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Completely Lost PLease help ASAp Vibrations & Waves

  1. May 5, 2006 #1
    One end of a horizontal string is attached to a small-amplitude mechanical 57 Hz vibrator. The string's mass per unit length is 4.4 10-4 kg/m. The string passes over a pulley, a distance L = 1.50 m away, and weights are hung from this end Assume the string at the vibrator is a node, which is nearly true.

    What mass m must be hung from this end of the string to produce a standing wave with the following number of loops?
    (a)one loop-----kg?

    (b) two loops----------kg?

    (c) five loops -------------kg?
     
  2. jcsd
  3. May 5, 2006 #2

    Hootenanny

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    Hi chazgurl, you should have realised by now that we do not hand out help or solutions unless you have shown some prior working.

    HINTS: (1) Equation for the velocity of a wave on a string.
    (2) Wave equation.

    ~H
     
  4. May 7, 2006 #3
    frequency= 1/2pi(g/L)^1/2= so.......-->4.02 Hz

    vo=2*pi*frequency*amplitude=

    v=+/-Vo ( 1-x^2/A^2)^1/2


    i dont know the wave equation
     
  5. May 7, 2006 #4

    Hootenanny

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    I was thinking more of;

    [tex]v = \sqrt{\frac{T}{\frac{m}{L}}}[/tex]

    I'm sure you do, [itex]v = f\lambda[/itex]

    ~H
     
  6. May 8, 2006 #5
    ok so....if i use v= (T/m/l)^1/2
    it should be v=(.07154/4.4x10^4 kg/m/.150m)^1/2

    which would turn out to be....32.7 m/s....im confused
     
  7. May 8, 2006 #6

    Doc Al

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    Figure it out, step by step. Answer these questions, in order:
    (1) What wavelength must the wave have to exhibit the required number of loops?
    (2) What must the speed of the wave be? (Since the frequency is given.)
    (3) What string tension is needed to produce such a speed?
    (4) What mass will be needed to provide that tension?
     
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