Traveling Wave Model: transverse wave on a string

  • #1

Homework Statement


A transverse wave on a string is described by the wave function: y=(0.120m)sin([tex]\frac{\Pi}{8}[/tex]x+4[tex]\Pi[/tex]t) Determine the transverse speed and acceleration of the string at t=0.200s for the point on the string located at x=1.60m. What are the wavelength, period, and speed of propagation of this wave?


Homework Equations


vy= -[tex]\omega[/tex]Acos(kx-[tex]\omega[/tex]t)
Since f=[tex]\frac{1}{T}[/tex] and [tex]\omega[/tex] is 4[tex]\Pi[/tex] does that mean T=0.500?


The Attempt at a Solution


I was able to come up with -1.51 for the transverse speed, and I am pretty sure I understand why the acceleration is 0. I am having problems coming up with the wavelength and the proagation speed.
 

Answers and Replies

  • #2
Doc Al
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Since f=[tex]\frac{1}{T}[/tex] and [tex]\omega[/tex] is 4[tex]\Pi[/tex] does that mean T=0.500?
Yes.

I am having problems coming up with the wavelength and the proagation speed.
Hint: How does k relate to the wavelength?
 
  • #3
Hi Doc,

I had tried [tex]\lambda[/tex]= [tex]\frac{k}{2\Pi}[/tex] with k=frac{\Pi}{8} and it didn't work out to the right answer.
 
  • #4
Doc Al
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What wavelength did you get?
 
  • #5
I got .0625m and the book says it should be 16m
 
  • #6
Doc Al
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45,184
1,509
Hi Doc,

I had tried [tex]\lambda[/tex]= [tex]\frac{k}{2\Pi}[/tex] with k=frac{\Pi}{8} and it didn't work out to the right answer.
That should be k ≡ 2π/λ.
 
  • #7
So how does that change the relationship between k and [tex]\lambda[/tex]? I couldn't write it as [tex]\lambda[/tex]= [tex]\frac{k}{2\Pi}[/tex]?
 
  • #8
Doc Al
Mentor
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So how does that change the relationship between k and [tex]\lambda[/tex]? I couldn't write it as [tex]\lambda[/tex]= [tex]\frac{k}{2\Pi}[/tex]?
No. If [itex]k = 2\pi / \lambda[/itex], then [itex]\lambda = 2\pi / k[/itex]. (Just algebraic manipulation--but make sure you understand it.)
 
  • #9
Oh alright I've got it now. It's those silly little math mistakes that get me every time. Thanks much Doc.
 

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