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Homework Help: Max.transverse tension in a wave

  1. Nov 5, 2009 #1
    1. The problem statement, all variables and given/known data
    A transverse sinusoidal wave is generated at one end of a long, horizontal string by a bar that moves up and down through a distance of 1.00cm. The motion is continuous and is repeated regularly 120 times per second. The string has linear density 120 g/m and is kept under a tension of 90.0N. Find (a) the maximum value of the transverse speed u and (b) the maximum value of the transverse component of the tension.

    2. Relevant equations
    y(x,t)=A sin(kx-wt) traveling wave
    k = [tex]\frac{2\pi}{\lambda}[/tex] angular wave number
    w = 2[tex]\pi[/tex]f angular frequency
    v = [tex]\sqrt{\frac{T}{\mu}}[/tex] wave speed
    [tex]\mu[/tex] linear density

    3. The attempt at a solution
    After calculating the values of k and w, and deriving the first equation, I found u = 3.77 m/s the answer in the book say it is 3.66 m/s is anything wrong?
    Letter b I have no idea on how to start solving. I tried to calculate it using F = ma, but I don't know how much mass I should consider, and probably this way is not right.
  2. jcsd
  3. Nov 6, 2009 #2


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    (a) Can you show your calculations of k and w? That would help us figure out what is wrong.

    (b) First you'll need to figure out what is the maximum angle the string makes with the horizontal direction.
  4. Nov 6, 2009 #3
    (a)The bar moves up and down 120 times per second, so the frequency f = 120 Hz.
    w = 2[tex]\pi[/tex]f replacing f = 120 Hz, w = 2*[tex]\pi[/tex]*120 = 753,98 s[tex]^{-1}[/tex]. How k = w/v and v = [tex]\sqrt{\frac{T}{\mu}}[/tex] replacing the value of T = 90 N and [tex]\mu[/tex] = 0,12 kg/m I received v = 27,39 m/s.
    k = 752,98/27,39
    k = 27,49 m[tex]^{-1}[/tex]
    With these values(and amplitude = 0.05) the equation is:
    y(x,t)=0,005sin(27,49x - 753,98t)
    Deriving in t:
    [tex]\partial[/tex]y/[tex]\partial[/tex]t = -0,005*753,98cos(27,49x - 753,98t) so the max. value of velocity happens when cos -1 and u = 3,77 m/s.

    (b)The tan of this max.angle:
    [tex]\partial[/tex]y/[tex]\partial[/tex]x = 0,138 cos(27,49x - 753,98t), which max.value is 0,138. arctan 0,138 = 7,85 degrees, so the angle between the tension and the vertical is 90 - 7,85 = 82,15 degrees. Using T[tex]_{v}[/tex]= cos [tex]\theta[/tex] * T
    T[tex]_{v}[/tex] = cos 82,15 * 90
    T[tex]_{v}[/tex] =12,3 N, book's answer is 12,2N, i think it's right :D
    Thanks a lot!
  5. Nov 6, 2009 #4
    Yet in this exercise, it asks me what's the maximum power transferred along the string. I know the formula of kinetic energy, but I can't figure out what is the formula for potencial energy on this string. Do I have to first find this formula before calculating the max.value of power? How can I do this?
  6. Nov 6, 2009 #5
    Sorry, I researched a little and see that:
    P = Force x Velocity
    Using this formula and replacing the values I get the result
  7. Nov 6, 2009 #6


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    Remember that k=753,98. This will not make a big difference though.
    Looks good to me, I don't know why your book got a different number. Note also that the answer is (amplitude)*(w), so k did not really enter into this calculation.

    Looks good, you're welcome!

    Try this page:

    Scroll about halfway down the page, to where it says "Energy and Power in a Traveling Harmonic Wave".
  8. Nov 6, 2009 #7
    Thanks,I find the solution for that problem:
    Power = Force * Velocity
    Power = (-[tex]\tau[/tex]*[tex]\partial[/tex]y/[tex]\partial[/tex]x)*[tex]\partial[/tex]y/[tex]\partial[/tex]t
    The negative it's because the force is contrary to the displacement?
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