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JEE Mains · Physics · STD 11 - 13. oscillations
A simple pendulum of length \(1\, m\) is oscillating with an angular frequency \(10\, rad/s\). The support of the pendulum starts oscillating up and down with a small angular frequency of \(1\, rad/s\) and an amplitude of \(10^{-2}\, m\). The relative change in the angular frequency of the pendulum is best given by
- A \(10^{-3} rad/s\)
- B \(1\,rad/s\)
- C \(10^{-1} rad/s\)
- D \(10^{-5} rad/s\)
Answer & Solution
Correct Answer
(A) \(10^{-3} rad/s\)
Step-by-step Solution
Detailed explanation
Angular frequency of pendulum \(\omega \propto \sqrt{\frac{g_{e f f}}{\ell}}\) \(\therefore \quad \frac{\Delta \omega}{\omega}=\frac{1}{2} \frac{\Delta g_{e f f}}{g_{e f f}}\) \(\Delta \omega=\frac{1}{2} \frac{\Delta g}{g} \times \omega\) \([ \omega_{s}=\) angular frequency of…
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