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  • Two simple pendulums of length 1 m and 4 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 compl

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

Option: 1

2


Option: 2

7


Option: 3

5


Option: 4

3


Answers (1)

best_answer

Angular frequency of SHM of the pendulum is given as
\omega=\sqrt{\frac{g}{L}}
so we have
\begin{array}{l} \omega_{1}=\sqrt{\frac{q}{1}}=\omega_{0} \\ \\ \omega_{2}=\sqrt{\frac{q}{4}}=\frac{1}{2} \omega_{0} \end{array}

 

\begin{aligned} &t=\frac{\theta}{\omega}\\ &\begin{array}{l} t=\frac{2 \pi}{\frac{1}{2} \omega_{o}} \\ t=2\left(\frac{2 \pi}{\omega_{o}}\right) \end{array} \end{aligned}

so in this time shorter pendulum will complete its 2 oscillations

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

Posted by

manish

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