Math CentralQuandaries & Queries


The period of a simple pendulum of length L feet is given by: T=2pi(sqrt(L/g))seconds. It is assumed that g, the acceleration due to gravity on the surface of the earth, is 32 feet per second per second. If the pendulum is a clock that keeps good time when L=4 feet, how much time will the clock gain in 24 hours if the length of the pendulum is decreased to 3.97 feet? (Use differentials and evaluate the necessary derivative at L=4 feet.) Answer is in seconds.


Hi Melissa,

The differential of T is an estimate of the change in T for a small change in L. The differential dT of T is given by

dT = dT/dL xdL

where dT/dL is the derivative of T with respect to L and dL is the small change in L. Find the derivative dT/dL and evaluate it at L = 4 feet. dL is the change in L which is -0.03 feet. Use these values to find dT. This is your estimate for the change in the period T in seconds caused by the change in L.

Since the period is smaller than it should be the clock will tick more times than it should in 24 hours.

I hope this helps,

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