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# How long must a pendulum be the Moon, where $g = 16Nk{g^{ - 1}}$, to have a period of 2.0 s?

Last updated date: 13th Jun 2024
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Hint: The time taken by the pendulum to move TO and FRO once is called its time period. The period of a simple pendulum depends upon various factors like the length of the suspension, gravitational acceleration, etc.

Complete step by step answer:
As we know that the time period of a simple pendulum is given by
$T = 2\pi \sqrt {\dfrac{l}{g}}$
Where T is the time period, l is the length of suspension of the pendulum and g is the acceleration due to gravity.
Now as we know the gravity of the moon is around $\dfrac{1}{6}$that of the earth so acceleration due to the gravity of the moon can also be assumed to be around $\dfrac{1}{6}$that of the earth.
As we know the acceleration due to gravity on earth is taken as $9.8m{\sec ^{ - 2}}$
So, the acceleration due to gravity on the moon can be taken as
${g_{moon}} = \dfrac{{{g_{earth}}}}{6}$
${g_{moon}} = \dfrac{{9.8}}{6} = 1.6m{\sec ^{ - 2}}$
Now we need to find the length (l) we are given that time period of oscillation to be 2 seconds and we have calculated acceleration due to gravity $\left( {{g_{moon}}} \right)$
Substituting these values, we get,
$2 = 2\pi \sqrt {\dfrac{l}{{1.6}}}$
$\to {\left( {\dfrac{2}{{2\pi }}} \right)^2} = \dfrac{l}{{1.6}}$
$\to {\left( {\dfrac{1}{\pi }} \right)^2} = \dfrac{l}{{1.6}}$
$\to l = \dfrac{{1.6}}{{{\pi ^2}}}$
$\to l = 0.162m$

Hence, the length of the pendulum should be 0.162 meters to have a time period of 2 sec on the moon.

Note:
The gravity of the moon is $\dfrac{1}{6}$that of the earth. The acceleration due to gravity on the moon can be also assumed as $\dfrac{1}{6}$that of the earth. The time period of the pendulum greatly depends upon its length and acceleration due to gravity.