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Homework Help: Standing wave

  1. Apr 14, 2007 #1
    1. The problem statement, all variables and given/known data

    Two transverse waves traveling on a string combine to a standing wave. The displacements for the traveling waves are Y1(x,t) =0.0160m sin(1.30m^-1x -2.50s^-1t + 0.30) and Y2(x,t) = 0.0160m sin(1.30m^-1x -2.50s^-1t +.070), respectively, where x is position along the string and t is time.

    1. Find location of the first antinode for the standing wave at x>0
    2. Find the t>0 instance when the displacement for the standing wave vanishes everywhere.

    2. Relevant equations
    f2pi = w
    T= 1/f

    3. The attempt at a solution

    I tried:

    y(x,t) = (Asw sinkx)cos wt

    with Asw= 2A = 2(0.0160)
    k=1.30 m^-1
    w= 2.50 s^-1

    then lambda= 2pi/k = 2pi/1.30 = 4.83
    first node that is x>0 is lambda/2 = 2.416
    but looking for first antidnode with x>0 so its location is half of first node
    which is = 1.208
    But my question for that is when I tried 1.208 m, it was incorrect...so should the answer's units be m^-1?

    That is my main problem...I think I am doing the process right but the units are confusing me.

    For part two, I tried:

    f=w/2pi = 2.50/2pi = .397
    T= 1/f = 1/.397 = 2.51

    I looked at a figure in my text book, and the instance the standing wave vanishes is when t= (8/16)*T

    so t=(8/16)*T = (1/2)*T=(1/2)*2.51= 1.255 s

    But the units for w (if w is correct) was in s^-1, but I thought the units of angular frequency is in rad/s...so is 2.50s^-1 the frequency?

    What am I doing wrong?:confused:
  2. jcsd
  3. Apr 14, 2007 #2
    I was a little confused by the terms w/in the arguments as well, generally isn't is:

    sin(kx+/-wt+phase) so for first part,

    at t=0, i would just try to set up the two so the difference in arguments as equalling pi, and solve for x to find antinode.
    Last edited: Apr 14, 2007
  4. Apr 14, 2007 #3
    Well in my textbook they have the equation as y(x,t)= (Asw sin kx) sin wt
    But, the transverse wave equations do include a phase, so I am just a bit confused:confused:
  5. Apr 14, 2007 #4
    ok, I'll try to what you suggest
  6. Apr 14, 2007 #5
    but what about the units m^-1 and s^-1?
  7. Apr 14, 2007 #6
    well those should be right for the constants which if I understand your original post are 1.3 and -2.5 so that when multiplied by length and time respectively the whole thing lacks dimensions (as in radians).
    Last edited: Apr 14, 2007
  8. Apr 14, 2007 #7

    PS: I ended up getting x=-0.1423+pi/1.3 for first positive zero (node), and 1.07 for first antinode. Is that what you got?
  9. Apr 14, 2007 #8
    denverdoc, I am really confused now. I thought the phase doesn't matter for standing waves. But I don't understand how you came up with your answer for part one. I'm trying to read more about standing waves online, because I am a bit lost now on this problem
  10. Apr 14, 2007 #9
    well the amplitude eqns both have phase offsets. For the first part, since the eqns are identical except for phase, the anti nodes and nodes occur when the phase offset is centered around pi/2,pi,3/2*(pi),etc. So what I did was calculate x to make the condition 1.3x+.3-pi/2=pi/2-(1.3x+0.07) In other words, if you think of phasor, the leading phasor leads 90 degrees by the same amt the second lags. Does that make any sense? maybe consider the nodes where the sine(180+some theta)+sine(180-some theta)=0
  11. Apr 14, 2007 #10
    hm, ok that explanation for phase offset makes sense...oh I'm sorry I wrote a typo down in problem. The Y2(x,t) phase should be .70 not .070. Sorry about that.

    However, when I try to get x= =-0.1423+pi/1.3, I keep getting an entirely different answer. Keeping .070, I get x= (pi + .37)/ 2.6

    if I set 1.3x + .30 - pi/2 = pi/2 -( 1.3x +.70)
    1.3x + .30 - pi/2 = pi/2 - 1.3x -.70
    1.3x + .30+ .70 = pi/2 +pi/2 - 1.3x
    1.3x +1= pi -1.3x
    2.6x= pi -1
    x= (pi - 1)/2.6

    Did I set the equation incorrectly?
  12. Apr 15, 2007 #11
    let me see, offhand I dont think it makes a difference as the phases can be added and put in one or the other eqn,
    1.3(x)+.7-pi/2=pi/2-((1.3x+.3) so yea, 2.6x=pi-1. Now this is a combination of transverse waves offset by a small phase difference but I don't see it as a standing wave exactly.

