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Maximum and minimum transverse speeds at an antinode

  1. Aug 2, 2016 #1
    1. The problem statement, all variables and given/known data
    Adjacent antinodes of a standing wave on a string are 15.0cm apart. A particle at an antinode oscillates in simple harmonic motion with amplitude 0.850cm and period 0.0750s. The string lies along the +x-axis and is fixed at x = 0.

    the speed of the two travelling waves are 4.00m/s

    Find the maximum and minimum transverse speeds of a point at an antinode.

    2. Relevant equations
    v=Aωcos(ωt)

    3. The attempt at a solution

    I was just reading this post: https://www.physicsforums.com/threa...erse-speeds-of-a-point-at-an-antinode.218615/ and I couldn't understand why they've taken cos(wt)=1 and cos(wt)=0 as the bounds. If it's an antinode, wouldn't it be max at 1 and min at -1?

    Secondly, their final answers were v max=0.712, vmin=4.36 x 10^-17 but if cos(pi/2)=0 then v=min will be 0, how did they get 4.36 x 10^-17?
     
  2. jcsd
  3. Aug 2, 2016 #2

    TSny

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    Speed is the magnitude of velocity. Speed = |v|. So, you need to consider the maximum and minimum values of |v|.

    You are right that the minimum speed would be zero. 10-17 is very small. I haven't gone through why they got this small number rather than zero. Probably some sort of "round off error".
     
  4. Aug 2, 2016 #3
    Thank you for your reply!
    Aw cos (pi/2) is directly 0 isn't it? because cos (pi/2) is 0?

    Also, could you guide please me on this question: explain why the relationship v=w/k for traveling waves also applies to standing waves? It was a later part of this question.
     
  5. Aug 2, 2016 #4

    TSny

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    Yes, that's right.

    Hmm, standing waves don't really have a speed (they just oscillate while "standing" in one place). So, I don't quite understand how you can say that v = w/k applies to a standing wave. However, standing waves do have an angular frequency w and a wavenumber k. So, I guess they want you to show that if you take the ratio w/k for a standing wave, then you get the wave speed for a traveling wave.

    One way to approach this is to remember that standing waves can be considered to be due to the superposition of two harmonic waves traveling in opposite directions, each with the same amplitude and the same w and k. So, you would need to show that the superposition of the two traveling waves produces a standing wave with the same w and k as the two traveling waves.
     
  6. Aug 2, 2016 #5
    Thank you!
     
  7. Aug 2, 2016 #6
    "There's two important 'features' of the standing wave. 1) wavelength - the distance between peaks. 2) period - the time it takes for the wave to go through one full oscillation (at constant x).

    You have 3 parameters (omega,v,k) and the equation omega=kv, so that means you have a choice for which two of these parameters you want to use to describe the two 'features'. Then the 3rd parameter is given by the equation.

    The natural choice is to use omega to describe 2PI/period and use k to describe 2PI/wavelength. And then v is given by the formula. But you don't have to do it this way. You could use v to describe wavelength/period and omega as 2PI/period, then k is given by the formula. So it is your choice really.

    In other words, the physical meaning of 'velocity' for a standing wave is wavelength/time"

    What does he mean by the last part?
     
  8. Aug 2, 2016 #7

    haruspex

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    I think the author meant to write wavelength over period.
    The idea is to retain a concept of wave velocity even though it is a standing wave. You can think of a simple standing wave as the sum of two identical waves with equal and opposite velocities. The magnitude of that velocity will be wavelength/period.
     
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