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Transverse Waves |
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| Apr18-07, 07:25 PM | #1 |
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Transverse Waves
Please help!!!
Having a little difficulty with a homework question and concept 1. The problem statement, all variables and given/known data y = 2.28sin(0.0276pi x + 2.42pi t) Find the amplitude, wavelength, frequency, speed and maximum transverse speed. 3. The attempt at a solution Definitely correct answers A = 2.28cm Wavelength = 72.46 cm Frequency = 1.21 Hz Questionable answers Speed = 87.68 cm/s Max transverse speed = ?????? Reasoning The speed should be w/k which would give 2.42pi/0.0276pi = 87.68 cm/s The max transverse velocity though....would that be when the trig function of cos is equal to 1 in this equation u = -wAcos(kx-wt) ???? |
| Apr18-07, 10:19 PM | #2 |
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The speed is correct. Transverse speed is when dy/dt is a max. I think you are pretty much nailing that as well.
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| Apr19-07, 03:04 AM | #3 |
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Apologies...still not quite sure.....would I use the w and A i know multiplied with cos(kx-wt) to get my answer.....so if cos(kx-wt)=1, then max transverse speed would be -wA???
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| Apr19-07, 03:59 AM | #4 |
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Transverse Waves
If your displacement function is a sin wave, it's not possible that your velocity "function" be a constant of 87.68cm/s as you calculated. Your speed would be a function as well.
As you know, velocity is the change of displacement over time (i.e. dy/dt), so by differentiating the function for y, you will get a the velocity function (which in this case is a cosine function). That cosine function will allow you to calculate the speed at any location x and time t. To calculate the maximum speed, we take a look at the velocity function. If you do your calculus right, the equation should look something like: v = B cos (Cx + Dt) There is an infinite possibility for the values of x and t, so we can assume that hthe cosine can return any value. The maximum value however of a cosine function is 1. So what does that tell you about your maximum velocity? |
| Apr19-07, 07:51 AM | #5 |
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| Apr19-07, 09:26 PM | #6 |
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Thanks much....makes a lot more sense.....appreciate the help
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