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A fixed string in the third harmonic

  1. Mar 1, 2008 #1

    TFM

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    [SOLVED] A fixed string in the third harmonic

    1. The problem statement, all variables and given/known data

    A string with both ends held fixed is vibrating in its third harmonic. The waves have a speed of 195m/s and a frequency of 230hz The amplitude of the standing wave at an antinode is 0.370cm.

    Calculate the amplitude at point on the string a distance of 20.0cm from the left-hand end of the string.


    2. Relevant equations



    3. The attempt at a solution

    I'm not sure how to go about this question? Any tips?

    TFM
     
  2. jcsd
  3. Mar 1, 2008 #2
    Firstly, You have to be asking yourelf, what does it mean for an oscillation to be in the third harmonic?
     
  4. Mar 1, 2008 #3

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    The formula for harmonics is:

    [tex] f_n = n\frac{v}{2L} [/tex]

    and the third harmonic is:

    [tex] f_3 = 3\frac{v}{2L} [/tex]

    The third Harmionc al,os has three antinodes, and 1.5 wavelengths

    TFM
     
  5. Mar 1, 2008 #4
    Now, using v=f\lambda, you can calculate the wavelength.

    The question asks for 20cm from a node. Since you know the amplitude at the antinode, can you now calculate the amplitude at a given point in the wave
     
  6. Mar 1, 2008 #5

    TFM

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    I have calculated the wavelength to be 1.18m

    This implies the length of string is 1.77m,

    TFM
     
  7. Mar 1, 2008 #6

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    I have now caluclated the nodes to be at:

    0, 0.59, 1.18, 1.77

    and the anti-nodes at

    0.295, 0.885, 1.475

    TFM
     
  8. Mar 1, 2008 #7

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    Have I correctly worked out what to do now?

    using:

    [tex] y(x,t) = A_S_Wsin(k_n x)sin(\omega _n t) [/tex]

    where:

    [tex] A_S_W = 0.0037 cm[/tex]

    [tex] k_n = 2\pi / 1.18 [/tex]

    [tex] \omega_n = 2\pi / (1/frequency) [\tex]

    giving:

    [tex] y(x,t) = 0.0037*sin((2\pi / 1.18) 0.2)sin(2\pi / (1/230) t) [/tex]

    I just need what t is, I think. Is this right?

    TFM
     
  9. Mar 1, 2008 #8
    You should be using [tex]v=f*\lambda[/tex] to find the wavelength.
     
  10. Mar 1, 2008 #9

    TFM

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    Rearranges to give wavelength = v/f

    gives wavelength of 0.848m

    nodes are at 0, 0.424m, 0.848m and 1.271m

    antinodes 0.212m, 0.635m, 1.06m

    length of string is 1.271m

    [tex] k_n = \frac{2\pi}{0.848} [/tex]

    am I using the correct formula for the amplitude

    [tex] y(x,t) = A_S_W sin(k_n x)sin(\omega _n t) [/tex]

    and if so, how do I work out the time, t

    TFM
     
  11. Mar 1, 2008 #10
    Hi,

    Your equation looks good, except for your value of the wavelength. If you recalculate your wavelength using [tex]v=f*\lambda[/tex] and then put this into your equation you will be good to go.

    In this question remember that you are only considering the position of the particle and in such an instance you can make a conclusion about the time dependence.
     
  12. Mar 1, 2008 #11
    Sorry, I didn't see that you had corrected the wavelength above. Yeah your formula is the right to use in this case.
     
  13. Mar 1, 2008 #12

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    Would it be best to presume that the time = 0?

    TFM
     
  14. Mar 1, 2008 #13
    No, if you do that the equation goes to zero and you will get 0 as the amplitude.

    The amplitude at any position on the string does not depend on the time (if you were interested in the vertical displacement of the particle then the time would be relevant). The amplitude of a particle on a transverse wave depends only on the position of that particle. (Remember the amplitude doesn't change with time unless the wave is being damped, which it isn't in this case.)
     
  15. Mar 1, 2008 #14

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    What would be a good time to use?

    TFM
     
  16. Mar 1, 2008 #15
    You just ignore the time part of the equation ie [tex]sin(\omega*t)[/tex] to give,

    [tex]y(x)=A_{SW}sin(k_{n}x)[/tex] which should be the amplitude.
     
  17. Mar 1, 2008 #16

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    I get an answer of 0.0036854, which seems too close to the antinodes 0.0037?

    TFM
     
  18. Mar 1, 2008 #17
    That should be the right answer. Don't forget that the first antinode occurs at a distance of 21.2 cm from the left of the rope, which is very close to the point under observation, thus their amplitudes will be quite close.
     
  19. Mar 1, 2008 #18
    Do you mind me asking what university you go to? It's odd that you seem to do all the same problems that I'm set. I can't work out how to do the second part of that question you are doing.
     
  20. Mar 2, 2008 #19

    TFM

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    The second parts okay (I am presuming you are referring tgo the time to get from its largest upward disp. to its largest downwards disp.), you have to think when it is at its highest point and when it is at it's lowest point. try drawing adiagram can help.

    TFM
     
  21. Mar 2, 2008 #20

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    I am now stuck on parts C and D, finding the maximum transverse velocity and the maximum transverse acceleration. The key points, I feel, though is that the velocity is greatest when the amplitude is 0 (when it crosses the normal position) and the acceleration is greatest when it reaches maximum/minimum amplitude.

    Any suggestions?

    TFM
     
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