Question

A simple pendulum is constructed by hanging a heavy ball by a 5.0 m long string. It undergoes small oscillations. (a) How many oscillations does it make per second? (b) What will be the frequency if the system is taken on the moon where acceleration due to gravitation of the moon is 1.67 m/s²?

Answer 1

It is given that:
Length of the pendulum, l = 5 m
Acceleration due to gravity, g = 9.8 ms^-2
Acceleration due to gravity at the moon, g' = 1.67 ms^-2

(a) Time period \[\left( T \right)\]  is given by,

\[T = 2\pi\sqrt{\frac{l}{g}}\]

\[= 2\pi\sqrt{\frac{5}{9 . 8}}\] 

\[ = 2\pi\sqrt{0 . 510} = 2\pi  \left( 0 . 71 \right)  s\]

i.e. the body will take  2 \[\pi\](0.7) seconds to complete an oscillation.

Now, frequency \[\left( f \right)\]is given by,

\[f = \frac{1}{T}\]

\[\therefore   f = \frac{1}{2\pi\left( 0 . 71 \right)}  \] 

\[             = \frac{0 . 70}{\pi}  Hz\]

(b) Let 

\[g'\] be the value of acceleration due to gravity at moon. Time period of simple pendulum at moon \[\left( T' \right)\],is given as:

\[T' = 2\pi\sqrt{\left( \frac{l}{g'} \right)}\]

On substituting the respective values in the above formula, we get:

\[T' = 2\pi\sqrt{\frac{5}{1 . 67}}\]

Therefore, frequency \[\left( f' \right)\]will be,

\[f' = \frac{1}{T'}\] 

\[       = \frac{1}{2\pi}\sqrt{\frac{1 . 67}{5}} = \frac{1}{2\pi}\left( 0 . 577 \right)\] 

\[       = \frac{1}{2\pi\sqrt{3}}    Hz\]