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Homework Help: Error in solutions or am I doing something wrong?

  1. Feb 6, 2010 #1
    1. The problem statement, all variables and given/known data

    This is quite simple, but for some reason I'm getting the wrong answer. I am given a longitudinal wave where the vertical displacement is given by [tex]z(x, t) = Acos(....)[/tex]. I must calculate the vertical acceleration at some time t and position x.

    2. Relevant equations

    3. The attempt at a solution

    I just differentiate twice with respect to time and get [tex][-\omega^{2}A][z(x,t))][/tex]. Correct?

    This is not the first time I have encountered a solution error (if it is the case).
  2. jcsd
  3. Feb 6, 2010 #2
    I think you have an extra A in that answer.
  4. Feb 6, 2010 #3
    Assuming the (...) argument of the cosine is just [itex]\omega t[/itex], then

    \frac{dz}{dt}=A\frac{d}{dt}\cos[\omega t]=-A\omega\sin[\omega t]

    \frac{d^2z}{dt^2}=-A\omega\frac{d}{dt}\sin[\omega t]=-A\omega^2\cos[\omega t]=-\omega^2 z

    So willem is correct, you multiplied by an extra [itex]A[/itex].
  5. Feb 6, 2010 #4
    Ah, yes sorry I did not mean to put that extra [tex]A[/tex] there.

    But it should read [tex]-\omega^{2}Acos(....)[/tex]

    This was my original answer and its still wrong.

  6. Feb 6, 2010 #5
    What is the argument of cosine? You keep putting "..." in there, just write it out for me to help you.
  7. Feb 6, 2010 #6
    The argument is [tex]cos(kx -\omega t)[/tex]
  8. Feb 6, 2010 #7
    Hmm, still doesn't seem to change anything

    \frac{dz}{dt}=A\frac{d}{dt}\cos[kx-\omega t]=\omega A\sin[kx-\omega t]

    because you have a [itex]-\omega[/itex] coming from the argument and then a negative sign coming from the derivative of cosine.

    \frac{d^2z}{dt^2}=\omega A\frac{d}{dt}\sin[kx-\omega t]=-\omega^2A\cos[kx-\omega t]=-\omega^2z

    What is telling you your answer is wrong, your textbook, website, intuition, etc??
  9. Feb 6, 2010 #8
    It is a website. I considered the possibility of my syntax being wrong, but I went over it and couldn't find any error. It tells us to express the answer in terms of specific variables, which I did.

    I dont know. This isn't the first time the website has made a mistake, so I wouldn't be surprised if its happened again.
  10. Feb 6, 2010 #9
    Did you write it as [itex]-\omega^2z[/itex] or [itex]-\omega^2A\cos[kx-\omega t][/itex]? If you wrote the latter, I haven't a clue as to what the website would be asking you, unless you were given some constants to use?
  11. Feb 6, 2010 #10
    Copied and pasted from the site:

    "Express the vertical acceleration in terms of [tex]\omega, g, k, A[/tex], and the independent variables [tex]x[/tex] and [tex]t[/tex]."

    Keep in mind that [tex]\omega = \sqrt{g k}[/tex]

    I came up with a[tex]_{z} (x, t) = - g k Acos(kx - \omega t) [/tex] (as my officially submitted answer)

    which is equal to [tex]-\omega^{2} z(x, t)[/tex]
  12. Feb 6, 2010 #11
    That looks like the correct formula to me, I do not know why you would be getting it wrong.
  13. Feb 6, 2010 #12
    Thanks. All I needed was a secondary supporting opinion. I will take it up with the instructor then.
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