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Oscillation of soundwave

  1. May 21, 2015 #1
    Given a soundwave with wavelength ##\lambda = 0,628 m## and a period ##T = 2 ms##.
    The stopper is started at the exact moment when the wave is at its minimum, call it ##-A##. After ##3.5ms## and ##0,157m## from the point of origin, the wave has reached its maximum, ##A##.

    Why is it so?
    According to my reasoning the wave has reached 0 at that time. It takes the ##2ms## to reach minimum again and then with the remaining ##1.5ms## it reaches 0 with the first ##0.5ms##, then the maximum after ##1ms## and then 0 again after ##1.5ms##, but this is incorrect.

    Edit: I would understand if we were given just the distance it has traveled in which case it would have travelled exactly a fourth from the point of origin and therefore reaching its maximum, but then why the ##3.5ms##?
     
    Last edited: May 21, 2015
  2. jcsd
  3. May 21, 2015 #2
    think of the appearance of the wave, say it moves towards you, particles are propagating along y direction. When you are at 0.157m away, at t=0, you should see this point is y=0. After 0.5ms, the particle will have y= -A, then after 1ms, the particle will have y=A. Then after one more period, 2ms, that is totally 3.5ms, that particle will have y=A, which is maximum.
    so it seems that it has no problem.

    Edit: sound wave is not transfer wave, but I used transfer wave to explain, but they should have same result.
     
  4. May 21, 2015 #3
    Doesn't it say at t=0 y=-A?
     
  5. May 21, 2015 #4
    that one is for particle position at stopper, but remember that 0.157m? the position 0.157m away from stopper has particle at y=0 when t=0
     
  6. May 21, 2015 #5
    You can just use the formula for the phase, no need to guess.
    You have [tex]\phi =\omega t - k x =2 \pi (\frac{t}{T}-\frac{x}{\lambda})[/tex].
    For t=0 and x=0 you have [tex]\phi_1 =0[/tex]
    For the values given in OP, you have [tex]\phi_2 = 2 \pi (\frac{3.5}{2}-\frac{0.175}{0.628})=3 \pi [/tex]
    Multiples of 2Π do not change the wave at all. So this is an actual change of Π which means that the two events are in opposition of phase. So from -A it goes to +A.

    If you still want to describe it "in two pieces", the x2=0.157m is one quarter wavelength. So at t=0, and x1=0, the wave at x2=0.175 m goes through zero, towards negative values.
    First time it will reach +A will be after 3/4 of a period. Next time it will reach the same value will be after 3/4+1 period or 1.75 periods or 3.5 ms.
     
  7. May 22, 2015 #6
    Like @nasu, your wave equation is:
    $$ y(x,t) = A\sin(1000\pi{t}-10x) $$

    Edit: if demand to calculate values for large time (t>10T), some programs makes truncation error mistakes.
     
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