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Nodes in a Standing Wave

  1. Mar 24, 2017 #1
    1. The problem statement, all variables and given/known data

    Two identical loudspeakers are driven in phase by a common oscillator at 800 Hz and face each other at a distance of 1.25 m. Locate the points along the line joining the two speakers where relative minima of sound pressure amplitude would be expected.

    2. Relevant equations

    v = λ*f

    3. The attempt at a solution

    First up: v = λ*f <=> 343 m/s = λ*800 Hz <=> λ = 0.429 m

    I just did what I learned back in high school, and though: x =n*λ/2 (see the PS), where x is the distance from the beginning of say the imaginary axis (let's say the first speaker). From the theory, we know that every particle that is a node, has a distance that is equal to a integer n, times half the length of the wave. So far so good. Therefore, since the maximum distance is 1.25m, we can say:

    0 <= x <= 1.25 m <=> 0 <= n*λ/2 <= 1.25 m <=> ... <=> 0 <= n <= 5.8 <=> n = 0, 1, 2, 3, 4, 5

    Then, I took those integers, put them into the formula, and got some certains. I went to check, and the manual's answers were completely different! Here's his solution:

    The wavelength is λ = v/f = 0.429m
    The two waves moving in opposite directions along the line between the two speakers will add
    to produce a standing wave with this distance between nodes: distance N to N = λ/2 = 0.214 m

    Because the speakers vibrate in phase, air compressions from each will simultaneously reach the point halfway between the speakers, to produce an antinode of pressure here. A node of pressure will be located at this distance on either side of the midpoint: distance N to A = λ/4 = 0.107 m
    Therefore nodes of sound pressure will appear at these distances from either speaker:

    1/2*(1.25 m) + 0.107 m = 0.732 m and

    1/2*(1.25 m) − 0.107 m = 0.518 m

    The standing wave contains a chain of equally-spaced nodes at distances from either speaker of

    0.732 m + 0.214 m = 0.947 m
    0.947 m + 0.214 m = 1.16 m
    and also at 0.518 m − 0.214 m = 0.303 m
    0.303 m − 0.214 m = 0.089 1 m

    The standing wave exists only along the line segment between the speakers. No nodes or antinodes appear at distances greater than 1.25 m or less than 0, because waves add to give a standing wave only if they are traveling in opposite directions and not in the same direction. In order, the distances from either speaker to the nodes of pressure between the speakers are 0.089 1 m, 0.303 m, 0.518 m, 0.732 m, 0.947 m, and 1.16 m.

    Now, I understand his logic. He finds the antinode at the midway, and then goes step-by-step, left and right to find the nodes. What I don't get, is why it matters that the waves meet at the same time, and why he's so sure they create an antinode.

    I'm probably missing something, but that's the way we did it at High School, so this is a completely new way to me, and I wonder why mine is wrong.

    Any help would be appreciated!

    PS: Back in HS, we were taught that x = (2*N +1)*λ/4 when it comes to nodes, and x = Ν*λ/2 when it came to antinodes. Now, in my new book (Serway's Physics for Engineers &...), it says that x = n*λ/2 is for nodes, and x = n*λ/4 for antinodes. Could someone explain to me why there's such a vast difference between the two, if they cover the same subject?
  2. jcsd
  3. Mar 24, 2017 #2


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    Meeting at the same time guarantees the existence of antinodes in the middle. He is so sure because what else could one have? The pressure is symmetric about the middle.
    Be careful what you are comparing here. Are you sure the example in Serway involves two speakers facing each other and not, say, sound waves in an open pipe? The boundary conditions (the pressure at x = 0 and x = L) are different if that's the case.
  4. Mar 24, 2017 #3
    Could you explain that bit more in-depth, please? What does the pressure have to do with it?

    I'm talking about the general "rule" that's in the theory part, it doesn't have to do with any special case. It's not part of the excercise, I'm just confused, because my HS book gave a different formula than Serway's.

    Granted, i just got into the chapter, and I'm probably missing something, but, is this a "special" case of standing waves with two open ends? Back in HS we had "just standing waves", no special cases and whatnot, so I'm probably confused by the two.
  5. Mar 24, 2017 #4


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    There is no single formula that pertains to standing waves. What you probably saw in high school is standing waves in pipes where you can have two ends open, one end open and one closed or both ends closed. Look at the "two ends open" case.
    There are displacement antinodes or pressure nodes at the two ends. You get standing waves of various harmonics only if you can fit an integer number of half wavelengths between the two ends of the pipe. Your independent variable, if you wish, is the pipe length and the dependent variable is the frequency. You give me the length and I will give you the frequencies of standing waves supported by a pipe of such length. The formulas that you saw in high school do exactly that.

    The question you posted specifies the frequency and the length. You cannot specify both and expect to use the same formulas. Something has to give and that is the requirement that you fit an integer number of half-wavelengths in the given length (unless the numbers are chosen so that this be the case.) So you start at the middle where you know you have an antinode and find nodes an antinodes working you way outwards until you reach the speakers. Note that t the speakers you have neither an antinode nor a node.
  6. Mar 24, 2017 #5
    Oh, okay, yeah, that makes sense! Thanks a ton for the help! I came across these cases (open/open, open/closed, closed/closed), but I was in a bit of a hurry, so I didn't give them much attention right now. The plan was to get through the curriculum fast once, and then go over it again, and explore it in detail.

    Again, thanks for taking the time and helping, I appreciate it!
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