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Fundamental standing wave problem

  1. Dec 27, 2008 #1

    A 40-cm long string, with one end clamped and the other free to move transversely, is vibrating in its fundamental standing wave mode. If the wave speed is 320 cm/s, the frequency is:
    A) 32 Hz
    B) 16 Hz
    C) 8 Hz
    D) 4 Hz
    E) 2 Hz

    2. Relevant equations

    [tex]v = \lambda[/tex][tex]f[/tex]
    [tex]L = n(\frac{1}{2}\lambda)[/tex]

    3. The attempt at a solution

    [tex].40 = \frac{1}{2}\lambda[/tex]
    [tex]\lambda = 0.8 [/tex]
    [tex] 3.2 = 0.8f[/tex]
    [tex]f = 4[/tex]

    Answer: E) 2 Hz

    I'm wondering if perhaps the fact that the string is clamped on one end means that it would be like a sound tube that is closed on one end, and then I would have to use 1/4lambda instead of 1/2?
  2. jcsd
  3. Dec 27, 2008 #2
    Hi clairez93,

    I think that's a good idea. Did you try using it? Did you get the correct answer?
  4. Dec 28, 2008 #3
    Yes, if you use 1/4*lambda instead of 1/2, it gets the correct answer of 2 Hz, but I'm wondering why that works. Does a clamped end really make it like a sound tube? Wouldn't the wave still start and end with a node?
  5. Dec 28, 2008 #4
    Try drawing a diagram of the situation. That's what I did to help me understand. You have a set length L as you said with this 40 cm string clamped at one end and free at the other. You have the string vibrating at the fundamental mode. This is the longest wavelength. Let's look at a diagram of the string as a function of x. Imagine the string clamped at x=0 and free at x=L. Assuming the string has the shape of a sine wave, in the fundamental mode, the shape of the string would start at x=0 and arch upwards and have its crest at x=L. This is 1/4 of a wavelength.

    Does this make sense to you? If not, I will try and make a diagram...... let me know.
  6. Dec 28, 2008 #5
    I meant to add that, no, by definition, there is a free end. That means that it would not end with a node.

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