Two simple pendulums of length 1, m and 4, m | vedclass

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Question

Let $\mathrm{T}_{1}$ and $\mathrm{T}_{2}$ be the time period of the two

pendulums $T_{1}=2 \pi \sqrt{\frac{1}{g}}$ and $T_{2}=2 \pi \sqrt{\frac{4}{

Vedclass · Accepted Answer

$2$

Vedclass · Answer

Let $\mathrm{T}_{1}$ and $\mathrm{T}_{2}$ be the time period of the two

pendulums $T_{1}=2 \pi \sqrt{\frac{1}{g}}$ and $T_{2}=2 \pi \sqrt{\frac{4}{g}}$

As $\ell_{1}<\ell_{2}$ therefore $\mathrm{T}_{1}<\mathrm{T}_{2}$

Let at $t=0$ they start swinging together. since their time periods are different, the swinging will not be in unison always. Only when number of completed oscillations

differ by an integer, the two pendulums will again begin to swing together

Let longer length pendulum complete $n$ oscillation and shorter length pendulum complete $(\mathrm{n}+1)$ oscillation. For unison swinging

$(n+1) T_{1}=n T_{2}$

$(n+1) 	imes 2 \pi \sqrt{\frac{1}{g}}=(n) 	imes 2 \pi \sqrt{\frac{4}{g}}$

$\Rightarrow n=1$

$	herefore n+1=1+1=2$

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

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