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Two simple pendulums of length 5m and 20m respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has completed ....... oscillations
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Correct Option: C
Let the pendulums be in phase after t sec of start. Within this time, if the bigger pendulum executes n oscillations, the smaller one will have executed (n + 1) oscillations.
Now, the time of n oscillation = 2π | ![]() | |
× n | ||
g |
& the time of (n + 1) oscillation = 2π | ![]() | |
× (n + 1) | ||
g |
To be in phase
2π | ![]() | ![]() | × n = 2π | ![]() | ![]() |
× (n + 1) | |||||
g | g |
or, 2n = n + 1
or, n = 1
Hence, the no. of oscillations executed by shorter pendulum = n + 1 = 1 + 1 = 2