First slide

Applications of SHM

Question

Two simple pendulums whose lengths are 100 cm and 121 cm are suspended side by side. Their bobs are pulled together and then released. After how many minimum oscillations of the longer pendulum, will the two be in phase again

Moderate

A

11
B

10
C

21
D

20

Solution

Let $T_{1}$ and $T_{2}$ are the time period of the two pendulums $T_{1} = 2 π \sqrt{\frac{100}{g}}$ and $T_{2} = 2 π \sqrt{\frac{121}{g}}$
$(T_{1} < T_{2}$ because $l_{1} < l_{2})$ .
Let at $t = 0$ , they start swinging together. Since their time periods are different, the swinging will not be in unision always. Only when number of completed oscillation differs by an integer, the two pendulum will again begin to swing together.
Let longer length pendulum complete n oscillation and shorter length pendulum complete (n+1) oscillation, for the unision swinging, then $(n + 1) T_{1} = {nT}_{2}$
$(n + 1) \times 2 π \sqrt{\frac{100}{g}} = n \times 2 π \sqrt{\frac{121}{g}} \Rightarrow n = 10$

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