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Homework Help: Spring Oscillation

  1. May 17, 2008 #1

    Ive been working on some oscillation problems and i got a couple of them correct (im suspecting flukes) but i cant get past this one:

    The position of a mass that is oscillating on a spring is given by (17.5cm)cos[(11.0s^-1)t].
    What is the speed of the mass when t = 0.820 s?

    I believe im supposed to use V = dx/dt = -A*w*sin(w*t + fi)

    where A is amplitude, w is angular frequency t is time and fi is the phase constant. i think im really having a problem understanding how to get the phase constant.

    p.s. this is my first post and i'm only looking for a point in the right direction, not a completed solution. :D

  2. jcsd
  3. May 17, 2008 #2


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    The original equation shows that the phase constant is zero.
  4. May 18, 2008 #3


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    It is always true that
    [tex]v = \frac{dx}{dt}[/tex]
    so given the formula for x(t) you can just derive it and get the velocity.

    In the general case
    [tex]x(t) = A \sin(\omega t + \phi)[/tex]
    the derivative indeed becomes
    [tex]v(t) = -A \omega \cos(\omega t + \phi)[/tex]
    but you don't really need that here.
    What you could observe from that formula, is that the phase does not matter in the result (e.g. it doesn't come in the pre-factor or anything, it just stays inside the trig function to keep it synchronized with the movement x(t)).
  5. May 18, 2008 #4
    ok, should i be converting cm to m during my calculations? (im assuming the final answer should be in m/s)
  6. May 18, 2008 #5


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    It is alsways a good idea to convert units into the SI units, (eg metres, kilogramps, etc) and then convert the final answer into the units required - If a constant is involved in an equation, the units if the constant are normally SI Units, so if you don't convert them, your answer will be wrong.

    The phase constant is simply used if the initial displacement is not 0 at t=0

    Hope this all helps,

  7. May 18, 2008 #6
    ok, all done. Thanks alot. (mark as solved) 75.8 cm/s
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