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Homework Help: Finding the amplitude in simple harmonic motion

  1. Dec 13, 2011 #1
    1. The problem statement, all variables and given/known data

    Consider a block-spring system that forms a linear simple harmonic oscillator, with the block undergoing SHM. The angular frequency is 9.78 rad/s, the frequency is 1.6 Hz and the period is 640 msec. The block is pulled from its equilibrium position at x=0 (at time = 0) to x=11 cm. It is released from x=11cm. What is the amplitude.

    2. Relevant equations

    I tried using x(t) = A[cos(ωt + ∅)]

    3. The attempt at a solution

    Ok, so I know that the answer this question is intuitively 11 cm, since the block can never go further than 11 cm from the equilibrium point. However, I thought I'd try testing my shaky understanding of the x(t) = A[cos(ωt + ∅)] formula and I hit a brick wall.

    I figured that it would take one quarter of a period for the block to get from x=11 cm to x = 0 cm, since that's one fourth of the journey through the period. So, I plugged the following numbers in:

    .11 m = (A)cos[(9.78rad/s)(.643sec/4) + 0]
    A = -82.17 m

    I don't understand why this method doesn't yield A = 11 cm (or anything remotely close to it).
    Last edited: Dec 13, 2011
  2. jcsd
  3. Dec 13, 2011 #2


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    Staff: Mentor

    A quarter period from maximum extension (t=0) should make the x(t) zero, not A.

    A half period after t=0 should make it -A...
  4. Dec 13, 2011 #3
    Your initial observation was good; using this equation is overkill. But, using it:

    The block starts at the farthest point, so the phase constant is zero. So

    x(t) = A cos(ωt)

    Now, if you wanted to find the position at time t = 0, you'd have

    x(0) = A cos(ω0) = A cos(0)

    They told you that x(0) = 11 cm, and cos(0) is just one. So A is 11 cm, as expected (and as you saw from the beginning; using this equation is completely unnecessary).

    You tried finding the position after a quarter of a period, where the block is back at the equilibrium point. First off, that can't tell you anything about A, because at that time, x = 0. You end up with an equation like

    0 = A cos(blah)

    where the cos(blah) also turns out to be zero. 0 = 0A is true for any value of A.

    When you did it, you messed up the left hand side. x is not 0.11 m one fourth of the way through the cycle; it must be 0. The reason you have such a strange answer is because, at this time, the value of the cosine is zero. Due to rounding errors, you get something really tiny like 0.005; dividing 0.11 by this gives you a huge answer.
  5. Dec 13, 2011 #4
    Thanks! That makes sense!
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