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Frequency of a simple harmonic oscillator

  1. Feb 14, 2016 #1
    1. The problem statement, all variables and given/known data
    The problem is attached

    2. Relevant equations
    f=2π/ω=2π√(m/k)

    3. The attempt at a solution
    My idea is that the mass doubles resulting in a √2 increase in the equation above. However, apparently the answer is (c). I have a strong feeling the book answer is wrong, but I wanted to confirm.

    Thank you,
     

    Attached Files:

  2. jcsd
  3. Feb 14, 2016 #2

    Simon Bridge

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    I draw your attention to: in the second case, the masses are on opposite ends of the spring.
    Imagine this is all horizontal on a frictionless track ... imagine two masses on half-length springs, both springs attached to the same nail fixed to the track.
     
  4. Feb 14, 2016 #3
    You are saying that this is equivalent to cutting the spring in half. This would lead to each mass having the same frequency. But how would you calculate the frequency when the masses are on the same spring using the above formula?

    Also, I would argue that your analogy is different because the fixed end is not moving while the scenario in the problem has one end moving farther away from the other end meaning the spring force on each block increases over time quicker as the displacement from the springs resting position is increasing quicker meaning the frequency increases. If possible, could you tell me what is wrong with this way of thinking?

    Thank you,
     
  5. Feb 14, 2016 #4

    Simon Bridge

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    I cannot do all your work for you. I can only point is something like a useful direction. You have to put the bits together. In the two-mass situation, the center of the spring is stationary. Each mass moves away from or towards that center as if they were on a half-length spring.

    If you cut the spring in half, what properties change?
     
  6. Feb 14, 2016 #5
    The mass of the system decreases in half for each spring system while the spring constant remains constant. This means each spring has a frequency √2 less than the spring with two masses on each end. But the two springs with one mass is the same as the original scenario so the new two mass system has a frequency √2 larger
     
  7. Feb 15, 2016 #6

    Simon Bridge

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    If you cut a spring in half, the spring constant of the half-springs is not the same as the spring constant of the original.
    The mass on the end of each half-spring is the same as the mass on the end of the original.

    Another way to look at it:
    The first situation is the same as having a spring with a mass on each end, but one mass is held fixed.
    The second, neither mass is held fixed.

    If you prefer to use equations, you can always look up the equation for the frequency of 2-masses either end of a spring (or derive it).
     
  8. Feb 15, 2016 #7

    SammyS

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    That looks more like the formula for the period, rather than the frequency:
    T = 2π/ω=2π√(m/k)​
     
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