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Homework Help: Time period for a Simple Harmonic Oscillator to go from 0-1m

  1. Jan 31, 2016 #1
    1. The problem statement, all variables and given/known data

    A particle with a mass(m) of 0.500kg is attached to a horizontal spring with a force constant(k) of 50.0N/m. At the moment t=0, the particle has its maximum speed of 20m/s and its moving to the left. Find the minimum time interval required for the particle to move from position x=0 to x=1.00m.

    2. Relevant equations
    General Equation of Motion: x(t)=(A)cos(ωt +Φ) where A is the amplitude or maximum position, ω is the angular frequency, and Φ is the phase constant

    3. The attempt at a solution
    ω=√(k/m)=√(50/0.5) = 10.0 rad/s
    vmax=ωA → A=(vmax/ω)=20/10=2.00m

    Since at t=0, the particle is at its maximum speed, the phase constant must make cosine=0 as the particles maximum speed is at position x=0, so the phase constant must be π/2.

    Equation of Motion: 2cos(10t+π/2)=x(t)

    I believe the solution from here it to make x(t)=1.00m as the particle is at x=0 at t=0, so the time at x(t)=1.00m is the time it takes to go from x=0 to x=1.00.

    2cos(10t+π/2)=1.00 → cos-1(1/2)=10t+π/2 → 10t= π/3 - π/2 → t= -π/60=-0.0524s.

    I believe the answer is just the absolute value of this. When I input t=0.0524s, i get x(t)=-1.00. Shouldn't this be equivalent to the time it takes to get to +1.00m.

    Apparently, the answer is 0.105s which is noticed is double the magnitude of my answer. I am pretty sure this is not a coincidence. It would be greatly appreciated if someone could tell me where I am going wrong.

    Thank You,
  2. jcsd
  3. Feb 1, 2016 #2


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    The way I interpret the problem statement, you don't want to start the "timer" (so to speak) at the moment [itex] t = 0 [/itex]. That's because at time [itex] t = 0 [/itex], the particle is moving to the left. The way I interpret the problem, it's okay to wait until the particle comes back to [itex] x = 0 [/itex] such that it's moving to the right; only then begin the time interval. But I suppose your approach should work too, leveraging symmetry.

    I think your logic is valid due to symmetry. I might have instead tried to used different points in time such that I didn't have to take an absolute value, but that's just me.

    All that said, I don't think the given answer is correct. I came up with the same answer that you did. Either we are right or we are misinterpreting the problem statement.
  4. Feb 1, 2016 #3


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    I also believe Ethan's answer is correct.

    If you are familiar with the idea that SHM is just a projection of uniform circular motion, then you can see that when the particle in SHM moves from x = 0 to x = 1 m, the point on the reference circle moves through 30 degrees, or 1/12 of a circle. So, the time required is 1/12 of the period of the SHM.
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