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Homework Help: Need help fast with transverse wave on string!

  1. Jan 12, 2009 #1
    A transverse wave is traveling on a string. The displacement y of a particle from its equilibrium position is given by y = (0.0230 m) sin (34.6t - 2.04x). Note that the phase angle 34.6t - 2.04x is in radians, t is in seconds, and x is in meters. The linear density of the string is 1.90 x 10-2 kg/m. What is the tension in the string?

    I figured since y=0.0230 sin (34.6t-2.04x) the frequency=34.6 wavelength=3.08 and Amplitude=0.0230
    Then I think I can solve for velocity using v=wavelength*frequency and get v=106.568
    Now I am stuck as to what I am supposed to do.. I think I have to multiply the amplitude by 4 and then use it somewhere.. Perhaps this equation: v=squareroot of F/(m/L)

    But now I am lost! Any help would be greatly appreciated!! Thank you!
  2. jcsd
  3. Jan 13, 2009 #2
    So your diplacement time-space function is of the form:

    [tex]y(x,t)=A\sin(\omega t -kx)[/tex]

    Our wave equation, for this case:

    [tex]\frac{\partial^2 y(x,t)}{\partial t^2}=c^2\frac{\partial^2y(x,t)}{\partial x^2}[/tex]

    So the dipersion relation of the simple wave equation is: [tex]\omega(k)=c k [/tex]
    So the phase velocity:


    So we see the phase velocity of the wave, is equal to the c parameter appearing in the wave equation.

    Of this parameter we know that it is (during the derivation of the wave equation for the transverse wave this was what we named c):

    [tex]c=\sqrt{\frac{F}{A \rho}}=\sqrt{\frac{F}{\mu}}[/tex]

    Where F is the tension, \rho is the volume mass density, A is the cross sectional area of the string, \mu is the linear mass density.

    So we have for the tension:

    [tex]F=\mu c^2=\mu \frac{\omega^2}{k^2}[/tex]

    Now put the numerical data in and you are done.
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