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Confused on an easy SHM problem.

  1. Dec 9, 2011 #1
    1. The problem statement, all variables and given/known data

    A 2.30 kg mass on a spring has displacement as a function of time given by the equation

    x(t) = (7.40 cm)cos[(4.16)t-2.42]

    a) Find the time for one complete vibration.

    The back of the book had this same equation but a different mass, so I was able to figure out the answer was 1.51 seconds that way.

    This was originally a 10 part question and I can do all the parts after a, but I can't find a.

    I tried plugging in "1.00" but that is wrong, and I used that for another part of the question. Equations such as T= 2pi*sqr(k/m) always have two variables, so I don't know what I am doing wrong.

    Please help. Thanks.

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 9, 2011 #2

    Stephen Tashi

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    You have to figure out how much time elapses when the argument (4.16)t - 2.42 of the cosine function changes by [itex] 2 \pi [/itex]. For example at t = 0, you have cos(-2.42).
    What would t be to make the value cos(-2.42 + [itex]2\pi[/itex]) ?
     
  4. Dec 9, 2011 #3

    cepheid

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    If you have some function x(t) that is periodic, then the period, T, is defined as the time interval such that:

    x(t + T) = x(t) (i.e. it is periodic)

    So, starting at any given time t, the period T is the time required for the value of the function to return back to the same value it had at time t. So, we want to figure out what value of T satisfies this for the particular function we've been given:

    x(t+T) = (7.40 cm)cos[(4.16)(t+T)-2.42] = x(t) = (7.40 cm)cos[(4.16)t-2.42]

    It must be true that cos[4.16(t+T) - 2.42] = cos(4.16t - 2.42).

    For simplicity, let's pick t = 0 as our example starting value. After T seconds, we want the function to have the same value that it had at t = 0. Therefore, it must be true that:

    cos(-2.42) = cos(4.16T - 2.42).

    Well, this is true if 4.16T is a multiple of 2pi, since after 2pi radians, the phase of the oscillation (which is the thing you're taking the cosine of) will have gone through one complete cycle, and you'll be back to the same phase angle you started at (-2.42 radians in this case). Therefore, for the first repetition:

    4.16T = 2pi

    T = 2pi/4.16

    So we learned something interesting: the factor which multiplies t in the cosine (which we call the angular frequency ω and which is equal to 4.16 rad/s in this example) is related to the period by the expression T = 2pi/ω. This makes sense, because ω is the rate at which you accumlate "phase" with time, and after one period T, you have accumulated a full cycle's worth of phase, hence ωT = 2pi. So, you can tell at a glance (just by looking at the SHM function) what ω is, and therefore what T is.

    For a spring, it turns out that ω = √(k/m), but you didn't need to use that information here, since you were already given ω.
     
  5. Dec 9, 2011 #4


    Thank you very much for your informative and helpful feedback. You also cleared up any confusion I had on the phases.
     
  6. Dec 9, 2011 #5
    That does make sense, thanks for your help.
     
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