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Help with wave problems

  1. Aug 7, 2004 #1
    Stuck on some wave problems now. I understand the basica ideas and formulas but these questions seem to be throwing me for a loop.

    Question 1:

    A massive aluminum sculpture is hung from a steel wire. The fundamental frequency for transverse standing waves on the wire is 200 Hz. The sculpture is then immersed in water so that 1/3 of its volume is submerged. The density of water is 1000 kg/m^3 and the density of aluminum is 2700 kg/m^3

    What is the new fundamental frequency?

    Since its a standing wave I'm applying lambda = 2*L and f = V/(lambda)

    they give us original f which shows:

    [tex] 200_{hz} = sqrt(T/\mu)/(\lambda) [/tex]

    where [tex] \mu = mass/length [/tex]

    [tex] 200_{hz} = sqrt(mass * grav/(mass /length))/(2*length) [/tex]

    we get length = .03942 meters

    Then for the new question we must change Tension to account for the new bouyant force.

    I get the new T to be:

    T = mass* gravity - density of water * gravity * Volume/3

    Soo I get to this point:

    [tex] f = sqrt((mass*g-density of water * gravity * Volume/3)/(mass/length))/(2*length) [/tex]

    Thats as far as I get, I can't get psat the 3 unknowns. I know I'm missing something as they give me the density of aluminum which I probably need somewhere.

    Question 2:

    Holding Up under Stress. A string or rope will break apart if it is placed under too much tensile stress. Thicker ropes can withstand more tension without breaking because the thicker the rope, the greater the cross-sectional area and the smaller the stress. One type of steel has density 7830 kg/m^3 and will break if the tensile stress exceeds 7 * 10^8. You want to make a guitar string from a mass of 4.10g of this type of steel. In use, the guitar string must be able to withstand a tension of 900 N without breaking. Your job is the following.

    A) Determine the maximum length the string can have

    B) Determine the minimum radius the string can have.

    C) Determine the highest possible fundamental frequency of standing waves on this string, if the entire length of the string is free to vibrate.

    I'm not sure where to start on this one. From what I understand it just wants us to find the the max length that can support a Tension of 900N.

    Only equation I can think of with T in it is V = sqrt(T/mu) but V is unknown.

    For B I thought I could just use a standar Tensile Stress = F_perp/Area but that doesnt seem to work.

    Any suggestions would be much appreciated.

  2. jcsd
  3. Aug 8, 2004 #2


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    The first "mass" in your last equation is the mass of the sculpture, [tex]M=\rho_{Al}*V [/tex],
    while the second "mass" is that of the wire. When immersed, the tension becomes
    [tex] T'=grav*V*(\rho_{Al}-\rho_{water}/3)[/tex].
    The length of the wire and [tex]\mu[/tex] stays the same.

    I hope this helps.

  4. Aug 8, 2004 #3


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    Dearly Missed

    Question 2:
    To get going on A+B, note that we must have:
    1. [tex]S_{max}>=\frac{F_{n,app}}{A}[/tex]
    Here, [tex]S_{max}[/tex] is the yielding stress,which you have been given.
    [tex]F_{n,app}[/tex] is the normal component of the applied force. Since this must be less than or equal to 900N, you may now solve for the minimum area by setting
  5. Aug 8, 2004 #4
    First, thanks for the initial error in my math. Second, that makes some sense but I dont' see how it helps me. With this change I can't even solve for L using the first equation because another unknown is thrown into the equation.

    Using what you said I get this for the initial equation:

    [tex] f = sqrt(\rho_{water}*V*g*l/mass_{wire})/(2*l) [/tex]

    Entering that into the equation I get[tex] l = .165375*V/mass_{wire} [/tex]

    I then change to my second equation with the new tension of [tex] T'=grav*V*(\rho_{Al}-\rho_{water}/3)[/tex]

    I get:

    [tex] Velocity = sqrt(9.8*V(\rho_{al} - \rho_{water}/3))/(mass_{wire}/l) [/tex]

    Try to solve for Velocity and I get Velocity = 61.93 * abs(V/m) and get stuck there again.

    Arildno, you siad use [tex]S_{max}>=\frac{F_{n,app}}{A}[/tex]
    which is what I've tried before.

    That formula gives me
    7*10^8 = 900N/A, so A = 900N/7*10^8 which is 1.28*10^-6 which doesn't seem to work.

    I'm still plugging away, any further help is much appreciated.
  6. Aug 8, 2004 #5

    Doc Al

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    hint for question #1

    For question #1, think in terms of proportions and ratios. The only thing that varies is the tension in the wire: [itex]f \propto \sqrt{T}[/itex]. And you already figured out the tensions: [itex]T_1 = \rho_a g V[/itex], [itex]T_2 = (\rho_a - \rho_w/3)g V[/itex].

    So figure out [itex]f_2/f_1[/itex].
  7. Aug 8, 2004 #6
    Hmm, totally didn't even think about that one. Thanks for the tip. Now I just need to figure out whats wrong with the other one.
  8. Aug 9, 2004 #7

    Doc Al

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    for question #2

    What makes you say that that doesn't work?
  9. Aug 9, 2004 #8
    What makes me say it doesn't work is that I have to do my homework on a website, www.masteringphysics.com, and when I input that for the answer it rejects it.
  10. Aug 9, 2004 #9

    Doc Al

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    That calculation gives you the minimum area--but that's just the first step in answering the questions in the problem. Post your calculations and I'll have a look.
  11. Aug 9, 2004 #10
    Blah, I realized it a bit ago, it gave me the area, not the radius, I feel lame now lol.

    Think I got this one solved now. Thanks
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