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Write expressions for simple harmonic motion

  1. Nov 11, 2008 #1
    1. The problem statement, all variables and given/known data
    Write expressions for simple harmonic motion (a) with amplitude 10 cm, frequency 5.0 Hz, and maximum displacement at t=0; and (b) with amplitude 2.5 cm, angular frequency 5.0 1/s, and maximum velocity at t=0.

    2. Relevant equations

    x(t) = A\cos (\omega t + \varphi ) \hfill \\
    \omega = 2\pi f \hfill \\
    f = \frac{1}
    {T} \hfill \\

    3. The attempt at a solution


    A=10 cm
    f=5.0 Hz

    Since the amplitude equals the max displacement at a given t(in this instance t=0), this tells us that the phase angle is zero. So our equation should be....?

    x(t) = (10cm)\cos \left[ {(10\pi s^{ - 1} )t} \right]

    A=2.5 cm
    w=5.0 s^-1

    V(x) = - A\omega \sin (\omega t + \varphi ) \hfill \\
    V(0) = A\omega = V_{\max } \hfill \\
    V(0) = - A\omega \sin (\varphi ) \hfill \\
    A\omega = - A\omega \sin (\varphi ) \hfill \\
    - 1 = \sin (\varphi ) \hfill \\
    \varphi = \tfrac{{3\pi }}
    {2} \hfill \\


    x(t) = (2.5cm)\cos \left[ {(5.0s^{ - 1} )t + \tfrac{{3\pi }}
    {2}} \right]

    Do these look right? Here is what my answer book gives:

    (a):x(t) = (10cm)\cos \left[ {(\pi s^{ - 1} )t} \right] \hfill \\
    (b):x(t) = (2.5cm)\sin \left[ {(5s^{ - 1} )t} \right] \hfill \\

    Any help is appreciated, thank you.
  2. jcsd
  3. Nov 11, 2008 #2
    At a glance, your part (a) should be right considering the frequency given. Your part (b) is equivalent to the answer they state.

    A phase shift corresponds to a shift in where the peaks occur. You can consider then, that the phase shift can shift a cosine into a sine or a sine into a cosine. In this sense,

    cos(x + \frac{3\pi}{2}) = sin (x)
  4. Nov 11, 2008 #3
    Thanks Coto. I had a feeling that the part a solution key was incorrect.
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