First slide

Simple hormonic motion

Question

A uniform rod of length 2.0 m is suspended through an end and is set oscillating with small amplitude under gravity. The time period of oscillation is approximately

Moderate

A

1.60 s
B

1.80 s
C

2.00 s
D

2.40 s

Solution

$T = 2 π \sqrt{\frac{I}{mgl}} = 2 π \sqrt{\frac{\frac{{mL}^{2}}{3}}{mg (\frac{L}{2})}} = 2 π \sqrt{\frac{2 L}{3 g}} = 2.40 \sec$

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