A simple pendulum has a period of 2.4 sec at a place where g = 9.81 m/s^2. What is the value of g at another place on the earth’s surface where this pendulum has period of 2.41 sec.

Question

A simple pendulum has a period of 2.4 sec at a place where g = 9.81 m/s^2. What is the value of g at another place on the earth’s surface where this pendulum has period of 2.41 sec.

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1 answer
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Answer 1

Explain Bot answered
8 months ago

To find the value of acceleration due to gravity (g) at another place on the earth's surface, where the period of the pendulum is 2.41 seconds, we can use the formula for the period of a simple pendulum:

T = 2π * √(L/g)

where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.

We can rearrange this formula to solve for g:

g = (4π^2 * L) / T^2

Given that the initial period of the pendulum is 2.4 seconds and g = 9.81 m/s^2, we can substitute these values into the formula to find the length of the pendulum (L) at the initial place:

9.81 = (4π^2 * L) / (2.4^2)

Simplifying this equation, we get:

L = (9.81 * 2.4^2) / (4π^2)

Now, we can use the new period (2.41 seconds) and the length of the pendulum (L) we just calculated to find the new value of g using the same formula:

g = (4π^2 * L) / (2.41^2)

Substituting the value of L from the previous equation, we can calculate the new value of g.