First slide

Applications of SHM

Question

Two simple pendulums of length 1 m and 16 m respectively are both given small displacement in the same direction at the same instant. They will be again in phase after the shorter pendulum has completed n oscillations. The value of n is

Moderate

A

1/3
B

2/3
C

1
D

4/3

Solution

$T_{1} = 2 π \sqrt{\frac{1}{g}}, T_{2} = 2 π \sqrt{\frac{16}{g}} = 4 T_{1}$
at any time t, phases of pendulums are:
$ϕ_{1} = ω_{1} t = \frac{2 π}{T_{1}} t, ϕ_{2} = ω_{2} t = \frac{2 π}{T_{2}} t$
First pendulum is faster. Both will be in same phase again when faster pendulum has completed one oscillation more than slower pendulum:
$\begin{array}{l} ϕ_{1} - ϕ_{2} = 2 π \\ \Rightarrow \frac{2 π}{T_{1}} t - \frac{2 π}{4 T_{1}} t = 2 π \Rightarrow t = \frac{4 T_{1}}{3} \end{array}$
Number of oscillations completed by shorter pendulum in time t:
$n = \frac{t}{T_{1}} = \frac{4}{3}$

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