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Homework Help: Wave equation confusion!

  1. Mar 25, 2006 #1
    Hello everyone! I'm having troubles understanding why i'm not getting this right...the problem is:

    A sinusoidal wave is traveling on a string with speed 10. cm/s. The displacement of the particles of the string at x = 25 cm is found to vary with time according to the equation y = (5.0 cm) sin[10.0 - (4.0 s-1)t]. The linear density of the string is 3.0 g/cm.

    What is the frequency of the wave?
    Well wouldn't just be 4.0 s^-1?

    Also it says What is the wavelength of the wave?
    I t hought it would be:
    You know the wave number is: (5.0 cm) and you know the relationship:
    Wave Length = 2PI/k, where k is the wave number: so i put:
    Wave Length = 2PI/5 = 1.2566 which was wrong any ideas why i'm not getting these right?

    THanks! :biggrin:
  2. jcsd
  3. Mar 25, 2006 #2


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    Find the period T first. After one period, the phase has advanced by [itex]2\pi[/itex]. Then use f=1/T.

    The wavenumber is not 5.0 cm, that's the amplitude.
    Use [itex]\lambda f=v[/itex] to find the speed of the wave.
  4. Mar 25, 2006 #3


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    4.0 is the angular frequency of the wave, [itex]\omega[/itex], in radians/sec; not the frequency, [itex]f[/itex], in cycles/sec. What do you need to multiply radians/sec by, in order to get cycles/sec?
  5. Mar 25, 2006 #4
    Thanks for the responce! I'm still kind of confusd on how you find the period with the given information, i'm very rusty on trig. I understand sin has a period of 2PI, but that doesn't mean that function's period is just 1/2Pi does it?

    I alwyas found hte period by first finding the angular frequency w, which is suppose to be the number infront of the t. So in this case, would it be 1/4.0, since it says 4.0s^-1?
    or is it suppose to be w = -1/4.0? Then i can find the period by using:
    f = w/2*pi
    T = 1/f

    oo my bad, the wave number should be what is infront of the x, but there is no x in the equation :\
    I really don't have anything to look at as a reference other then the internet becuase my book doesn't have chapter 16, yes!

    Didn't see your message until now jtbell, well 2PI = 360 degree's, so i'm asumming, i would take 4.0 rad/sec *360/2PI ?
    But it does say s^-1, so again does this mean its angular frequency is 1/4?
    Last edited: Mar 25, 2006
  6. Mar 25, 2006 #5


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    Right. The angular frequency [itex]\omega[/itex] is 4.0/s. And this is related to the frequency by [itex]f=2\pi \omega[/itex].

    You can get this simply by considering when the phase advances by [itex]2\pi[/itex]. Consider [itex]\sin(\omega t+\phi)[/itex]. One period passes if [itex]\omega T=2\pi[/itex] or [itex]T=2\pi/\omega=1/f[/itex].

    And don't fuss with degrees. Use radians.
  7. Mar 25, 2006 #6
    Thanks for the help!
    f = w/2*pi i think though!
    but i got it right once i flip it around. Now i found all the parts to the question but i'm confused on how i'm suppose to find the tension of this string.
    I found the following:
    What is the frequency of the wave?
    What is the wavelength of the wave?
    Give the general equation giving the transverse displacement of the particles of the string as a function of position and time.
    y(x,t) = (5cm)*sin[ (.3999cm^-1)x -(4s^-1)t]
    which was correct. Now they said:
    What is the tension in the string?

    I don't have the book chapter, and in the notes it shows no equation dealing with tension at all. Any ideas? THanks!
  8. Mar 25, 2006 #7


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    Yes, you're right :blushing:

    A disturbance in a stretched string satisfies the wave equation (for small amplitudes). The speed is:

    Where T is the tension in the string and [itex]\mu[/itex] the mass per unit length.
    Last edited: Mar 25, 2006
  9. Mar 25, 2006 #8
    Thanks again!
    hm..i don't see where i'm messing this one up at:
    v = 10cm/sec
    T = ?
    [tex]\mu= 3.0 g/cm[/tex]
    if they want newtons i converted mu to kg/m
    3.0 g/cm (1 kg/1000g) (100cm)/1 m = .3 kg/m;
    v = .1 m/sec
    v^2/[tex]\mu[/tex] = T;
    T = .1^2/.3 = .03333N

    but that was wrong, i also tried 333.33N by forgetting to convert the 10cm/sec to .1m/sec, any ideas where i f'ed it up?

    Oops i'm a tard, it should be v^2*mu

    Yep that fixed it! w00t!
    Last edited: Mar 25, 2006
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