Question

Consider the acceleration due to gravity on the
surface of the moon to be 2 m s2. If the time
period of a simple pendulum on the surface of
earth is 3.5 s, then its time period on the surface
of moon is
(g = 10 m/s2)​

Answer 1

Given:

Time period of simple pendulum on the surface of the Earth is 3.5 seconds. Gravity on moon is 2 m/s².

To find:

Time period of the simple pendulum on the surface of the moon.

Calculation:

General Formula for time period of a simple pendulum is given as :

$\boxed{ \sf{T = 2\pi \sqrt{ \dfrac{l}{g} } }}$

Here , l is length of the string , g refers to gravity.

On Earth surface:

$\sf{ \therefore \: T = 2\pi \sqrt{ \dfrac{l}{g} } }$

$\sf{ = > \: T = 2\pi \sqrt{ \dfrac{l}{10} } }$

$\sf{ = > \: 2\pi \sqrt{ \dfrac{l}{10} } = 3.5 \: sec }$

On moon surface:

$\sf{ \therefore \: T2 = 2\pi \sqrt{ \dfrac{l}{(g2)} } }$

$\sf{ = > \: T2 = 2\pi \sqrt{ \dfrac{l}{(2)} } }$

$\sf{ = > \: T2 = 2\pi \sqrt{ \dfrac{l \times 5}{(2 \times 5)} } }$

$\sf{ = > \: T2 = 2\pi \sqrt{ \dfrac{l \times 5}{10} } }$

$\sf{ = > \: T2 = \sqrt{5} \times 2\pi \sqrt{ \dfrac{l }{10} } }$

$\sf{ = > \: T2 = \sqrt{5} \times 3.5}$

$\sf{ = > \: T2 = 7.82 \: sec}$

So, final answer is:

Time period on moon surface is 7.82 sec.