 |
| ||
| |||
|
| Math Central | Quandaries & 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. Melissa |
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
dL
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,
Penny
| ||
| ||
| ||
 |
 |
|
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.