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Transverse Wave Velocity

  1. Apr 9, 2009 #1
    1. The problem statement, all variables and given/known data
    The left-hand end of a long horizontal stretched cord oscillates transversely in SHM with frequency f = 250 Hz and amplitude 2.6 cm. The cord is under a tension of 140 N and has a linear density [tex]\mu[/tex] = 0.12 kg/m. At t = 0, the end of the cord has an upward displacement 1.6 cm and is falling. What is the velocity and acceleration at t = 2.0 seconds at a point on the string that is 1.00 m from the left-hand end?

    2. Relevant equations
    [tex]v_{y} = -D_{M} \omega \cos (kx - \omega t)[/tex]

    [tex]v = \sqrt{\frac{F_{T}}{\mu}}[/tex]

    [tex]v = \lambda f[/tex]

    3. The attempt at a solution
    I determined the wave velocity using the equation,
    [tex]v = \sqrt{\frac{F_{T}}{\mu}} = \sqrt{\frac{140 N}{.12 kg/m}} = 34 m/s[/tex]

    I used this to find the wavelength [tex]\lambda[/tex],

    [tex]\lambda = \frac{v}{f} = \frac{34 m/s}{250 Hz} = 0.14 m[/tex]

    So, in the equation for the velocity of the particle,

    [tex]k = \frac{2\pi}{\lambda} = 45/m[/tex]

    [tex]\omega = 2\pi f = 1570/s[/tex]

    Therefore, the equation for the particle velocity should be,

    [tex]v = -(0.026 m)(1570/s)\cos[(45/m)(1.00 m) - (1570/s)(2.0 s)][/tex]
    [tex]v = 35 m/s[/tex]

    However, the book says that the answer is 41 m/s. Can someone tell me what I am doing wrong? Any help would be appreciated.
  2. jcsd
  3. Apr 9, 2009 #2


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    Staff Emeritus
    Science Advisor
    Homework Helper

    Since initially, at x=0, the displacement is neither zero nor at a maximum, this equation should be modified to include a phase, i.e.

    cos(kx - ωt + φ)

    Use the initial displacement and velocity at x=0 to determine φ.

    Looks like you have the right idea otherwise.
  4. Apr 10, 2009 #3
    Yes, adding the phase angle made everything work out correctly.

    Thanks for your help!
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