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3 years ago

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The average period of pendulum clock is found to be 1.2s at sea level. The period of the same pendulum on a mountain top is foun

d to be 1.18s. If the acceleration due to gravity at sea level is 9.790m/s2, what is the acceleration due to gravity at the mountain top?

1 answer:

Kipish [7]3 years ago

5 0

Answer:

g' = 10.12m/s^2

Explanation:

In order to calculate the acceleration due to gravity at the top of the mountain, you first calculate the length of the pendulum, by using the information about the period at the sea level.

You use the following formula:

$T=2\pi \sqrt{\frac{l}{g}}$ (1)

l: length of the pendulum = ?

g: acceleration due to gravity at sea level = 9.79m/s^2

T: period of the pendulum at sea level = 1.2s

You solve for l in the equation (1):

$l=\frac{gT^2}{4\pi^2}\\\\l=\frac{(9.79m/s^2)(1.2s)^2}{4\pi^2}=0.35m$

Next, you use the information about the length of the pendulum and the period at the top of the mountain, to calculate the acceleration due to gravity in such a place:

$T'=2\pi \sqrt{\frac{l}{g'}}\\\\g'=\frac{4\pi^2l}{T'^2}$

g': acceleration due to gravity at the top of the mountain

T': new period of the pendulum

$g'=\frac{4\pi^2(0.35m)}{(1.18s)^2}=10.12\frac{m}{s^2}$

The acceleration due to gravity at the top of the mountain is 10.12m/s^2

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