    See here: http://www.kettering.edu/~drussell/Demos/superposition/superposition.html [Broken]

    since the waves are moving in the same direction with same everything and fixed phase difference, the resulting expansion is not so simple as
    cos(wt)*sin(kx+theta) which for standing wave gives both the spatial info we used in part a to compute nodes,antinodes, etc and the modulating cos wt which when cos(wt)=0 makes everything 0 irrespective of x. Are you sure the signs are the same for the 2.5 coefficient?
    Last edited by a moderator: May 2, 2017
  13. Apr 15, 2007 #12
    No, I was wrong. I'll type down the full thing again, since I keep making errors.
    Y1(x,t)= 0.0160 sin (1.30m^-1x -2.50s^-1t + 0.30)
    Y2(x,t) = 0.0160 sin(1.30m^-1x + 2.50s^-1t +0.70)
  14. Apr 15, 2007 #13
    Also, when you wrote it doesn't make a difference since the phases can be added in one or the other equations, did you mean that 2.6x= pi - 1 is not the equation needed to find x, but that either one of the above equations that we set equal to one another is sufficient to solve? I think that is what you said, but I just want to make sure. I'm going to read the info you linked
  15. Apr 15, 2007 #14
    Oh thats good news:

    no I think first part fine, it was with the sin(a+/-b) expansion, when they are in different directions, two of 4 terms drop out, in this case, I think it is legit just to say that Y(resultant)=2*A*cos(wt)(sinkx) since we are interested in time when amplitude vanishes everywhere along x axis. so that just means solving for t when cos(wt)=0
  16. Apr 15, 2007 #15
    denverdoc, now that I clarified the transverse wave equations, is the 2.6x=pi-1 not matter anymore? Can I apply this to find the node and therefore antinode?
    also for the second part for the time. What about the x in the equation? I've read the info you linked to, but I don't see how it applies to time.
  17. Apr 15, 2007 #16
    Not aure exact ?, now that we know that the two waves represent a "collison" in opposite directions, part b became a whole lot easier. In fact the time at which displacement for all particles is zero, becomes trivial. This is only because the cosine term can be cleanly factored from the rest, and so when cos(theta)=0, everything is zero, irrespective of the exact shape of whatever waveform we had as initial time. In other words if you have a situation where r(x,t)=a(x)*b(t);
    r(x,t) must be zero when b(t)=0, no matter what the value of a(x).

    So for part, b you don't need the answer from part a. But to do part a, when t=0 and we were only worried about the spatial (dependency on x)aspects, I believe the answer remains the same. I hope that answers question.

    (BTW, deveolping fuller understanding actually easier, not harder when you have seen the complete derivation from the PDE's, boundary, and initial conditions.)
  18. Apr 15, 2007 #17
    I worked on part two and got the correct time. I just didn't immediately understand what you were explaining for part two.
    However I'm still working on part one.
  19. Apr 15, 2007 #18

    standing wave--apparent illusion where displacements in transverse direction move in SHM with respect to time and the spacing of nodes/antinodes, appear to be SHM frozen in time, depending on boundary conditions. I thought part 1 was done frankly.
    Last edited: Apr 15, 2007
  20. Apr 17, 2007 #19
    I know you said you thought you were quite frank for part one,
    but I couldn't get the correct answer. I understand that since there is a phase, then the nodes, etc. will not simply be 0,pi, etc. There will be a phase along with the node.

    since we are looking for first antinode when x>0, then the first node when x>0 is pi/k; with the phase does it [and this is the confusing part] (pi - phase)/k
    or pi/k - phase?
  21. Apr 18, 2007 #20


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    Homework Helper

    Y1(x,t) =0.0160m sin(1.30m^-1x -2.50s^-1t + 0.30) and Y2(x,t) = 0.0160m sin(1.30m^-1x -2.50s^-1t +.070). This is not a standing wave. Get over it. Standing waves components travel in opposite directions. Please restate the problem.
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