Oscillations

Question : 15 of 25

Marks: +1, -0

The acceleration due to gravity on the surface of moon is

1.7\,\text{m s}^{-2}

. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s? (g on the surface of earth is

9.8\,\text{m s}^{-2})

Solution:

Using the formula

T=2\pi\sqrt{\frac{l}{g}},

we get

For earth

T_{e}=2\pi\sqrt{\frac{l}{g_{e}}},\dots(i)

and for moon,

T_{m}=2\pi\sqrt{\frac{l}{g_{m}}}\dots(ii)

Dividing (ii) by (i), we get

\frac{T_{m}}{T_{e}}=\sqrt{\frac{g_{e}}{g_{m}}}=\sqrt{\frac{9.8}{1.7}}

T_{m}=3.5\times\sqrt{\frac{98}{17}}

=3.5\sqrt{5.7647}

=3.5\times2.4

T_{m}=8.4\,\text{s}

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