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Homework Help: Speed of a Transverse Wave

  1. May 30, 2010 #1
    1. The problem statement, all variables and given/known data
    A wire that is 10 m long and has a uniform density(u) of 7.75 g/cm^3 is pulled to a tension of F=80 N. The wire, however, does not have a uniform thickness; rather, it varies uniformly from an initial radius of 1mm to a radius of 3mm where it is attached to a wall. If you send a wave pulse down the length of the string, how long does it take to reach the wall?

    2. Relevant equations
    v = sqrt(F/u)
    u = m/L (m= mass, L= length)

    3. The attempt at a solution
    If I could find the volume of the wire, then I could determine the mass of the string, and the velocity of the pulse using the equations above. From there, once I had the velocity, since the units were m/s, I would divide by the distance to find the time it would take for the pulse to reach the wall. (Does this logic make sense?)

    But how would I determine the volume?
  2. jcsd
  3. May 30, 2010 #2
    That would be by integration.
  4. May 30, 2010 #3
    What would the integration look like? Would the limits be from 1mm (.001m) to 3mm (.003m) and the integrand be pi*(r^2)*10m (volume of a cylinder)?
  5. May 31, 2010 #4
    No. One method of how to obtain the integrand would be the classic "divide-and-conquer". We will divide the wire into infinitely many thin slices. We then consider the volume of each thin slice.

    The volume of each thin slice would be [tex]\pi r^{2} dx[/tex]. However, we realise also that the radius of each slice varies, and can be expressed as [tex]0.01 + 0.002x[/tex], where x represents the distance of the slice from the starting point (taken to be the free end). So, we have the volume of each slice being [tex]\pi (0.01 + 0.002x)^{2} dx[/tex].

    The total volume of the wire would thus simply be a summation of the volume of each slice across the entire length of the wire: [tex]\int^{10}_{0}\pi (0.01 + 0.002x)^{2} dx[/tex]
  6. May 31, 2010 #5
    Wouldn't r be equal to .001+.0002x (in terms of millimeters) and (1*10^-6) + (2*10^-7)x (in meters)? Since my limits are in meters, wouldn't my integrand need to be in the same units?
  7. Jun 1, 2010 #6
    Oops, it should be r=0.001+0.0002x in terms of meters. Accidentally left off a factor of 10.
    Why do you think that would be in milimeters?
  8. Jun 1, 2010 #7
    I think that you are correct. I had my units mixed up. Thank you for the help on this problem, I truly appreciate it.
  9. Jun 2, 2010 #8
    would this happen to be for a phys 132 class
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