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NEET 2022 (Phase 1) · Simple Pendulum

Simple Pendulum question from NEET 2022 (Phase 1)

Q55NEET 2022 · PHASE 1Medium

Simple Pendulum

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:

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Answer: (a)

  1. T∝lT \propto \sqrt{l}: TlongTshort=121100=1110\frac{T_{long}}{T_{short}} = \sqrt{\frac{121}{100}} = \frac{11}{10}.
  2. They are in phase again when the shorter one makes one more vibration: n⋅10=(n−1)⋅11n\cdot 10 = (n-1)\cdot 11.
  3. n=11n = 11 vibrations of the shorter pendulum (10 of the longer).
  4. So the answer is (a) 11.
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