Oscillations

JEE Main level, single correct, medium

60 of 194

Question 60 of 194

Two simple pendulums of length 1 m1 \mathrm{~m} and 4 m4 \mathrm{~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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Answer: (A) 2

Let T1T_1 and T2T_2 be the time period of the two pendulums T1=2π1 g\mathrm{T}_1=2 \pi \sqrt{\frac{1}{\mathrm{~g}}} and T2=2π4 g\mathrm{T}_2=2 \pi \sqrt{\frac{4}{\mathrm{~g}}} As ℓ1<ℓ2\ell_1 \lt \ell_2 therefore T1<T2\mathrm{T}_1 \lt \mathrm{T}_2 Let at t=0\mathrm{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\mathrm{n} oscillation and shorter length pendulum complete (n+1)(n+1) oscillation. For unison swinging
(n+1)T1=n2(n+1)×2π1g=(n)×2π4g⇒n=1∴n+1=1+1=2\begin{aligned} & (n+1) T_1=n_2 \\ & (n+1) \times 2 \pi \sqrt{\frac{1}{g}}=(n) \times 2 \pi \sqrt{\frac{4}{g}} \\ & \Rightarrow n=1 \\ & \therefore n+1=1+1=2 \end{aligned}

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