Question

A uniform thin rod of length L is oscillating in vertical plane about a fixed horizontal axis passing through one of its end. Frequency of small oscillations of the rod is Va. If bottom half of the stick were cut off, then its new frequency would become​

Answer 1

The new frequency would become √2 ν₀

Therefore, option (3) is correct.

Explanation:

We know that time period of oscillation of a pendulum of length l is given by

$T=2\pi\sqrt{\frac{l}{g}}$

Initial time period

$T_0=2\pi\sqrt{\frac{L}{g}}$

Thus, the frequency

$\nu_0=\frac{1}{T_0}$

$\implies \nu_0=\frac{1}{2\pi}\sqrt{\frac{g}{L}}$

When the length is cut to half, the new length l' = L/2

Then the final frequency

$\nu_f=\frac{1}{2\pi}\sqrt{\frac{g}{L/2}}$

$\implies \nu_f=\sqrt{2}\times \frac{1}{2\pi}\sqrt{\frac{g}{L}}$

$\implies \nu_f=\sqrt{2}\nu_0$

Hope this answer is helpful.

Know More:

Q: A simple pendulum is suspended from the ceiling of a lift. when the lift is at rest its time period is t. with what acceleration should the lift be accelerated upwards in order to reduce its period to t ∕2? (g is acceleration due to gravity).

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