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Simple Harmonic Motion of a cart

  1. Nov 19, 2007 #1
    1. The problem statement, all variables and given/known data
    A 0.620-kg cart is moving down an air track with a speed of 2.33 m/s when it collides with a spring. The spring is initially at its equilibrium position, but the cart compresses the spring. It takes 0.0780s after the cart hits the spring for the cart to come to a stop. The cart then remains attached to the spring and oscillates back and forth in simple harmonic motion.
    (a) What is the period of this oscillation?
    (b) Find the force constant of the spring.
    (c) What is the amplitude of the oscillation?
    (d) What is the maximum acceleration experienced by the cart as it oscillates?
    (e) How far is the cart from the spring’s equilibrium position when it is moving with a speed of 1.00m/s?

    2. Relevant equations
    x = Acos([tex]\omega[/tex]t)
    v = −A[tex]\omega[/tex]sin([tex]\omega[/tex]t)
    a = −A[tex]\omega[/tex]^2cos([tex]\omega[/tex]t)

    3. The attempt at a solution
    I don't know how to find the period without angular velocity, and I'm not sure how to find the angular velocity with the information I have. I did find the force constant to 101.8N/m using conservation of energy, which then allowed me to find the amplitude to be 0.182m. But the rest have me confused because I'm not sure how to find angular velocity with the information I have. I could determine it with the force constant since [tex]\omega[/tex]=[tex]\sqrt{}k/m[/tex] but I'm supposed to find period without knowing the force constant
     
  2. jcsd
  3. Nov 20, 2007 #2
    Are you sure k is correct? How did you find it? I have no calculator to hand (don't like the compueter one...and it's late, but I don't think 101.8 is correct, but I may be doing it wrong in my head - it is VERY late!)

    The mass hits the spring in equilibrium position and then comes to a stop 0.0780 seconds later...what fraction of an oscillation does that represent?

    Then use the period to calculate the spring constant...

    Hope that helps
     
  4. Nov 20, 2007 #3


    It takes 0.0780s after the cart hits the spring for the cart to come to a stop.Then it takes 0.0780s after the cart stops for the cart to come back.It's half of one complete oscillation.
     
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