What happens to the time period of a simple pendulum when it is taken from earth to the moon?

In moon, if you take a pendulum of earth［keeping the system unchanged,you will see that the pendulum swings slowly...

The system is unchanged, that is the possible friction at the knot of support and thread hanging from it is unchanged, so from the formula of time period , you can say easily that "L" remains constant and so "g" must decrease...

Besides, the air of moon is not "strong" enough to hold back the slowly-moving pendulum...This it's air resistance is negligible or there is hardly any air...

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50 Questions 150 Marks 60 Mins

CONCEPT:

Simple Pendulum:

An ideal simple pendulum consists of a heavy point mass body (bob) suspended by a weightless, inextensible, and perfectly flexible string from rigid support about which it is free to oscillate.
For a simple pendulum, the time period of swing of a pendulum depends on the length of the string and acceleration due to gravity.

\(T = 2\;{\rm{\Pi }}\sqrt {\frac{l}{{\rm{g}}}}\)

The above formula is only valid for small angular displacements.

Where, T = Time period of oscillation, l = length of the pendulum, and g = acceleration due to the gravity

EXPLANATION:

From the above equation, it is clear that the time period of a pendulum is inversely proportional to the square root of acceleration due to gravity.
As we know, that the acceleration due to the gravity of the moon is one-sixth of that of the earth that means on moon the value of acceleration due to the gravity decreases.
Therefore when acceleration due to the gravity decreases the time period of a pendulum increases. Hence, option 3 is correct.