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Interference waves

  1. Jun 25, 2013 #1
    1. The problem statement, all variables and given/known data

    Two waves propagate opposite directions along a string one is

    [itex]y_{1}[/itex] = 0.6 cos[ [itex]\frac{π}{2} (\frac{x}{2.0m} - \frac{t}{8.0s}[/itex]) ]

    and the other is

    [itex]y_{2}[/itex] = 0.6 cos[ [itex]\frac{π}{2} (\frac{x}{2.0m} + \frac{t}{8.0s}[/itex]) ]

    find the frequency [itex]\nu[/itex], wavelength λ and speed v of each wave.

    2. Relevant equations

    y = [itex]y_{m}[/itex]sin[ [itex]2π(\frac{x}{λ} - \frac{t}{T}[/itex]) ], a simple harmonic traveling wave

    3. The attempt at a solution

    y = [itex]y_{m}[/itex]cos[ [itex]\frac{π}{2}(\frac{x}{λ} - \frac{t}{T}[/itex]) ] = [itex]y_{m}[/itex]sin[ [itex]\frac{π}{2}(\frac{x}{λ} - \frac{t}{T}) + \frac{π}{2} [/itex] ], expressing the given wave(s) in terms of a sine function
    y = [itex]y_{m}[/itex]sin[ [itex]2π(\frac{x}{4λ} - \frac{t}{4T}) + \frac{π}{2} [/itex] ], the given wave(s) in the form of a simple harmonic traveling wave

    so by inspection

    [itex]\frac{π}{2}(\frac{x}{2.0m} - \frac{t}{8.0s}) = 2π(\frac{x}{4λ} - \frac{t}{4T}) [/itex]

    4λ = 2.0 m
    4T = 8.0 s

    then

    λ = 0.5 m
    T = 2.0 s
    [itex]\nu[/itex] = 1/T = 0.5 Hz
    v = [itex]\nu[/itex]λ = 0.5 Hz * 0.5 m = 0.25 m/s

    But the answer key for the text (Halliday/Resnick 2nd Ed, yes I'm that old) is [itex]\nu[/itex] = 1/T = 2.0 Hz, λ = 2.0 m, v = 4.0 m/s. Can anyone tell me waht I am doing wrong? I seem to be missing a fundamental concept here; I am stumped.
     
    Last edited: Jun 25, 2013
  2. jcsd
  3. Jun 25, 2013 #2

    haruspex

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    Right.
    Wrong. You've turned 2π into π/2.
     
  4. Jun 25, 2013 #3
    Thnx for the response haruspex, but can you be more specific? I think I should explain my reasoning more for the form

    [itex]y_{m}[/itex] sin [itex][ \frac{π}{2}(\frac{x}{λ} - \frac{t}{T}) + \frac{π}{2} ] [/itex]

    occurring. I know how to obtain the frequency, wavelength and speed from a simple harmonic wave of the form

    [itex]y_{m}[/itex] sin [itex] [ 2π(\frac{x}{λ} - \frac{t}{T} )], \hspace{40mm} [/itex](1)

    The wavelength and frequency are given in the sine's argument, and the speed is simply [itex]\lambda \nu = v[/itex]. But in the problem, I was given

    [itex] y = y_{m}[/itex] cos [itex][ \frac{π}{2}(\frac{x}{2.0m} - \frac{t}{8.0s} )] [/itex],

    so I wanted to express it in the simple form (1) above. I know cos[itex]\theta[/itex] = sin[itex](\theta+\frac{\pi}{2})[/itex], so from that I wrote

    [itex]y_{m}[/itex]cos [itex] [\frac{π}{2}(\frac{x}{λ} - \frac{t}{T} )][/itex] = [itex]y_{m}[/itex] sin [itex] [ \frac{π}{2}(\frac{x}{λ} - \frac{t}{T}) + \frac{π}{2} ], \hspace{40mm} [/itex] (2)

    but for a harmonic wave the sine argument needs to be of the form [itex][2π(\frac{x}{λ} - \frac{t}{T}) + \frac{π}{2} ][/itex], so I factored out 4 from the argument that occurred in the given function (now expressed as a sine in (2) ) as follows

