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Finding the phase angle in simple harmonic motion

  1. Oct 25, 2005 #1

    A frictionless block of mass 2.35 kg is attached to an ideal spring with force constant 310 N/m. At t=0 the spring is neither stretched nor compressed and the block is moving in the negative direction at a speed of 12.6 m/s.

    1. Find the amplitude.

    I got this without a problem. I used the formula A = square root of [tex]x_i^2[/tex] + [tex]v_i^2[/tex]/[tex]\omega^2[/tex]. [tex]\omega = \sqrt{k/m} = 11.49[/tex] rad/s. Plugging in known values results in 1.10 m, which is correct.

    2. Find the phase angle.

    Here's where I don't know why my answer is not correct.

    I use the equation [tex]v_i = -\omega * A * sin(\omega*t + \phi)[/tex]. I know the initial velocity, I know the angular frequency, and I know the amplitude. I'm solving for the angle in radians.

    [tex]-12.6 = -(11.49)(1.10)(sin(\phi)[/tex] This is at time t, so [tex]\omega * t[/tex] = 0.

    I get [tex]\phi = 85.5 ^\circ[/tex]. Converting it into radians, it's 1.49 (rad).

    This isn't right, though.
    Last edited: Oct 25, 2005
  2. jcsd
  3. Oct 25, 2005 #2

    Doc Al

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    Staff: Mentor

    If your numbers were computed a bit more accurately, you'd get:
    [tex]-12.6 = - (12.6) \sin \phi[/tex], or
    [tex] 1 = \sin \phi[/tex], thus [tex]\phi = \pi /2[/tex] (radians)
    Last edited: Oct 25, 2005
  4. Oct 25, 2005 #3
    My response is, "My answer is off by an additive constant."

    I'm solving for a phase angle here, [tex]\phi[/tex]. What additive constant could possibly exist?
  5. Oct 25, 2005 #4

    Doc Al

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    Staff: Mentor

    I messed up the sign in my previous post (D'oh!); your first answer was close. (See my correction.)
  6. Oct 25, 2005 #5
    Thanks. My idea of direction was screwed up, but it didn't matter for the first part of the question because velocity is squared.

    Round once too many and the error becomes greater than two percent. Argh.
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