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
Answer
(1) 11
Solution
T = 2 lg T l
TlongTshort = 121100 = 1110
Let Tshort = 10t and Tlong = 11t. They are again in phase at the mean position after the least common multiple of the two periods, i.e. after 110t.
number of vibrations of the shorter pendulum = 110t10t = 11 (the longer one makes 10).
Answer: (1)