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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) ????
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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.
Apr19-07, 03:04 AM   #3
 
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???
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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Quote by croyboy View Post
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???
Yes. Except I'd call the maximum speed +wA. Why would you say minus?
Apr19-07, 09:26 PM   #6
 
Thanks much....makes a lot more sense.....appreciate the help
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