Suppose a simple pendulum is used to measure the acceleration due to gravity at various points on the Earth. If g varies by 0.145% over the various locations where it is sampled, what is the corresponding variation in the period of the pendulum? Assume that the length of the pendulum does not change from one site to anothe

The period of a simple pendulum is given by the formula T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity.

If g varies by 0.145%, we want to find out how much the period T varies.

First, we need to differentiate the period equation with respect to g. Using the chain rule, we get dT/dg = -T/(2g).

This gives us the rate of change of T with respect to g. To find the percentage change in T, we multiply this by the percentage change in g, which gives us (-T/(2g)) * 0.145% = -0.145T/(2g)%.

Since the minus sign just indicates that T and g are inversely proportional (i.e., if g increases, T decreases), the percentage variation in the period of the pendulum is 0.145T/(2g)%.

So, if the acceleration due to gravity varies by 0.145% over the various locations, the corresponding variation in the period of the pendulum is 0.145T/(2g)%.