Discussion
A simple pendulum of mass m and length l oscillates with a frequency \(\omega_{0}\). A piece breaks off from the pendulum and the mass is reduced to m/2. The frequency of oscillation ω then becomes :
Answer: C
The frequency is unchanged: \(\omega = \omega_{0}\).
For small oscillations a simple pendulum has
\[ \omega_{0} = \sqrt{\frac{g}{l}} \]
which depends only on the length and the acceleration due to gravity. The mass cancels, because both the restoring force \(mg\sin\theta\) and the inertia \(ml\) are proportional to m. Halving the mass, with the length the same, therefore leaves the frequency as it was.
Doubling, halving or multiplying by \(\sqrt{2}\) would all require the frequency to depend on m, which it does not; such mass dependence belongs to a spring-mass oscillator, where \(\omega = \sqrt{k/m}\).
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