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Standing wave question

  1. Apr 27, 2006 #1
    [​IMG]
    Here is my question (2 parts)

    a) Draw the standing wave that occurs if the string
    tension is quadrupled while the frequency is held
    constant.
    im not sure how these are related, any equation?

    b) Suppose the tension is doubled while the
    frequency shaking the string is held constant. Will there be a standing wave? If so how many antinodes will it have?
    I know an antinode is an area of maximum amplitude, but not sure if there will be a standing wave, how do i determine this.

    thanks.
     
  2. jcsd
  3. Apr 28, 2006 #2

    berkeman

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    Staff: Mentor

    How is the velocity of propagation of a wave down a string related to the tension in the string?
     
  4. Apr 28, 2006 #3

    andrevdh

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    Homework Helper

    Once you have found the equation berkeman is referring to it is a simple step to go from the velocity to frequency when dealing with wave motion. You just need to relate these two quantities via the amount of nodes and antinodes in the standing wave.

    If the tension in the string is increased, that is the string is pulled tighter between the two end points, the speed of propagation of the disturbances (shaking the string at a certain frequency...) will increase. This means that the individual outgoing and reflected pulses will interfere at a lower rate (they will be further apart since they are now travelling faster). The endpoint conditions will stay the same - nodes. It is just that less nodes and antinodes will form between the endpoints since in both cases since the frequency of generation of disturbances in the string is kept constant - that is the phase relationship between the outgoing and reflected pulses are kept the same in time, but in space the meet at a lower rate since they are further apart along the string.
     
    Last edited: Apr 28, 2006
  5. Apr 29, 2006 #4

    andrevdh

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    A standing wave will form if the resultant waveform will fit on the string. If the waveform fits at a frequency [itex]f_o[/itex] and the speed of propagation changes then the wavelength needs to change by the same factor since
    [tex]f_o=\frac{v}{\lambda}[/tex]
    stays constant. If the wavelenght were equal to [itex]L[/itex] before the speed change and then changed to [itex]2.0\ v_o[/itex] then the wavelength also needs to change by the same factor.
     
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