Question

A simple pendulum of length 40 cm is taken inside a deep mine. Assume for the time being that the mine is 1600 km deep. Calculate the time period of the pendulum there. Radius of the earth = 6400 km.

Answer 1

It is given that:
Length of the pendulum, l = 40 cm = 0.4 m
Radius of the earth, R = 6400 km
Acceleration due to gravity on the earth's surface, `g = 9.8 "ms"^(- 2)`

Let

\[g'\] be the acceleration due to gravity at a depth of 1600 km from the surface of the earth. 
Its value is given by,

\[g' = g\left( 1 - \frac{d}{R} \right)\] \[\text{where  d  is  the  depth  from  the  earth  surfce, }\] \[\text { R  is  the  radius  of  earth,   and }\] 

\[\text {g  is  acceleration  due  to  gravity .}\]

\[\text{ On  substituting  the  respective  values,   we  get: }\] \[  g' = 9 . 8\left( 1 - \frac{1600}{6400} \right)\] 

\[= 9 . 8\left( 1 - \frac{1}{4} \right)\] 

\[= 9 . 8 \times \left( \frac{3}{4} \right) = 7 . 35   {\text{ms}}^{- 2}\]

Time period is given as,

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

\[\Rightarrow T = 2\pi\sqrt{\left( \frac{0 . 4}{7 . 35} \right)}\] 

\[ \Rightarrow T = 2 \times 3 . 14 \times 0 . 23\] 

\[             = 1 . 465 \approx 1 . 47  s\]