Caculus Help [Pendulum]

[Kushballo7]

1. A simple pendulum consists of a mass m swinging at the end of a
massless string of length L. The time T for one complete swing of the
pendulum is known as the period, and if the pendulum swings through
only a small angel, the period is given by T=2(pi)sqrt(L/g) where g is the
acceleration due to gravity.

(a) Show that a small change dL, in the length produces a chane in the
period dt satisfying dT/T = dL/2L

(b)Suppose that the pendulum looses 15 seconds per hour. How should
the length of the pendulum be adjusted?

(c)The preceeding formula for the period can be used to measure the
acceleration due to gravity. Assume that the error in measuring the
length L is negligible, and express the error dg in the acceleration of
gravity in terms of the error dT in measuring the period.

 

[soroban]

Hello, Kushballo7!

Here are the first two parts . . .

1. A simple pendulum consists of a mass m swinging at the end of a massless string of length \(\displaystyle L.\)
The time \(\displaystyle T\) for one complete swing of the pendulum is known as the period,
and if the pendulum swings through only a small angle,
the period is given by: \(\displaystyle T\;=\;2\pi\sqrt{\frac{L}{g}}\), where \(\displaystyle g\) is the acceleration due to gravity.

(a) Show that a small change \(\displaystyle dL\), in the length produces a change in the
period dt satisfying: \(\displaystyle \frac{dT}{T}\;=\;\frac{dL}{2L}\)

We have: .\(\displaystyle \L T\;=\;\frac{2\pi}{\sqrt{g}}\cdot L^{\frac{1}{2}}\qquad\Rightarrow\qquad dT\;=\;\frac{2\pi}{\sqrt{g}}\cdot\frac{1}{2}L^{-\frac{1}{2}}dL\;=\;\frac{\pi\,dL}{\sqrt{g}\sqrt{L}}\)

Then: . \(\displaystyle \L \frac{dT}{T}\;=\;\frac{\frac{\pi\,dL}{\sqrt{g}\sqrt{L}}}{\frac{2\pi\sqrt{L}}{\sqrt{g}}} \;= \;\frac{dL}{2L}\)

(b)Suppose that the pendulum loses 15 seconds per hour.
How should the length of the pendulum be adjusted?

This is a trick question . . . beware!

"Loses 15 seconds/hour": .\(\displaystyle \frac{dT}{T}\;=\;\frac{-15\text{ sec}}{3600\text{ sec}}\;= \;-\frac{1}{240}\)

From \(\displaystyle (a)\) we have: . \(\displaystyle \frac{dL}{2L}\;=\;-\frac{1}{240}\qquad\Rightarrow\qquad \frac{dL}{L}\;=\;-\frac{1}{120}\)

In English: The reason the pendulum is 15 seconds short is that the pendulum is too short.
. . . . . . . . . It is only \(\displaystyle \frac{119}{120}\) of its proper length.

The proper length is: \(\displaystyle \frac{120}{119}L\)
. . The pendulum must be lengthened by \(\displaystyle \frac{1}{119}\) *

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

* . That is the "trick" part.

Suppose you needed a 125-foot pipe, but your pipe is only 100 feet.
. . Your pipe is only 80% of the desired length, or 20% short.
. . But you must increase its length by 25 feet or 25%.

 

[wjm11]

Hello, Soroban,

Nicely done, but I have a question: if we are losing 15s each hour, don't we need to shorten the period (to speed up the clock) by shortening the pendulum by 1/120L as indicated by the negative sign?