Rinkel's modification of Ruchhardt's method

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SUMMARY

The discussion focuses on Rinkel's 1929 modification of Ruchhardt's method for measuring the distance a ball falls before it starts to rise again in a gas-filled tube. The derived equations include the period of simple harmonic motion (SHM) given by ##T = 2\pi \sqrt{\frac{mV}{\gamma p A^2}}## and the distance fallen, expressed as ##mgL = \frac{\gamma pA^2 L^2}{2V}##. The conversation emphasizes the importance of accurately applying energy conservation principles and understanding the forces acting on the ball, specifically gravitational force and excess pressure due to gas dynamics.

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Homework Statement


upload_2015-6-4_10-38-43.png

In Ruchhardt’s method of measuring ##\gamma##, illustrated in Fig. 12.2, a ball of mass ##m## is placed snugly inside a tube (cross-sectional area ##A##) connected to a container of gas (volume ##V##). The pressure ##p## of the gas inside the container is slightly greater than atmospheric pressure ##p_0## because of the downward force of the ball, so that
##p = p_0 + \frac{mg}{A}##
Show that if the ball is given a slight downwards displacement, it will undergo simple harmonic motion with period ##T## given by
##T = 2\pi \sqrt{\frac{mV}{\gamma p A^2}}##
[You may neglect friction. As the oscillations are fairly rapid, the changes in ##p## and ##V## which occur can be treated as occurring adiabatically.]
In Rinkel’s 1929 modification of this experiment, the ball is held in position in the neck where the gas pressure ##p## in the container is exactly equal to air pressure, and then let drop, the distance ##L## that it falls before it starts to go up again is measured.
Show that this distance is given by
##mgL = \frac{\gamma pA^2 L^2}{2V}##

Homework Equations


##pv^\gamma = const. ##

The Attempt at a Solution


I have correctly derived the first equation of period of the SHM.
From this equation, I got,
##\omega ^2 = \frac{pA^2\gamma}{mV}##
upload_2015-6-4_10-54-49.png

As the initial velocity is 0, so
## L = 2 \cdot amplitude##
Applying energy conservation for the point A & B, we get:
##mgL = \frac{1}{2} m \omega^2 (\frac{L}{2})^2##
##mgL = \frac{pA^2 \gamma L^2}{8V}##
 
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Hi arpon. You got so involved with presenting this that you forgot to add your question?
 
NascentOxygen said:
Hi arpon. You got so involved with presenting this that you forgot to add your question?
Here is the question:
arpon said:
Show that this distance is given by
##mgL = \frac{\gamma pA^2 L^2}{2V}##
 
forum rules stipulate that complete solutions to homework problems are not allowed
In the second part, I am afraid your application of energy conservation is not accurate; since in both points A and B you should be getting:
Kinetic energy = 0, Potential energy = ##\frac{1}{2} k A^2## for each of them, which do not give the expression required.

So, my approach would be to break down the different force mechanisms here:
There are 2 forces acting upon the ball: The downward gravitational force ##mg## and the upward excess pressure ##\Delta P## which is given by $$\Delta P = \gamma \frac{P_0}{V_0}\Delta V.$$
Here, ##P_0=p_0=p## is the pressure and ##V_0=V## is the volume at point A in the Rinkel's modification set up of the experiment. (Note that the above formula for excess pressure follows directly from the adiabatic nature of the whole process. Just take the derivative of ##PV^\gamma =## constant.)

[ mentor note: some content in this post has had to be edited out to make it compliant with PF homework help rules ][/color]
 
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