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Particle in Simple Harmonic Motion

  1. Feb 28, 2013 #1
    1. The problem statement, all variables and given/known data
    Not exactly sure why a time value of 0.500s is given, but I am positive it is why my answer isn't correct:

    Q. a 1kg object is attached to a horizontal spring. The spring is initially stretched by 0.100m and the object is released from rest there. It proceeds to move without friction. The next time the speed of the object is zero is 0.500s later. What is the maximum speed of the object?

    2. Relevant equations
    Vmax = Aω
    ω=√(k/m)
    Fs=-kx


    3. The attempt at a solution
    Given:
    t = 0.5s
    Amplitude: 0.100m
    m: 1kg

    I can find k:
    Fs=-kx
    mg = -kx
    (1kg * -9.8) = -k (0.100m)
    k = 98 N/m

    Find ω
    ω=√(k/m)
    = √(98/1)
    = 9.90

    Vmax = Aω
    =(0.1)(9.90)
    =0.990 m/s.

    However, the answer is 0.628m/s. I don't understand the purpose of "the next time the velocity is zero is 0.5s later" I interpret that as:

    "it to 0.5seconds to reach negative amplitude", which to my understanding isn't needed?
     
  2. jcsd
  3. Feb 28, 2013 #2
    wait I see my problem, it is moving horizontally not vertically so mg = -kx doesn't apply here.
    v = m / s
    = 0.2m / 0.5s
    = 0.4 m/s

    don't know how to go about finding acceleration from here.
     
  4. Feb 28, 2013 #3
    Think. If the next time the speed is zero is in 0.5 second, then what is the frequency of the oscillation?
     
  5. Feb 28, 2013 #4
    to anyone wondering in the future....

    the mass was actually the unneeded info:
    ω=2∏f

    frequency is = # cycles / s
    = 1 hz in this situation

    then use Vmax

    SOLVED. Thanks.
     
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