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Waves and Wave equation

  1. Jan 23, 2013 #1
    1. The problem statement, all variables and given/known data

    Hello all,

    stuck on a question involving a formula for a wave that doesn't make much sense to me.

    Assuming that a wave on a string is represented by:

    y(x,t) = y_i*sin((2∏/λ)(vt-x))

    Where y is transverse displacement at time t of the piece of string at x. The other symbols have their "usual meaning". Find the velocity and acceleration of the small piece of string at x = 10m, as a function of time.

    Making use of the fact that the piece of string satisfies Newton's second law, show that the piece of string is acted on by a Hooke's Law force.

    2. Relevant equations

    y(x,t) = y_i*sin((2∏/λ)(vt-x))

    y(x,t) = A*sin(kx - ωt)

    k = 2∏/λ

    3. The attempt at a solution

    I'm a bit confused by the format of the formula...

    I know that to find velocity and acceleration, I can take the derivative of the equation once for velocity and again for acceleration. Looking at the set up, it looks like if we replace 2∏/λ with k and expand, we see in the brackets

    (kvt - kx)

    where kv could equal ω?

    I don't know if this is at all valid. I also don't understand why the variable "v" is used in the equation, but from the question i can assume that it is analogous to omega ω.

    So

    y(10,t) = y_i*sin((2∏/λ)(vt-10))

    v(10,t) = y_i*cos((2∏/λ)(vt-10))*???

    I'm having trouble knowing how to take the derivative of this when I don't know λ and v.
    Unless they are constants and I can ignore them?

    Any help will be appreciated... Thanks.
     
  2. jcsd
  3. Jan 23, 2013 #2

    cepheid

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    The usual form for a wave is something like sin(kx - ωt). The phase is given by kx - ωt, so that if x is constant (i.e. you're sitting at a particular point in space as the wave passes by you) the variation of wave intensity with time is sinusoidal with angular frequency ω. When t = T, where T is one period of the oscillation (##T \equiv 1/f = 2\pi /\omega##), then ωt = ωT = 2π. The phase increments by 2π as expected after one cycle. Note: f is the frequency of the wave, and ω = 2πf is the angular frequency.

    You can rewrite kx - ωt as k[x - (ω/k)t] $$= k\left(x - {2\pi f}\frac{\lambda}{2\pi}t\right)$$ $$= k(x -f\lambda t)$$

    Of course, there is just the standard relation that wave speed = frequency*wavelength: ##v = f\lambda## so this becomes k(x - vt). This form of the wave function is actually kind of intuitive, because it says that at time t, the amplitude of the wave at a distance vt ahead of a point x is the same as what the amplitude was at point x back at time 0.

    EDIT: And yes, of course f and λ (and hence v) are constant for a given wave. Don't confuse "v" (the wave propagation speed, in the x direction) with dy/dt, the speed of a point oscillating in the y direction as the wave passes by, which you are asked to compute.
     
    Last edited: Jan 23, 2013
  4. Jan 24, 2013 #3

    tms

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    [itex]v[/itex] is just the speed of the wave, as distinct from the transverse speed of the bits of string.
    For a given wave, they are constant.
     
  5. Jan 24, 2013 #4
    Okay..

    So considering that λ and v are constants, the equations should look, hopefully, something like:

    y(10,t) = y0*sin[k(vt-10)]

    v(10,t) = y0*cos[k(vt-10)]*k

    a(10,t) = -k2y0*sin[k(vt-10)]


    I assume this is as far as I can go? I am given nothing else but distance x = 10m, so I cannot find the amplitude, or any of the constants, can I?

    I also don't understand what the second part of the question is asking, when it asks to show the string is acted on by a Hooke's Law force knowing that the string follows Newton's second law.

    What do the two equations have in common, and how can I show that the string is modeled by F = -kx, from F = ma?
     
  6. Jan 24, 2013 #5

    tms

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    Check your derivatives again. Remember the chain rule.

    As for the second part, you have just calculated the acceleration, and you already know the displacement. Just set [itex]ma = -kx[/itex], or [itex]y[/itex] in this case. And be sure to keep the two [itex]k[/itex]s straight; use a different letter for one of them, perhaps.
     
  7. Jan 24, 2013 #6
    Is the chain rule not just multiplying the initial derivative by the derivative of what is in the brackets of the sin function?

    where k(vt-10) is a product rule?

    if k is a constant, derivative is zero, derivative of (vt - 10) should be v (which I forgot to put in... argh)

    So equations should be:

    y(10,t) = y0*sin[k(vt-10)]

    v(10,t) = y0*cos[k(vt-10)]*k*v

    a(10,t) = -k2v2y0*sin[k(vt-10)]


    and if a(10,t) = -k2v2y0*sin[k(vt-10)]

    I set that equation, multiplied by m, to y? y being the initial displacement equation..?

    Also what do you mean by keeping the k's straight, are they not the same value as a constant..?

    thanks
     
  8. Jan 24, 2013 #7

    tms

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    Easier to just multiply it out in this case: [itex] kvt - 10k[/itex].
    That's it.
    Start with [itex]F = ma[/itex], with [itex]a[/itex] as derived above. Then look at that equation and compare it to your equation for [itex]y[/itex]. The answer should jump out at you.
    The [itex]k[/itex] in Hooke's law and the [itex]k[/itex] in the equation for the wave are entirely different entities that just happen to use the same letter.
     
  9. Jan 24, 2013 #8
    Ah, thought you were referring to the k's in the derived equation. Sorry!

    Anyway, if

    F = ma = -kx

    Would x in this case be y? and thus y0sin[k(vt-10)]

    so ultimately we see

    -mk2v2y0*sin[k(vt-x)] = -Ky0sin[k(vt-x)]

    where uppercase K is spring constant, the amplitudes cancel, the sin functions cancel, and were left with

    mk2v2 = K

    But I don't see how that actually says anything..? Nothing jumped out at me. :(
     
  10. Jan 24, 2013 #9

    tms

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    You've got all the parts, you just have to put them together correctly.

    Think of the form of Hooke's law: the force equals -1 times a constant times the displacement. Now take your force: [itex]F = ma = m[/itex] times the acceleration you've derived. How can you put that in the form of Hooke's law? Remember that you also have an expression for the displacement, that is [itex]y[/itex].
     
    Last edited: Jan 24, 2013
  11. Jan 24, 2013 #10
    Well, if we replace the y0sin[k(vt-10)] portion of the F = ma equation, that would give

    F = -mk2v2y

    This somewhat models Hooke's law, if we can consider m, k, and v to be all constants and serve as the constant in the Hooke's law equation.

    Is this more on the right track..?
     
  12. Jan 24, 2013 #11

    tms

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    You've got it. [itex]m[/itex], [itex]k[/itex], and [itex]v[/itex] are all constants in this situation.
     
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