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QNEET-2022-Ph-431: NEET 2022 Physics - Class 11 NEET-2022-Ph-431 Solution: Two pendulums of length 121 cm and 100 cm start vibrating in…
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: 11 9 10 8
Multiple Choice Options
- A 11
- B 9
- C 10
- D 8
Step-by-Step Verified Solution
T = 2 � L g
Let n 1 and n 2 be integer.
n 1 T 1 = n 2 T 2
2 � n 1 1.21 g = 2 � n 2 1.00 g
_ n 2 n 1 = 11 10
_ After completion of 11 th oscillation of shorter pendulum, it will be in phase with longer pendulum.