A pendulum clock keeps correct time on the surface of the Earth where

PHYSICS · Multi-part · CBSE · Physics · Oscillations · Oscillations

Problem

A pendulum clock keeps correct time on the surface of the Earth where gE=9.80 m s−2g_{\text E} = 9.80\,\text{m}\,\text{s}^{-2}. The pendulum has a length such that its time period is exactly 2.00 s2.00\,\text{s}. (i) Calculate the length of the pendulum on the Earth. (ii) The same clock is taken to the Moon where gM=1.63 m s−2g_{\text M} = 1.63\,\text{m}\,\text{s}^{-2}. To keep the same time period of 2.00 s2.00\,\text{s}, find the length the pendulum must have on the Moon. (iii) Compute the ratio LMLE\dfrac{L_{\text M}}{L_{\text E}} and comment briefly on whether the pendulum should be made longer or shorter on the Moon to keep the clock accurate.

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