First slide

Simple hormonic motion

Question

Time period of a simple pendulum of length l is T₁ and time period of a uniform rod of the same length l pivoted about one end and oscillating in a vertical plane is T₂. Amplitude of oscillations in both the cases is small. Then $\frac{T_{1}}{T_{2}}$ is :

Moderate

A

$\frac{1}{\sqrt{3}}$
B

$1$
C

$\sqrt{\frac{4}{3}}$
D

$\sqrt{\frac{3}{2}}$

Solution

For simple pendulum, $T_{1} = 2 π \sqrt{\frac{l}{g}}$
For rod, $T_{2} = 2 π \sqrt{\frac{I}{mgl}} = 2 π \sqrt{\frac{\frac{{ml}^{2}}{3}}{mg \frac{l}{2}}} = 2 π \sqrt{\frac{2 l}{3 g}} \frac{T_{1}}{T_{2}} = \sqrt{\frac{3}{2}}$

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