Transverse Velocity of Wave, given position & time

  • #1
50
2
Can't see where I'm going wrong here - would greatly appreciate if anyone can point it out!

I've gotten the other parts of the question right, so I know that:
ω = 125.66 rad/s
A = 2.50 * 10-3m
k = 3.49 rad/m
The wave is moving in the +x direction.

The general equation for the position of a particle on the wave is:
y(x,t) = Acos(kx - ωt)

So this means (please correct me if I'm wrong) that the transverse velocity of a particle on the wave is:
Vy(x,t) = ωAsin(kx - ωt) = (125.66)((2.50 * 10-3m)sin(3.49x - 125.66t)

I'm asked to find Vy at x = 1.35m, t = 0.0625s.
For this I got 2.07*10-4m/s, but I'm told this is incorrect. Where am I going wrong?
 

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
12,624
3,018
Hello. Check to see if you have the overall correct sign for your answer.

The most general equation for a harmonic wave traveling in the positive x direction is y = Acos(kx-ωt+φ) where φ is a phase constant. Do you know if you are supposed to assume that the phase constant is zero in this problem? It's always a good idea to state the problem word-for-word so that we can be sure that we have all of the information.
 
  • #3
50
2
Thanks for the response TSny. Here's the full text of the question:

"A transverse sine wave with an amplitude of
2.50 mm and a wavelength of 1.80 m travels from left to right along a long, horizontal stretched string with a speed of 36.0 m/s. Take the origin at the left end of the undisturbed string. At time t=0 the left end of the string has its maximum upward displacement."

I had to 'give up' on the question & the answer given was v = 0m/s - I could then see that the part of the question immediately before - which asked for the displacement of a particle at the same position & time - was exactly equal to the amplitude, so I probably should have realised that the velocity should be 0 there. I'm still puzzled as to why my calculation didn't give that answer though - rounding error, I suppose?
 
  • #4
TSny
Homework Helper
Gold Member
12,624
3,018
OK. The information given in the problem implies that the phase constant φ is zero. So, you have the correct mathematical form for the wave. Yes, I think your answer differs from zero due to round-off error. Try working the problem by expressing k and ω in terms of ##\pi##. Then when you evaluate the argument of the cosine function for the specific values of x and t given, you can get the argument in terms of ##\pi##.

Anyway, your method was correct. :smile:
 
  • Like
Likes Ryaners
  • #5
50
2
Thank you :) I'll give that a go when I've gotten through all of my other assignments...
 

Related Threads on Transverse Velocity of Wave, given position & time

  • Last Post
Replies
1
Views
2K
Replies
3
Views
1K
  • Last Post
Replies
4
Views
12K
  • Last Post
Replies
10
Views
3K
Replies
2
Views
3K
Replies
0
Views
11K
Replies
2
Views
2K
  • Last Post
Replies
1
Views
9K
  • Last Post
Replies
7
Views
67K
Replies
1
Views
2K
Top