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Question

The ratio between the height H of a mountain and the depth h of a mine if a simple pendulum swings with the same period at the top of the mountain and at the bottom of the mine is?


Solution

Time period of simple pendulum, 
T = 2π√(l/g)

At height H above surface of the earth, value of g is
g' = g(1 - 2H/R)
Time period at height H,
T1 = 2π√ [ l/g(1 - 2H/R) ]

At depth h of a mine, value of g is
g' = g(1 - h/R)
Time period at depth h,
T2 = 2π√ [ l/g(1 - h/R) ]

Given,
T1 = T2
2π√ [ l/g(1 - 2H/R) ]
= 2π√ [ l/g(1 - h/R) ]

√ [ l/g(1 - 2H/R) ]
= √ [ l/g(1 - h/R) ]

​​​​​​l/g(1 - 2H/R) = l/g(1 - h/R) 
1 - 2H/R = 1 - h/R
2H/R = h/R
2H = h
H/h =1/2

So, the required ratio is
1 : 2
 

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