Question

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 theearth, 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 ofthe pendulum is decreased to 3.97 feet? (Use differentials and evaluate the necessary derivative at L=4 feet.) Answer is in seconds.

My study group has been at this for 4 hours! PLEASE HELP US, what do we do??

Answer 1

We are told that we have a simple pendulum with the following characteristics:

Period given by the equation: seconds.

Length is L feet.

g is the acceleration due to gravity which is given as

We are told that the pendulum is part of a clock that keeps good time when L=4.

We are asked to determine the amount of time the clock will gain in 24 hours if the length of the pendulum is changed to 3.97 feet.

We begin by substituting our constant value for g into the equation:

Now, we take the derivative with respect to L:

Now we multiply both sides by dL:

Now we substitute and into the equation:

s

So we have arrived at the difference in time for one period. Recall that the period is defined as the amount of time it takes to complete one cycle. So each cycle of the pendulum will be shorter by.

We were told that the clock kept good time with the pendulum at 4 feet. When the pendulum was at 4 feet, it had a period of:

s

We know that:

So 24 hours, we have a number of periods equal to:

So we normally have:

With the shorter pendulum, we complete one period in:

So in 24 hours, we will complete a number of periods equal to:

The difference is:

Going back to the original pendulum, we can setup a proportion to find the amount of additional time the clock will read in 24 hours:

And we can convert this into minutes as:

So the clock will run fast by roughly 4 ¾ minutes.