# Pulled string

1. Nov 19, 2014

### Ange98

1. The problem statement, all variables and given/known data
A string of length L, which is clamped at both ends and has a tension T , is pulled aside a distance h at its center and released. a) What is the energy of the subsequent oscillations? b) How often will the shape shown in the figure reappear? (Assume that the tension remains unchanged by the small increase of length caused by the transverse displacements.) [Hint: In part (a), consider the work done against the tension in giving the string its initial deformation.]
Picture here:
http://ocw.mit.edu/courses/physics/...nit-ii-waves/pset-5/MIT8_03SCF12_OCW_PS05.pdf

2. Relevant equations:
This is part of my problem.. I have no idea what equations I should be working with.

3. The attempt at a solution
Tried using work and energy but I couldn't get any where with it. I have the answer though it is:
TL[ sqrt(1+ (2h/L)^2) + 1 ]
and b is every 2(ML/T)^(1/2) seconds.
Any help is greatly appreciated!!

Last edited by a moderator: Nov 19, 2014
2. Nov 19, 2014

### Orodruin

Staff Emeritus
How did you try to apply work and energy? In order for us to help you, you need to provide the details for us to see where you went wrong. Otherwise we do not know exactly where you got stuck or if you are thinking in completely wrong terms. (It is also required by the forum rules - see Guidelines for students and helpers.)

3. Nov 19, 2014

### Ange98

OKay so my textbook doesn't have examples and I have a very hard time understanding my prof so these are the formulas that I'm thinking may apply somehow:
E=KE+U
but it says work done against tension so W=Fd=Th? and thats all I can pull out of that.
Then I have the standing wave equations so PSI(x,t)=Acos(wt-kx) and mu=M/L and v = sort (T/mu) but theres no M here , andU=1/2 kx^2 , KE=1/2mv^2 , I tried using PSI's derivative with respect to t but the thing is I dont actually understand this question. There was only two pages of reading for this section and energy was not mentioned.

4. Nov 19, 2014

### Orodruin

Staff Emeritus
Careful here, the tension direction is not in the same way you are pulling.

If you have the work done to pull the string out, what is the kinetic energy at the moment of release?

5. Nov 19, 2014

### Ange98

So the tension is directed along the string away from where its being pulled? And W=KEfinal-KEinitial. Here I feel like you would have potential energy but no kinetic, but in my diagram from class at the maximum displacement its all kinetic ( if I'm reading it right) so is the kinetic equal to the work which is the total energy?

6. Nov 19, 2014

### Orodruin

Staff Emeritus
In the moment you release the string, it is at rest so the kinetic energy is ....?

There are essentially two different ways for arguing for the work done on the string. Naturally, they should both give the same answer. The easiest is probably to consider the (negative of the) work done by the string on whatever is providing the external force that is extending the string.

7. Nov 19, 2014

### BiGyElLoWhAt

yes
Not quite. Conservation of energy: $KE_i + U_i \pm \Delta W = KE_f + U_f$
Solve for W
I think you're misreading it. Think about the motion of the string. What's it's velocity as it approaches the peak? What's it's velocity as it approaches the center from the peak? Now what values must it cross in order to satisfy the fact that it's oscillation is a continuous function?

8. Nov 19, 2014

### Ange98

Okay I couldn't understand why the diagram showed that because i was quite sure KE would be 0, but my prof is foreign and very tough to understand so I must have mixed that up. I'll just hope my prof is there tomorrow because I truly do not understand this and I feel like I'm just getting jerked around on here which is not helping whatsoever. I'm going into this problem not having a clue where to go with it and everyones just tossing a few fragments in which doesn't help. You dont have to do all the math but walking me through it wouldn't hurt. I just want to understand.

9. Nov 19, 2014

### BiGyElLoWhAt

People tend to understand things better when they solve them by themselves, with people guiding them in the proper direction. It's the socratic method. No ones jerking you around, but if you don't want us to help you and you'd rather your prof tell you how to do it, that's perfectly fine.

10. Nov 19, 2014

### Orodruin

Staff Emeritus
On the contrary I would say that you have been given a few very explicit cornerstones to build a solution on. Take this one for example:
There are several questions you can ask yourself about this statement, mainly "What is the force from the tension on an object holding the string extended in the middle?" and "What is the relevant displacement?". You will learn at a much more profound level if you take some time to think about these questions and try to deduct an answer than if we simply give those answers to you. If you try and fail, that is not a problem, give us your attempt and we can tell you where you go wrong.