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Harmonic wave functions

  1. Nov 2, 2004 #1
    Ok heres the deal. The problem gives me this wave function:

    y= (15cm) cos[(pi/5cm)x - (pi/12s)t]

    I undserstand that 15cm represents the amplitude, 5cm represents the wavelength, and 12s represents the T. What I dont get is that the standard wave function is in this form:

    y(x,t)= A cos[(2pi/lambda)x - (2pi/T)t]

    I thought I was good at algebra, but I can't seem to figure out how to the the pi's in the given equation to be at a factor of 2 with the rest of it being the same. The problem asks me to find: amplitude (which I got 15cm), the wavelength (I got 2.25cm), the T (I got 6s), and speed and direction. Please help if you can!

  2. jcsd
  3. Nov 2, 2004 #2
    y= (15cm) cos[(pi/5cm)x - (pi/12s)t]

    can be written as

    y= (15cm) cos[(2*pi/10cm)x - (2*pi/24s)t]

    So, the wavelength is 10 cm and T is 24s.

    Speed is 10/24 cm/s and direction is positive x.
  4. Nov 2, 2004 #3


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    Homework Helper

    Or use Partial derivatives for speed.

    [tex] \frac{\partial y}{\partial t} = v [/tex]


    [tex] v = f \lambda = \frac{\omega}{k} [/tex]
    Last edited: Nov 2, 2004
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