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Determine the amplitude of a mechanical wave given only it's velocity and period

  1. Sep 1, 2009 #1
    1. The problem statement, all variables and given/known data

    A transverse sinusiodal wave on a string has a period T= 25.0 ms and travels in the negative x direction with a speed of 30.0 m/s. At t=0, an element of the string at x=0 has a transverse position of 2.00 cm and is traveling downward with a speed of 2.00 m/s.

    2. Relevant equations

    I know that the general equation for a waveform is:

    y(x,t)= Asin(kx-(omega)t+(phi))

    {omega is the angular frequency and phi is the phase constant}

    I also know that at t=0 and x=0

    y(x,t)=.02 m --> .02=Asin(phi) --> A=.02/sin(phi)

    I can also determine omega given the period:

    omega=(2*pi)/.025=251.3274 rad/s

    3. The attempt at a solution

    Even with my relevant equations, I cannot figure out what I am supposed to do with the extra known factor, velocity. I have scrubbed the current chapter and can't find where I can use the transverse position, angular frequency and velocity in the y direction to determine a waves amplitude. What am I missing? Please help! I have so many problems to do but I just can't do another one until I figure this out! Thanks so much.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 2, 2009 #2
    There`s certainly a way of getting A with that data. I´ll be glad to explain it to you but it's a little heavy, I mean all the calculations.

    Let's start:

    we are given the constants of the wave T which, as you said, give us w (omega) and v which give us k (by the relationship v=(lamda)/T and from that k=2·(pi)/(lamda) ). We also have the elongation y(0,0)=Yo (i know the data is numerical but in order to have it more simple we´ll work with letters) and the velociti of that point at that time Vy(0,0)=Vo. (I think this is the fact that you misregarded in your first post)

    Ok then we can write the eq. like this (as you did):

    y(x,t)=A·sin(kx-wt+&) [& is (phi)]

    the first condition we know is y(0,0)=Yo , lets write it:

    y(0,0)=A·sin(k·0-w·0+&)=A·sin&=Yo -> A·sin&=Yo

    first relationship between unknown & and A.

    Now we use the fact that velocity of the point x=0 at t=0 is 2m/s downwards which makes:

    Vo=-2 m/s so we´ll work with Vo.

    You should remember or know that the velocity of a point at a moment is given by the partial derivative with respecto to time of the wave function y(x,t)=... So lets differenciate:

    [tex]\partial[/tex]y/[tex]\partial[/tex]t=-Aw·cos(kx-wt+&)

    so,

    Vy(0,0)=-Aw·cos(K·0-w·0+&)=-Aw·cos&=Vo -> -Aw·cos&=Vo

    Second relationship between unkanown A and &.

    Now you just have to solve the eqs. system:

    A·sin&=Yo

    A·w·cos&=Vo
     
  4. Sep 2, 2009 #3
    My personal advice with the system is to divide first eq. by the other and write the new eq. (without A) like this:

    sin&=(w·Yo/Vo)·cos&

    And the use the trigonometric relationship sin2&+cos2&=1 in order to get the expression of sin& (as we want A, not &) then once you've done all the calculations just put the expression you get for sin& in the one of the first eqs. which had sin& on it:

    A·sin&=Yo -> A=Yo/sin&

    And there you have your solution.

    Hope it is worth and hope I made myself clear. If you have any other trouble just tell me.

    Salutations, zaphys. :)
     
  5. Sep 2, 2009 #4

    rl.bhat

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    Homework Helper

    When a transverse sinusoidal wave travels on a string, every element of the string executes simple harmonic motion. Its equation is given by
    y = A*sin(ωt), where y is the transverse displacement, A is the amplitude and ω is the angular velocity = 2π/T.
    The transverse velocity is given by
    dy/dt = A*ω*cos(ωt) = ω*sqrt[ A^2 - A^2*sin^2(ωt)] = ω*sqrt[ A^2 - y^2]
    Angular velocity, transverse velocity and transverse displacement is given. Find the wave amplitude.
     
  6. Sep 2, 2009 #5
    Why didn't I see that? Thank you so much...(solving for A reveals the amplitude to be 2.15 cm).

    I think I just hit the wall in my head where I am not thinking holistically about the problem.

    Thanks again.
     
  7. Sep 2, 2009 #6
    Wait a second, rl.

    How did you come up with that equality for the derivative of the waveform?

    I think that was what I was actually looking for!
     
  8. Sep 2, 2009 #7

    rl.bhat

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    Homework Helper

    If y is the displacement, then dy/dt will be the velocity. Using simple trigonometry I got the equality.
     
  9. Sep 24, 2009 #8
    Thank you so much for your help with this. While I did find the solution, I am having trouble visualizing what this means. Do you know, or can you draw, the graphical representation of how to solve for A? Many thanks.
     
  10. Sep 29, 2009 #9
    guess i get no love with this request---
     
  11. Sep 29, 2009 #10

    rl.bhat

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    Homework Helper

    Sorry. I don't how to do that.
     
  12. Sep 29, 2009 #11
    No worries! In my mind, I picture a circle that has a radius equal to A. Where the angular frequency is the rate at which A fluctuates from +A to -A in radians per second. I am just not sure what that really looks like or if I am way of base. Your equation says that if I know the transverse position and the angular frequency then I can determine the amplitude. I am just trying to figure out how this all ties together because I think it's an important fundamental concept. Thanks for the try.
     
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