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The Transverse Speed

  • Thread starter roam
  • Start date
  • #1
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Hello!
The following is my problem (a question from an old exam paper);
http://img341.imageshack.us/img341/5142/69855901wv3.gif [Broken]




Well, I'm stuck on part c and d.
I already got the correct answers for a & b;
[tex]v = \frac{\omega}{k} => \frac{4}{3} = 1.3 m/s[/tex]

[tex]K = \frac{2\pi}{\lambda} => \lambda = 2.09 m[/tex]

[tex]v = f\lambda => f = 0.65 Hz[/tex]

[tex]v = \sqrt{\frac{T}{\mu}} => 1.3 = \sqrt{\frac{T}{4}} => T = 7.1[/tex]


I checked the answers and it is correct up to this point. Now I don't know how to do question c. What is it meant by "transverse speed". What formula should I use?

(The right answer is -0.022 m/s is the transverse speed at x = 1 & t = 1)

I don't know which formula to use...

Thanks.

 
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Answers and Replies

  • #2
dynamicsolo
Homework Helper
1,648
4
The transverse velocity is the speed at which a point on the string is moving perpendicularly to the length of the string. It is the time derivative of the displacement y. Keep in mind, though, that here y is a function of two variables, so you will want the partial derivative of y with respect to t. You would then evaluate this derivative function at
x = 1, t = 1.
 
Last edited:

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