Q 11-13-155NEETNEET 2022Top questionMedium
Two pendulums of length $121$ cm and $100$ cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is:
Answer: (B) $11$
$T \propto \sqrt{l}$, so $\dfrac{T_{long}}{T_{short}} = \sqrt{\dfrac{121}{100}} = \dfrac{11}{10}$.
The shorter pendulum is faster. They are next in phase when it has made exactly one vibration more than the longer one:
$$nT_{short} = (n - 1)T_{long} \;\Rightarrow\; 10n = 11(n - 1) \;\Rightarrow\; n = 11$$
Solution by Sreeraj P, M.Sc Physics