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Amplitude of 2 waves at a point

  1. Feb 11, 2015 #1
    1. The problem statement, all variables and given/known data
    Two sound sources oscillate in phase at 500Hz with the same amplitude sm. At a point 5m from source 1 and 5.17m from source 2, what is the amplitude?

    2. Relevant equations
    s(x,t)=sm*cos(kx-wt)
    w=2πf
    Speed of sound=340m/s

    3. The attempt at a solution
    v=340m/s
    w=2πf=1000π rad/s
    k=w/v=50π/17
    t1=5m/340m/s=1/68s
    t2=5.17m/340m/s=0.0152s

    s1(5,1/68)=sm*cos(50π/17*5-1000π*1/68)
    s2(5.17, 0.0152)=sm*cos(50π/15*5.17-1000π*0.0152)

    Adding the two together I get ~2sm, the correct answer is 1.41sm
     
  2. jcsd
  3. Feb 11, 2015 #2

    Nathanael

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    Homework Helper

    How come you're using two different times? The waves act at the same time right? It doesn't make much sense for one wave to interfere with the future of the other wave.

    Anyway you don't want to plug a specific time-value in for the waves. Doing that gives you the displacement of the wave at a fixed position in space and a fixed instant in time.
    You only want to fix the spatial position, but leave the time variable there because you want to watch how the wave acts over time (in that specific spot).

    Fix the spatial position so that the wave displacement only depends on time, then find the amplitude of the resulting wave.

    (The times you calculated are how long it will take for the sound to reach the point after the speaker is turned on. But that's not relevant, as long as we assume it's been long enough for both waves to reach the point.)
     
  4. Feb 11, 2015 #3
    Okay so I rewrote the two equations as:
    s(5.00, t)=sm*cos(50π/17*5-wt)
    s(5.17, t)=sm*cos(50π/17*5.17-wt)
    I set t=0 after this and added the two cos functions together which gave the correct answer, is this correct?
     
  5. Feb 12, 2015 #4

    gneill

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    Staff: Mentor

    How did you "add the cos functions"? Remember, their value varies over time so picking one instant and summing the then-current values of the cos functions won't give you the magnitude of the wave unless you're very lucky indeed and happened to pick an instant n time that corresponds to the maximum.

    What will work is solving for a value of t that corresponds to a maximum of the sum.
     
  6. Feb 12, 2015 #5
    I added them like: cos(a)+cos(b)=constant
    I did this because the 2 waves are in-phase so the 2 amplitudes should add together. I'm not sure how to solve for a value of t that will give a maximum since if I take the derivative I will need the speed of the function.
     
  7. Feb 12, 2015 #6

    gneill

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    Staff: Mentor

    Your cos(a) and cos(b) will not be constant over time. At any given location in space they will both vary through the maximum range of their amplitudes. So you can't just pick an arbitrary time t and hope that it corresponds to a maximum.

    You have a function ##f(t) = cos(k_1 x_1 - ω t) + cos(k_2 x_2 - ω t)##, where since you have a fixed location the x's are constants. You can differentiate this w.r.t. t to find a maximum (or minimum, either will yield the amplitude).
     
  8. Feb 12, 2015 #7
    After differentiating this with respect to time I got t=(k2x2+k1x1)/(2w)=0.00476s
    When I plugged this time back into my equations I got the wrong answer so I'm guessing I'm doing something wrong.
     
  9. Feb 12, 2015 #8

    gneill

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    Staff: Mentor

    Can you show your work?

    EDIT: Actually, you're expression looks okay. Check your arithmetic (assuming you plugged in the correct values for the x's and w).
     
    Last edited: Feb 12, 2015
  10. Feb 12, 2015 #9
    Y
    Yea sorry I would've posted all the work but I'm on my phone so it's kind of difficult. I messed up plugging the values in to solve for t, it's actually equal to 0.0149s. It worked! Thanks for the help I needed quite a lot of it on this question haha
     
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