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USAPhO 2009 F=MA exam #13, (spring SHM)

  1. Jan 25, 2014 #1
    1. The problem statement, all variables and given/known data

    Late correction, this is the 2008 F=ma exam: http://www.aapt.org/Programs/contests/upload/olympiad_2008_fnet_ma.pdf

    A mass is attached to the wall by a spring of constant k. When the spring is at its natural length,
    the mass is given a certain initial velocity, resulting in oscillations of amplitude A. If the spring
    is replaced by a spring of constant 2k, and the mass is given the same initial velocity, what is the
    amplitude of the resulting oscillation?

    2. Relevant equations
    Uh... This is the problem here. I might just be forgetful, but I cannot recall any equation relating amplitude and f, T, k, or v.

    T = 2∏√(m/k) is the most relevant formula I could recall.

    3. The attempt at a solution

    Intuition led me to choose the correct answer: (1/√2)A

    I know if k is doubled, the amplitude will decrease. I had a feeling there is a relationship k ∝ A^2 given a constant v_initial and mass. Can anyone shed some light on this?
  2. jcsd
  3. Jan 25, 2014 #2


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    Think about KE and PE. What did doubling the spring constant do to the total energy of the system?
  4. Jan 26, 2014 #3
    PE = (kx2)/2

    Doubling k doubles the PE at any point x.

    Thank you. But now I'm trying to think of an equation relating PE and Amplitude and I can't think of anything.
  5. Jan 26, 2014 #4
    What is amplitude?
  6. Jan 26, 2014 #5
    Voko, I don't know what you're asking. If you want the definition of amplitude, it is the distance (in this case) from equilibrium to the maximum compression/ extension of the spring.

    I'm still thinking of an equation relating PE with amplitude.
  7. Jan 26, 2014 #6
    So amplitude has something to do with distance. So does PE. Can you connect the two?
  8. Jan 26, 2014 #7
    Oh wow! I must have been suffering from a massive brain fart.

    PE = (kx2)/2

    Given the same kinetic energy (v stays constant), and a doubled spring constant, PE1 = PE2 = (2k(x/√2)2)/2

    x2 = (1/√2)A
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