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Archived Period, amplitude, maximum speed, and total energy

  1. Dec 3, 2007 #1
    1. The problem statement, all variables and given/known data
    A 175g mass attached to a horizontal spring oscillates at a frequency of 2.0Hz. At one instant, the mass is at x = 5.0cm and has Vx = -20cm/s. Determine the following.

    a.) the period
    b.) the amplitude
    c.)the maximum speed
    d.)the total energy


    2. Relevant equations



    3. The attempt at a solution
    I solved part a by using T = 1/f = .5s. I'm not sure how I find the amplitude though. To find the maximum speed, I think I would just use Vmax = 2(pi)fA or (2 pi A)/T. I'm also not sure how I would find the total energy.
     
  2. jcsd
  3. Feb 8, 2016 #2

    gneill

    User Avatar

    Staff: Mentor

    A complete solution is offered:

    Let's take the parts in reverse order, beginning with the energy.

    Part (d): The total energy of the system
    We first determine the spring constant using the mass-spring frequency equation:

    ## f = \frac{1}{2 \pi} \sqrt{\frac{k}{M}}##

    ## k = 4 \pi^2 f^2 M = 27.64~N/m##

    The total energy of the system comprises kinetic and potential energy associated with the mass and the spring respectively.

    ##E = \frac{1}{2} M v^2 + \frac{1}{2} k x^2##

    where v is the velocity of the mass and x the displacement from equilibrium (stretch or compression of the spring). We are given the velocity and position (presumably a displacement from equilibrium) at a particular instant.

    ##M = 175~gm = 0.175~kg##
    ##v = -20~cm/s = -0.20~m/s##
    ##x = 5.0~cm = 0.050~m ##

    So that:

    ##E = \frac{1}{2} (0.175~kg) (-0.20~m/s)^2 + \frac{1}{2} (27.64~N/m)(0.050~m)^2##
    ##E = 0.0380~J##

    Part (c): The maximum speed
    We can use the expression for the total energy. The speed will be maximum when all the energy is due to the speed of the mass and the spring is at its equilibrium position with no stored energy. So:

    ##E = \frac{1}{2} M V_{max}^2##

    ##V_{max} = \sqrt{\frac{2E}{M}} = 65.9~cm/s##

    Part(b): The amplitude
    Again using the energy expression, the maximum amplitude occurs when the mass is stationary and the spring is at its maximum extension, so all the system energy is in the spring. So:

    ##E = \frac{1}{2} k x_{max}^2##

    ##x_{max} = \sqrt{\frac{2E}{k}} = 5.25~cm##

    Part(a) The period of oscillation
    This is simply the inverse of the frequency (2.0 Hz), so

    ##T = \frac{1}{f} = 0.50~s##
     
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