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Graph of simple harmonic motion

  1. Aug 22, 2010 #1
    1. The problem statement, all variables and given/known data

    A particle moves with amplitude A and period T (see figure). Express the following in terms of A and T and numberical constants.
    a) The time at which the phase is pi/2
    b) maximum v
    c) max a
    d) time at which acceleration is a maximum

    2. Relevant equations



    3. The attempt at a solution
    I got b and c, but i'm not sure about a and d.

    For a, they ask for the time at which phase is pi/2. I know T = 2pi. For d, when acceleration is max, is time 0?

    Can someone help me please?

    THANKS
     

    Attached Files:

  2. jcsd
  3. Aug 22, 2010 #2
    T does not equal 2pi

    [tex]\omega T = 2 \pi[/tex]
    and
    [tex]\omega = \frac{2\pi}{T}[/tex]
    where omega is the radian frequency, because when you evaluate the sinusoid at t = T, the argument equals 2pi.

    [tex]x(t) = cos(\omega t)[/tex]
    [tex]x(t) = cos(\frac{2\pi}{T}t)[/tex]

    Using this last equation, we can solve both a and d. For a, find a t that makes the argument pi/2. For d, differentiate twice to find acceleration and find the argument that maximizes acceleration.

    you can use a more conceptual approach for both, however. For example, if you know how a cosine behaves, you can instantly know A just by looking at the graph.
     
  4. Aug 22, 2010 #3
    I'm not getting what you mean by finding a t that makes it pi/2. We don't have any values of t
     
  5. Aug 22, 2010 #4
    for a, time has to be 1/4t.

    is that right?
     
  6. Aug 22, 2010 #5
    differentiating the equation twice gives acceleration = -A omega^2 cos (omega t)

    amax = Aomega^2, so t has to be equal to 1?
     
  7. Aug 23, 2010 #6
    no. the time at which a phase of pi/2 happens, t, will have T in it.
    no. wt must equal 0 or pi for |a(t)| to be maximum.
     
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