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A Sinusodial wave is propagating

  1. Jan 20, 2012 #1
    A sinusoidal wave is propagating along a stretched string that lies along the x-axis. The displacement of the string as a function of time is graphed in the figure (attachment) for particles at x=0m and x=0.0900m.

    A) What is the amplitude of the wave? (solved)
    B) What is the period of the wave? (solved)
    C) You are told that the two points x=0 an x=0.09m are within one wavelength of each other. If the wave is moving in the +x-direction, determine the wavelength and wave speed. (solved)

    If the wave is moving in the -x-direction, determine the wavelength and wave speed. What should i do here ?


    http://s716.photobucket.com/albums/...rent=5508afb2-b0b2-404b-aa14-09f9c8e37e4a.jpg
     
  2. jcsd
  3. Jan 20, 2012 #2

    Simon Bridge

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    Exactly the same as before except the velocity is the other way.
    What difference does the direction the wave travels make?
     
  4. Jan 20, 2012 #3
    Yes, but how can it be 6.0 m/s?
    Thats is whats stands in the answer sheet
     
    Last edited: Jan 20, 2012
  5. Jan 20, 2012 #4
    Answer for C) v=0.04s*0.090m=3.6*10^-3 mm/s
     
  6. Jan 20, 2012 #5

    Simon Bridge

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    this cannot possibly be correct
    - when you changed units from m to mm you forgot to multiply by 1000
    - you cannot get length/time from multiplying a length and a time.
    ... try again.

    show me how you worked out A, B, and C - what is your reasoning?
     
  7. Jan 20, 2012 #6
    A) A=4mm
    B) T=0.040s (red wave)
    C) it should be 3.6 m/s
    What should i do to find v?
     
    Last edited: Jan 20, 2012
  8. Jan 20, 2012 #7

    vela

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    Pick a peak on the x=0 wave and, from the diagram, figure out how long it takes to reach x=0.090 m. The speed of the wave will then be the distance that point traveled, 0.090 m, divided by the time it took to get there.
     
  9. Jan 20, 2012 #8

    Simon Bridge

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    Wot he said - easier to find a place where y=0, but you also need the same slope.
    You just need the position of the wave at two different times.
    speed is distance over time :)

    To be clear - and because this is not always taught:

    An if arbitrary distortion of the string is: y=f(x)
    If it propagates to the right (+x) direction at speed v, then the equation becomes y(x,t)=f(x-vt)

    In this case f(x)=Asin(kx): k=2π/λ ...(k has the effect of turning the distance x into an angle)

    Therefore: y(x,t)=Asin[k(x-vt)]=Asin(kx-ωt). You can rewrite this in terms of λ and v - it will be more useful to this problem that way.

    In your case this translates into the time-waves at two different x's: y(0,t) and y(0.0900,t) ... which are the red and blue lines respectively.

    generally: y(x,t)= 4.sin(kx-2πt/0.04)

    so: y(0,t)=-4sin(2πt/0.04) and y(0.09,t)=4.sin(0.09k - 2πt/0.04)

    it is a good idea to keep the pi's separate for as long as you can.
    you can use the graph to solve for k in the second one and so get the wavelength.
    and get the speed from v=fλ ... but it is easier to do distance over time from the graph.

    changing the direction of travel is a matter of changing the sign of v in the derivation.
     
    Last edited: Jan 20, 2012
  10. Jan 24, 2012 #9
    0.09k=2pi*t/0.04
    What is the value of t?
     
  11. Jan 24, 2012 #10

    Simon Bridge

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    woah: where did that come from?
    Is this still the same problem?
    Did you try out post #7 first?
     
  12. Jan 26, 2012 #11
    Yes. How can i find the spped from the graph ?
     
  13. Jan 27, 2012 #12

    Simon Bridge

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    The two graphs show you the height-vs-time of different positions on the string.
    These graphs are due to a wave f(x) moving in the x direction.
    So we needs some way to find out how long it takes f(x) to move a fixed distance.
    The distance we have is 0.09m - all the points in f(x) move at the same speed, so we just need to know the time one point in f(x) crosses x=0, and the time the same point in f(x) crosses x=0.09. An easy to identify point is when f(x) is a maximum.

    So look at the time when the x=x0=0 graph first reaches maximum - record that at t0
    Then look for the time the x=x1=0.09m graph first reaches it's maximum - record it as t1
    t1-t0 is how long it took for f(x) to move from x0 to x1.
    Now you have a distance, and the time it took to travel that distance - you can find the speed.

    (Note: you don't have to use peaks - I used the points when the graph crosses from -y to +y - easier to pinpoint but also easier to stuff up.)

    The method in post #8 involves two simultaneous equations and two unknowns.
     
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