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String wave problems

  1. Jan 3, 2014 #1
    1. The problem statement, all variables and given/known data
    I am posting three problems together as I believe they are quite short.

    1.Show that the particle speed can never equal to the wave speed in a sine wave if the amplitude is less than wavelength divided by ##2\pi##.

    2.Two wave pulses identical in shape but inverted with respect to each other are produced at the two ends of a stretched string. At the instant when the pulses reach the middle, the string becomes completely straight. What happens to the energy of two pulses?

    3.Show that for a wave travelling on a string
    $$\frac{y_{max}}{v_{max}}=\frac{v_{max}}{a_{max}}$$​
    where the symbols have usual meanings. Can we use componendo and dividendo taught in algebra to write
    $$\frac{y_{max}+v_{max}}{y_{max}-v_{max}}=\frac{v_{max}+a_{max}}{v_{max}-a_{max}}$$​

    (By usual notation, y is the displacement, v is transverse velocity and a is the acceleration of any particle on the string)

    2. Relevant equations



    3. The attempt at a solution

    Attempt for 1.
    Let the string wave be represented by the equation ##y=A\sin(\omega t-kx)##. The transverse velocity of particle at position at any instant of time t is given by ##∂y/∂t=A\omega \cos(\omega t-kx)##. The speed is hence given by:

    $$\sqrt{A^2\omega^2\cos^2(\omega t-kx)+\frac{\omega^2\lambda^2}{4\pi^2}}$$

    where the second term is the wave speed given by ##\omega/k=\omega\lambda/(2\pi)## and ##\lambda## is the wavelength. Since ##A \leqslant \lambda/(2\pi)##, we have

    $$\sqrt{A^2\omega^2\cos^2(\omega t-kx)+\frac{\omega^2\lambda^2}{4\pi^2}}\leqslant \frac{\omega \lambda}{2\pi}\sqrt{1+\cos^2(\omega t-kx)}\leqslant \frac{\sqrt{2}\omega \lambda}{2\pi}$$
    The above does not agree with the problem. :confused:

    Attempt for 2.
    I honestly have no idea about this one. :(

    Attempt for 3.
    It is very easy to find the relations presented in the problem statement by assuming the wave equation to be ##y=A\sin(\omega t-kx)##. I am not sure why the author asks the second part of the problem. I don't see why it would be wrong to do some simple algebra on the relation obtained.

    Any help is appreciated. Thanks!
     
  2. jcsd
  3. Jan 3, 2014 #2
    For #2

    I think that all the kinetic energy of the particles will be converted into potential energy. The particles of the string will move a bit away from each other.

    For #3- I am not sure but-

    One thing to know is that the dimension of Vmax in LHS is [L]. Also the equation isn't valid (for obvious reasons) when Ymax=Vmax (similar thing for RHS)
     
  4. Jan 3, 2014 #3

    TSny

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    Can you describe the motion of a particle of the string?
     
    Last edited: Jan 3, 2014
  5. Jan 3, 2014 #4
    Hi consciousness and TSny! :)

    Yes, the dimensions don't match and I think this must be the reason, thank you! :)

    Umm...I am not sure but do you ask the following?

    $$y(x,t)=A\sin(\omega t-kx)$$
    $$x(t)=\frac{\omega}{k}t$$
     
  6. Jan 3, 2014 #5

    TSny

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    Set aside the equations for the moment. Can you describe in words the motion of a point of the string? In particular, can you describe (at least roughly) the vertical component of its motion? the horizontal component of its motion?
     
  7. Jan 3, 2014 #6
    The vertical component of motion is simple harmonic with the mean position on the x-axis.

    The horizontal component of motion is simply translational motion with a constant velocity in horizontal direction.
     
  8. Jan 3, 2014 #7

    TSny

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    Last edited: Jan 3, 2014
  9. Jan 4, 2014 #8
  10. Jan 4, 2014 #9
    That's a fundamental property of wave propagation in matter: the only thing that propagates is the wave itself; the matter oscillates only.
     
  11. Jan 4, 2014 #10
    Thanks voko and TSny, I think I get the first problem. :)

    What about problem 2? :confused:
     
  12. Jan 4, 2014 #11
    Think about a small element on the string. When it oscillates, when is its potential energy maximal? Its kinetic energy?
     
  13. Jan 4, 2014 #12
    The potential energy is maximum when it is at the maximum amplitude and kinetic energy is maximum when it is at its mean position.
     
  14. Jan 4, 2014 #13
    So when the string is straight, where is the energy of the pulses?
     
  15. Jan 4, 2014 #14
    It is converted to kinetic energy as the particles are on their mean position, right?
     
  16. Jan 4, 2014 #15
    Can it be anywhere else?
     
  17. Jan 4, 2014 #16
    Nope. :D

    Thanks a lot voko! :smile:
     
  18. Jan 4, 2014 #17
    When a wave propagates on a string the velocity of the particles is maximum when they are in their mean position. But when two waves having the same amplitude destructively interfere, the string will become completely straight (principle of superposition is valid) so the kinetic energy becomes zero! A similar thing cant happen here?

    Edit- After thinking some more I think that the velocity of the particles is indeed highest when they superimpose but this problem is mathematically solved IMO.
     
    Last edited: Jan 4, 2014
  19. Jan 4, 2014 #18
    What does that mean?
     
  20. Jan 4, 2014 #19
    When two travelling waves interfere, it is possible that the string becomes straight and remains that way. All the particles are at rest. The same is not true for this situation but I got confused somehow.
     
  21. Jan 4, 2014 #20
    No, that is not possible. Both waves carry energy. It cannot just disappear.
     
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