- C \(\frac{\pi a}{\mathrm{~T}}\)
- D \(\frac{\sqrt{3} \pi a}{T}\)
Answer & Solution
Correct Answer
(D) \(\frac{\sqrt{3} \pi a}{T}\)
Step-by-step Solution
Detailed explanation
\(v = \omega \sqrt{a^2 - X^2}\) \(v = \frac{2\pi}{T} \sqrt{a^2 - \left(\frac{a}{2}\right)^2}\)
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