Question

A pendulum clock giving correct time at a place where g = 9.800 m/s² is taken to another place where it loses 24 seconds during 24 hours. Find the value of g at this new place.

Answer 1

According to the question, the pendulum loses 24 s in 24 hrs.

This means, In 1 hrs, time loss = 1 seconds.

The correct clock has a time period T which will be of  2 s, since we are talking about pendulum clock.

Thus, no. of oscillations made in hrs =   3600/2 = 1800

But due to loss of 1 seconds, total time taken by it will be 3601 seconds. This means, this clock is making 1800 oscillations just in 3601 seconds.

Thus, the new time period T' = 3601/1800 = 2.000555 s

Now, $T' = 2\pi \frac{l}{g}$

$T = 2\pi \frac{l}{9.8}$

Thus, T'/T = 9.8/g

2.00055/2 = 9.8/g

g = 9.79 m/s².

Hence, the acceleration due to gravity at an new place is 9.79 m/s².

Hope it helps.

Answer 2

According to the question, the pendulum loses 24 s in 24 hrs.

This means, In 1 hrs, time loss = 1 seconds.

The correct clock has a time period T which will be of  2 s, since we are talking about pendulum clock.

Thus, no. of oscillations made in hrs =   3600/2 = 1800

But due to loss of 1 seconds, total time taken by it will be 3601 seconds. This means, this clock is making 1800 oscillations just in 3601 seconds.

Thus, the new time period T' = 3601/1800 = 2.000555 s

Now,

Thus, T'/T = 9.8/g

2.00055/2 = 9.8/g

g = 9.79 m/s².