Plucked string, potential energy

In summary, the conversation discusses a homework problem involving a string of length L that is clamped at both ends and has a tension T. The question asks for the energy of subsequent oscillations as well as an approximation for small oscillations. The first part of the question is answered, but there is confusion about the second part and how it relates to a later question asking to show that the potential energy is the sum of potential energy for each mode. The conversation also discusses neglecting the extension of the wire and the change in tension.
  • #1
Unicorn.
41
0
Hello, I don't understand the second question, i don't know what I have to do:

Homework Statement


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) Find the approximative expression of this energy for small oscillations


Homework Equations





The Attempt at a Solution


In a/ I found that U=2Th²/L
For the second question I took a segment dx, so we have that
dU=T(ds-dx)
ds=dx(1+1/2(dy/dx)²)
I replaced it so that
U= T/2∫(dy/dx)²dx from 0 to L
I'm stuck here, how can I find the exact potential energy for small oscillations, I'm not even sure that I'm using the good method since in the next questions, they ask to show that the energy in b is the sum of potential energies of each mode, so I can't use y(x,t) to find dy/dx in this question, right ?
Thanks
 
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  • #2
Looks to me that you have answered (b), not (a). I would have taken your answer to (a) as correct if it were not for the fact that it goes on to ask (b). This makes me think that (a) is asking for an exact expression, with no approximations.
 
  • #3
I don't understand, in the question (a) it's asking for the exact expression, and b for an approximation of U for small oscillations
 
  • #4
Unicorn. said:
I don't understand, in the question (a) it's asking for the exact expression, and b for an approximation of U for small oscillations
We agree on that, but your answer to (a) is not exact, in several ways. It is only an approximation valid for small h.
 
  • #5
I'm really confused.
So the answer to (a) must be U=Tn²pi²A²/4L we use y(x,0)
And for (b) it must be U=2Th²/L
Then, for c/, they ask to show that the potential energy is the sum of potential energy of each mode by finding the same expression as b/
So I have to find U=2Th²/L, right ?
The problem now I don't know how to find b/ by using "small oscillation argument" without answering to the question c/ at the same time.
 
  • #6
Unicorn. said:
So the answer to (a) must be U=Tn²pi²A²/4L we use y(x,0)
The initial half length is L/2. If the centre point is displaced h orthogonally to that, what is the exact extension?
And for (b) it must be U=2Th²/L
That's my interpretation.
Then, for c/, they ask to show that the potential energy is the sum of potential energy of each mode by finding the same expression as b/
So I have to find U=2Th²/L, right ?
I'm not sure what c is asking for. It sounds like they want you to do the Fourier analysis to find the energy at each harmonic, but I wouldn't know how to do that other than by assuming the total of those is what you calculated in b.
 
  • #7
I did c/ and I found U=2Th²/L
I don't understand what you're asking for a/ ..?
 
  • #8
Unicorn. said:
I don't understand what you're asking for a/ ..?
The initial length is L. If the centre point is displaced h orthogonally to that, what is the exact length of the wire now? How much longer is that than L?
 
  • #9
It says that we neglect the extension of the wire.
Now, I don't know how to deal with the a/ question , is it a n function ?
 
Last edited:
  • #10
Unicorn. said:
It says that we neglect the extension of the wire.
Not in the OP, it doesn't . Please post the question exactly as given.
 
  • #11
Sorry, I didn't notice that i forgot it. "We neglect the extension of the wire and the change of the tension."
 
  • #12
Unicorn. said:
Sorry, I didn't notice that i forgot it. "We neglect the extension of the wire and the change of the tension."
That changes things. Now I agree with your answer for (a) but I'm mystified as to what (b) is asking. Note that you've referred to potential energy a few times, but in the question as you posed it it just says energy.
 

FAQ: Plucked string, potential energy

What is a plucked string?

A plucked string is a string that is set into motion by being pulled and released, typically by hand or with a pick or plectrum. This action causes the string to vibrate and produce sound.

2. How does a plucked string store potential energy?

When a string is plucked, it is stretched and then released. This stretching creates potential energy in the string, which is then converted into kinetic energy as the string vibrates and produces sound waves.

3. What factors affect the potential energy of a plucked string?

The potential energy of a plucked string is affected by several factors, including the tension of the string, the length of the string, and the amplitude of the vibration. The type of material the string is made of can also impact its potential energy.

4. How is the potential energy of a plucked string related to its pitch?

The pitch of a plucked string is directly related to its potential energy. A string with higher potential energy will produce a higher pitch, while a string with lower potential energy will produce a lower pitch.

5. Can potential energy in a plucked string be converted into other forms of energy?

Yes, potential energy in a plucked string can be converted into other forms of energy. As the string vibrates, the potential energy is converted into kinetic energy, which then creates sound waves. Additionally, some of the energy is also dissipated as heat due to friction within the string and its surroundings.

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