Physics Help Forum two-mass oscillation
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 Aug 31st 2008, 10:52 AM #1 Junior Member   Join Date: May 2008 Posts: 12 two-mass oscillation Hey all My first query on the forum. I am trying to relearn a little physics. Here is a problem I am unsure of. Here we have a mass, M1, attached to a spring (pardon my sloppy 'Paint' spring) of unstretched length d and pulled by a constant force of $\displaystyle F_{2}=m_{2}g$. Suppose the system is in equilibrium when x=L. Is L>d or Ld. Equilibrium is $\displaystyle y=\frac{-m}{k}g$ due to M2. Therefore, x=L=y. That is, the mass M2 moves the mass M1 horizontally the same amount as it displaces when it is hanging down. Since at equilibrium, the forces balance and we have $\displaystyle \frac{d^{2}y}{dt^{2}}=0$, then $\displaystyle y=\frac{-m}{k}g$ Could we then solve for $\displaystyle k=\frac{g}{y}m$?. Period of oscillation is $\displaystyle T=2{\pi}\sqrt{\frac{m}{k}}$ and amplitude A in $\displaystyle x=Asin({\omega}t+{\phi})$ How can I relate these and solve the problem. I have many thoughts but am unsure about them. Attached Thumbnails
Aug 31st 2008, 11:47 AM   #2

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 Originally Posted by galactus Hey all My first query on the forum. I am trying to relearn a little physics. Here is a problem I am unsure of. Here we have a mass, M1, attached to a spring (pardon my sloppy 'Paint' spring) of unstretched length d and pulled by a constant force of $\displaystyle F_{2}=m_{2}g$. Suppose the system is in equilibrium when x=L. Is L>d or Ld. Equilibrium is $\displaystyle y=\frac{-m}{k}g$ due to M2. Therefore, x=L=y. That is, the mass M2 moves the mass M1 horizontally the same amount as it displaces when it is hanging down. Since at equilibrium, the forces balance and we have $\displaystyle \frac{d^{2}y}{dt^{2}}=0$, then $\displaystyle y=\frac{-m}{k}g$ Could we then solve for $\displaystyle k=\frac{g}{y}m$?. Period of oscillation is $\displaystyle T=2{\pi}\sqrt{\frac{m}{k}}$ and amplitude A in $\displaystyle x=Asin({\omega}t+{\phi})$ How can I relate these and solve the problem. I have many thoughts but am unsure about them.
You've got most of it except that you are looking for x(t), not y(t) (especially after the string has been cut!) You have the equilibrium position correct as
$\displaystyle x = \frac{mg}{k} + d$
(In order to write this in terms of x you need to include the unstretched length d to this.)

You have the period of the motion correct.

For the amplitude, think of this in terms of an energy argument. Can the (ideal) spring go any further beyond d than mg/k? Then that must be your amplitude.

-Dan
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 Aug 31st 2008, 02:32 PM #3 Junior Member   Join Date: May 2008 Posts: 12 Thanks TQ for the reply. As for the amplitude, could we just set: $\displaystyle \frac{mg}{k}+d=Asin({\omega}t+{\phi})$ and solve for A?: We could say at t=0 is when the line is cut, giving $\displaystyle \frac{mg}{k}+d=Asin({\phi})$ Intuitively, I wouldn't think there would be much amplitude. After the string is cut, M1 just goes back to d without oscillating. No, the spring can not go any further beyond d then mg/k. Does that make $\displaystyle A=\frac{mg}{k}+d$ as well.
Sep 1st 2008, 05:49 AM   #4

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 Originally Posted by galactus Thanks TQ for the reply. As for the amplitude, could we just set: $\displaystyle \frac{mg}{k}+d=Asin({\omega}t+{\phi})$ and solve for A?: We could say at t=0 is when the line is cut, giving $\displaystyle \frac{mg}{k}+d=Asin({\phi})$ Intuitively, I wouldn't think there would be much amplitude. After the string is cut, M1 just goes back to d without oscillating. No, the spring can not go any further beyond d then mg/k. Does that make $\displaystyle A=\frac{mg}{k}+d$ as well.
Bear in mind that the oscillation will be about the end of the spring, the new equlibrium point. So this equation will be
$\displaystyle \frac{mg}{k}+d=Asin({\omega}t+{\phi}) + d$

So at t = 0 we have
$\displaystyle \frac{mg}{k}=Asin({\phi})$
where $\displaystyle \phi = \pi / 2$ (since the motion is at its extreme end at t = 0.)

-Dan
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 Sep 1st 2008, 07:26 AM #5 Junior Member   Join Date: May 2008 Posts: 12 Thanks very much. That was what I was thinking, but was unsure of myself. Take care and thanks.

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