1. Not finding help here? Sign up for a free 30min 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!

Writing a function for a wave.

  1. Jan 18, 2013 #1
    1. The problem statement, all variables and given/known data

    A wave travels along a string in the positive x-direction at 27 m/s. The frequency of the wave is 46 Hz. At x = 0 and t = 0, the wave velocity is 2.6 m/s and the vertical displacement is y = 3 mm. Write the function y(x, t) for the wave. (Use the following as necessary: x and t. Assume x and y are in m, and that t is in s.)

    2. Relevant equations

    y(x,t) = Asin(κx - ωt + ∂)

    ω = 2pif = 2pi(46) = 289.03

    κ = ω/v = 289.03/27 = 10.70

    3. The attempt at a solution

    I determined that:

    y(x,t) = Asin(10.70x - 289.03t + ∂)
    y(0,0) = Asin(∂) = 0.003 m

    Here is where I have trouble. I tried to take a partial derivative of the first function in order to solve for the velocity, since velocity is given. I've never encountered partial derivatives before, so I am not sure if what I did is correct. It doesn't give me the right answer.

    yt= -289.03Acos(10.70x - 289.03t + ∂)

    If (x,t) = (0,0)
    yt = -289Acos(∂) = 2.6 m/s

    I tried to combine this equation with

    Asin(∂) = 0.003 m

    to solve for ∂ and A, but I wasn't close to the final answer.

    Is this a matter of taking the partial derivative wrong, or am I making some other mistake?
     
  2. jcsd
  3. Jan 19, 2013 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    the wave velocity is 2.6 m/s
    ... the wave velocity is the speed of the wave in the x direction.

    the partial derivative: ∂y(x,t)/∂t is the speed of point x in the y direction.

    If the wave at t=0 has form ##y(x,0)=f(x)## and it is travelling in the +x direction with speed c,
    then at time t it will have form: ##y(x,t)=f(x-ct)##.

    In your case, f(x) is the wave equation at t=0.
     
  4. Jan 19, 2013 #3
    Thanks for the response.

    At t = 0, x = 0.

    y(0,0) = Asin(∂)

    According to the problem, at this point, y(0,0) = 0.003 m.

    So:

    Asin(∂) = 0.003 m.

    I now have two variables, A and ∂. How do I solve for them?
     
  5. Jan 19, 2013 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You did it right. Show your further work.

    ehild
     
  6. Jan 19, 2013 #5

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Looking at it again and squinting this time:
    ...this could be interpreted either way ... this wave velocity could be the rate of change of displacement y of whatever is at point x or it could be the speed of the whole wave-form (and if the latter, why mention the x position?) ... but it has already been said that:
    ... OK, so c=27m/s and vy(0,0) = 2.6m/s

    So what was the final answer supposed to be?
    What did you get?
     
  7. Jan 19, 2013 #6
    Equation #1: -289Acos(∂) = 2.6 m/s
    Equation #2: Asin(∂) = 0.003 m

    Equation #1 (solved for A): A = 2.6/[-289.03cos(∂)]

    Equation #2 (with substitution): (-2.6/289.03)tan(∂) = 0.003
    (cont.) tan(∂) = -(0.003*289.03)/2.6
    ∂ = tan-1[-(0.003*289.03)/2.6]
    ∂ = -0.32

    Back to Equation #2: Asin(-0.32) = 0.003
    A = -0.0095 (Amplitude is always positive, so can I just disregard the negative sign?)

    The correct answer is: 0.0095 sin (10.70x - 289.03t + 2.82)

    My answer as it is currently: 0.0095 sin (10.70x - 289.03t - 0.32)

    My phase does not equal 2.82. Yet, using -0.32 as the phase resulted in the correct amplitude (assuming that it is alright to disregard the negative sign.)
     
  8. Jan 19, 2013 #7

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You got A correctly. The easier way to calculate it is:

    Acos(δ)=-2,6/289=-0.009
    Asin(δ)=0.003

    Square both equations and add them together: A2cos2δ+A2sin2δ=A2=90*10-6, A=0.0095.

    From the first equations you see that sin(δ) >0 and cos(δ)<0. So δ is in which quadrant?

    ehild
     
  9. Jan 19, 2013 #8
    I see now. It is in the second quadrant. I need to add pi to -0.32 which results in 2.82.

    Thanks! I think I can go to bed now...
     
  10. Jan 19, 2013 #9

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Good night! :)

    ehild
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Writing a function for a wave.
  1. Wave function (Replies: 4)

  2. Wave Functions (Replies: 3)

  3. Wave function (Replies: 2)

Loading...