Question:
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:
(1) 10 (2) 8
(3) 11 (4) 9
Answer:
Time period of pendulum of length 121 cm is
T1=2π√121g,
Time period of pendulum of length 100 cm is
T2=2π√100g
Let shorter pendulum makes n vibrations; then the longer will make one less than n to come in phase again.
nT2=(n−1)T1
n.2π√100g=(n−1)2π√121g
n=11
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