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Periodic Motion

The time peri...

Question

The time period of a second's pendulum on the surface of moon is found to be $5 s$ . The acceleration due to gravity on the surface of moon is:

A

3.2 m / s^{- 2}

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B

1.6 m / s^{- 2}

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C

0.8 m / s^{- 2}

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D

0.6 m / s^{- 2}

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Solution

The correct option is D $1.6 m / s^{- 2}$
Time period, $T \propto 1 \sqrt{g}$
Thus, $\frac{T_{m}}{T_{e}} = \sqrt{\frac{g_{e}}{g_{m}}}$
Thus, $2.5 / 1 = \sqrt{10 / g_{m}}$ or $g_{m} = 1.6$

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