Vibrations

A hint I rec'd:

ideal springs, point masses cannot collide, y1 and y2 are the distances of the bottom end of the springs from the top, so that the length of the second spring is y2-y1. As for gravitational effects, gravity pulls on the weights to start the springs moving, but you don't need to deal with gravity in the calculations

Any ideas?

total k = 1/(1/k_1+1/k_2)

total m = m_1 + m_2

I've been trying this, no luck. Is there anyone that can help me?

Try cross-posting to science and mathematics.

That's where (s)he started. They were less than welcoming over there...

I'll try it there too, maybe now that I have a lot less questions and some more input, they can help.

official, I think the problem needs a little bit more definition...

questions are

1) Are we supposed to assume this is an undamped system? In other words, the springs will never return to equilibrium? Because you have not provided a damping constant...

2) Are we supposed to assume that t=0, the strings are unstretched? in other words, the unstretched lengths of both springs = 1?

3) It is possible for m2 to provide an upward force on m1...which I think is really quite complicated...is that supposed to be ignored, so that from the perspective of the top spring, it just has a mass (m1+m2) attached to it?

1) Are we supposed to assume this is an undamped system? In other words, the springs will never return to equilibrium? Because you have not provided a damping constant...

They are ideal springs, without mass, collisions, etc.

2) Are we supposed to assume that t=0, the strings are unstretched? in other words, the unstretched lengths of both springs = 1?

Yes

3) It is possible for m2 to provide an upward force on m1...which I think is really quite complicated...is that supposed to be ignored, so that from the perspective of the top spring, it just has a mass (m1+m2) attached to it?

Right, we can ignore this part.

Hmm, I tried this problem for some hours today, and I just don't see how to derive any sort of expression/answer out of it. I think the answer is not a number, but rather a mathematical expression.

Just throwing around ideas, I doubt they are correct:

I think it's something along these lines:

For the mass m1 the acceleration is:
a1 = -k1*s/m1 [s is the extension of the spring]

Integrate twice with respect to time to get y1(t), adding the two constants given.

Similarly for m2:
a2 = -k2*s/m2 + a1

Then integrate, blah de blah blah...

Hmm

So nobody knows how to do this one??!!

Was my answer wrong?

I dunno, maybe I just don't get what you mean?

Ok, for the mass m1 the acceleration is:
a1 = -k1*s/m1

Because F = ma = -k*s (-spring constant*extension).

Integrate once to get the velocity:

v1 = (-k1*s/m1)*t + c

c is the constant of integration. We are given
y1'(0) = -2*sqrt(6)
which is the constant we are looking for (since y1' = v1), so

v1 = (-k1*s/m1)*t -2*sqrt(6)

To get the position we integrate again with respect to time and add the constant of integration y1(0) = 1 , giving

y1(t) = (-0.5*k1*s/m1)*t^2 - 2*sqrt(6)*t + 1

which gives, plugging in the values for k1 and m1:

y1(t) = -1.5*s*t^2 - 2*sqrt(6)*t + 1

You might use a different symbol for extension, which I give as s.

Follow the same method for m2, where

a2 = -k2*s/m2 + a1

(see the eqn. for a1 above). Note: what will happen with the term a1 above? Can you see a simpler way than integrating that part again?

I hope that helps, and I hope the answer is correct!

It is like this page:

http://othello.mech.northwestern.edu/ea3/book/modes1/modes.html

So a different idea. I am lost however.

Sorry, I just looked over my notes w/r/t springs and I see my solution is complete and utter rubbish, so please completely ignore my last post. The 's' that appeared in the equations is actually the change in the position, i.e. y(x2) - y(x1), which is the bit that needs to be integrated, so you end up with sines and co-sines in the solution. I'll get back to this problem later, because now I can see the solution by using calculus. But you mention there are four ways to do it, I'd like to see it being doing in other ways.

Thanks