Question

Two simple pendulums of length 1m and 4m 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 number of oscillations equal to:

Answer 1

Answer:

They will be again in phase after the shorter pendulum has completed 2 oscillations.

Explanation:

Angular frequency of SHM of the pendulum is given as

$\omega = \sqrt{\frac{g}{L}}$

so we have

$\omega_1 = \sqrt{\frac{g}{1}} = \omega_o$

$\omega_2 =\sqrt{\frac{g}{4}} = \frac{1}{2}\omega_o$

now the relative angular frequency of two pendulum

$\omega_1 - \omega_2 = \frac{1]{2}\omega_o$

so in order to come back in same phase again the time taken by it is given as

$t = \frac{\theta}{\omega}$

$t = \frac{2\pi}{\frac{1}{2}\omega_o}$

$t = 2(\frac{2\pi}{\omega_o})$

so in this time shorter pendulum will complete its 2 oscillations

#Learn

Topic : SHM

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