Two simple pendulums of length 0.5 m and 20 m respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has
completed x oscillations, where k is

1

3

2

5

Solution

Time period of simple pendulum

For 1st pendulum T₁ = $2 π \sqrt{\frac{5}{g}}$

= $2 π \sqrt{\frac{20}{9.8}}$

= $\sqrt{20} s$

For IInd pendulum T₂ = $2 π \sqrt{\frac{20}{g}}$

= 2 $π \sqrt{\frac{20}{9.8}}$

= 2 $\sqrt{20} s$

Suppose the time will be in same phase is t then

$\frac{1}{t} = \frac{1}{T_{1}} - \frac{1}{T_{2}}$

= $\frac{1}{\sqrt{20}} - \frac{1}{2 \sqrt{20}}$

= $\frac{1}{2 \sqrt{20}}$

$t = 2 \sqrt{20}$

Hence the number of oscillation of simple pendulum of shorter wavelength

(x) = $\frac{2 \sqrt{20}}{\sqrt{20}}$

(x) = 2

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