    [itex] [ \frac{π}{2}(\frac{x}{λ} - \frac{t}{T}) + \frac{π}{2} ] = [ \frac{π}{2}(\frac{4x}{4λ} - \frac{4t}{4T}) + \frac{π}{2} ] [/itex]

    = [itex] [ \frac{4π}{2}(\frac{x}{4λ} - \frac{t}{4T}) + \frac{π}{2} ] [/itex]

    = [itex] [ 2π(\frac{x}{4λ} - \frac{t}{4T}) + \frac{π}{2} ] [/itex].

    Now (or so I'm thinking) the argument has the form of the simple harmonic wave, so I was thinking that by inspection

    [itex]2π(\frac{x}{4λ} - \frac{t}{4T})[/itex] = [itex]\frac{π}{2}(\frac{x}{2.0m} - \frac{t}{8.0s} ) [/itex]

    (ignoring the constant phase term π/2) and I got

    [itex]4\lambda = 2.0m[/itex]

    [itex]4T = 8.0m[/itex]

    and derived my final answers from these two equations. But the answers don't jibe. Can you tell me how I can solve this problem?
     
  5. Jun 25, 2013 #4

    haruspex

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    All fine up to there, but your next step is illogical. If you want [itex][ \frac{\pi}{2}(\frac{x}{2.0m} - \frac{t}{8.0s} )] [/itex] in the form [itex][ 2\pi(\frac{x}{λ} - \frac{t}{T}) ] [/itex] (sine versus cosine doesn't matter here) then just equate them: [itex]\frac{\pi}{2}(\frac{x}{2.0m} - \frac{t}{8.0s} )= 2\pi(\frac{x}{λ} - \frac{t}{T}) [/itex]
    Since this has to be true for all x and t, you can deduce λ and T.
    In what you did, you effectively changed the meanings of λ and T.
     
  6. Jun 25, 2013 #5
    Can I ask a simple question here about the frequency of the waves.
    Take the equation. y2 = 0.6 cos((pi/2)[x/2+t/8]). Lets let x = 0 and forget about the amplitude then we have.
    y2 = cos((pi/2)(t/8)) = cos ((pi/16)t)
    From this, I thought the period of the wave would be 32 seconds giving a frequency of 0.03 Hz.
    Where am I wrong?
     
  7. Jun 25, 2013 #6

    haruspex

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    You aren't. Either the problem has been misstated or the the answer has.
     
  8. Jul 29, 2013 #7
    problem misstated

    I have been away from my physics studies for awhile, but wanted to say thnx to both haruspex and barrj for checking in on this post. The problem in its entirety is quoted verbatim from the Halliday & Resnick Fundamentals text (2nd ed), pr 27, pg 314:

    Two waves are propagating on the same very long string. A generator on the left of the string creates a wave given by [itex]y_{1} = 0.06 cos[ \frac{π}{2}(2.0m^{-1}x - 8.0s^{-1}t)][/itex] and one at the right end of the string creates the wave [itex]y_{2} = 0.06cos[ \frac{π}{2}(2.0m^{-1}x + 8.0s^{-1}t)][/itex]. Calculate the frequency, wavelength and speed of each wave.

    I erroneously (ok, it was stupid) equated

    [itex]\frac{π}{2}(\frac{x}{2.0m} - \frac{t}{8.0s}) = {2π}(\frac{x}{λ} - \frac{t}{T})[/itex]

    and not

    [itex]\frac{π}{2}(2.0\frac{x}{m} - 8.0\frac{t}{s}) = {2π}(\frac{x}{λ} - \frac{t}{T})[/itex]

    as I should have; this led to the answers given in the text. haruspex is correct in betting that the problem was misstated. Thnx for responding; I appreciate the help A LOT.
     
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