1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Transverse Velocity of Wave, given position & time

  1. Nov 13, 2015 #1
    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?
     
  2. jcsd
  3. Nov 13, 2015 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    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.
     
  4. Nov 14, 2015 #3
    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?
     
  5. Nov 14, 2015 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    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:
     
  6. Nov 14, 2015 #5
    Thank you :) I'll give that a go when I've gotten through all of my other assignments...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted