1. ## 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.

2. 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.

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# the value of g (acceleration due to gravity) at a place is measured with the help of a simple pendulum by using the formula T=2π(l/g)^1/2.find the percentage error in the calculated value of g, if the error in measuring time is 0.5%.

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