Pendulum on Planet X

A simple pendulum with time period of oscillation equal to 17.32 seconds on the Earth is set into Simple Harmonic motion on Planet X where it weighs one third of its weight on the Earth. What is the Time period of oscillation of the pendulum on Planet X. Assume that there is no loss of energy due to Friction or air resistance where applicable. Round off your answer to the nearest integer


The answer is 30.

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1 solution

Srinivasa Gopal
Mar 10, 2020

T = 2π × √(L/g) g on Planet X is g/3 , so T on Planet X is V3 times T on Earth or T on Planet X = 17.32* 1.732 =30 seconds

The actual time period is not an integer. It is 29.999119987093 29.999119987093 sec. The question should be "give your answer as an integer nearest to the time period". Or the time period on the earth should be given as 10 3 10\sqrt 3 seconds.

A Former Brilliant Member - 1 year, 3 months ago

Thanks for pointing this out. Have modified the problem statement.

Srinivasa Gopal - 1 year, 3 months ago

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