# Differentiation Application

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• Jan 10th 2007, 04:56 PM
aussiekid90
Differentiation Application
This problem has been proving extremely tough for me. Thanks a lot for the help.

The period of a pendulum is given T=2(Pi)Squareroot(L/G) where L is the length of the Pendulum in feet, g is the acceleration due to gravity and T is the time in seconds. Suppose that the pedulum has ben subjected ton an increase in temeprature such that the length has increased by 1/2%.
a) Find the approximate percent change in the period.
b) Using the result in part (a), find the approximate error in this pendulum clock in one day.
• Jan 11th 2007, 01:20 AM
ticbol
Quote:

Originally Posted by aussiekid90
This problem has been proving extremely tough for me. Thanks a lot for the help.

The period of a pendulum is given T=2(Pi)Squareroot(L/G) where L is the length of the Pendulum in feet, g is the acceleration due to gravity and T is the time in seconds. Suppose that the pedulum has ben subjected ton an increase in temeprature such that the length has increased by 1/2%.
a) Find the approximate percent change in the period.
b) Using the result in part (a), find the approximate error in this pendulum clock in one day.

Here is one way.

T = 2pi*sqrt(L/g)
Differentiate both sides,
Since g is constant,
T = [2pi / sqrt(g)]*sqrt(L)
dT = [2pi / sqrt(g)][(1/2) / sqrt(L)]dL
dT = [pi / sqrt(gL)]dL -------------------(1)

When dL = (1/2)percent of L, or 0.005L, the L becomes (L +0.005L) = (1.005L).
Plug those into (1), to get the change in T,
dT = [pi / sqrt(g*1.005L)](0.005L)
dT = [pi / sqrt(g)]*[(0.005L) / sqrt(1.005L)] ------(2)

So the percent change in T is
(dT / T) *100percent
Where dT /T =
= [pi / sqrt(g)]*[(0.005L) / sqrt(1.005L)] / [2pi*sqrt(L/g)]
= [(pi)sqrt(g) / sqrt(g)*2pi]*[(0.005L) / sqrt(1.005L)*sqrt(L)]
= [(1/2)]*[0.005 / sqrt(1.005)]
= 0.0024938
or,
= 0.0025

And that is 1/4)percent.
Therefore, when the L was increased by (1/2) percent, the period T was increased by (1/4) percent ------------answer.

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b) Using the result in part (a), find the approximate error in this pendulum clock in one day.

T is in seconds.

1 day = 24hrs*(60min/hr)(60sec/min) = 24*3600 = 86,400 sec
(0.0025)(86,400) = 216 sec = 3min and 36sec.

Therefore, in one day the pendulum clock is faster by 3 minutes and 36 seconds than the correct time. ----------answer.