Question:medium

A bar magnet has a period of oscillation $T$. If a similar brass piece of the same mass is placed over it, then the number of oscillations it makes in one second is

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Frequency $\propto \frac{1}{\sqrt{I}}$ → more inertia = slower oscillation.
Updated On: May 2, 2026
  • $\frac{1}{\sqrt{2}T}$
  • $\frac{\sqrt{2}}{T}$
  • $\frac{1}{2T}$
  • $\frac{2}{T}$
  • $\frac{1}{T}$
Show Solution

The Correct Option is A

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