First slide

Simple hormonic motion

Question

Two simple pendulums of lengths I m and 16 m respectively are both given small displacements in the same direction at the same instant. They will again be in phase after the shorter pendulum has completed n oscillations where n is:

Easy

A

1/4
B

4/3
C

5
D

4

Solution

Let shorter pendulum makes n vibrations then longer makes (n-1) vibrations to come in phase again then ${nT}_{1} = (n - 1) T_{2}$
$\frac{n}{n - 1} = \frac{T_{2}}{T_{1}} = \sqrt{\frac{16}{1}} = 4 \Rightarrow n = \frac{4}{3}$

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