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B Simple harmonic motion springs

  1. Dec 28, 2017 #1

    Jas

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    I have a spring with mass M attached, and leave it at equilibrium. Then I displace it some more by stretching it down a bit more. Displacement due to the mass= X, displacement due to stretching it even more=Y.

    Why isn't the amplitude of oscillation= X+Y, but is only actually only Y? This is really confusing me!

    Relevant equations: (L is the natural length of the spring, and x is the extension, λ is the modulus of elasticity
    Tension=λx/L
    Energy stored=λx^2/2L
     
  2. jcsd
  3. Dec 28, 2017 #2

    PeroK

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    What if ##Y = 0##?
     
  4. Dec 28, 2017 #3

    Jas

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    Then it would just stay in equillibrium
     
  5. Dec 28, 2017 #4

    PeroK

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    But you would have it oscillate with amplitude ##X+ Y = X##.
     
  6. Dec 28, 2017 #5

    sophiecentaur

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    The nature of SHM is that the restoring force is proportional to the displacement from the equilibrium (or rest) position. Y will be positive or negative, depending on whether the mass is above or below position X. We, of course, assume that the k of the spring is the same over the whole range of positions of the mass.
    Here's another idea. Assuming the spring is partly open went unloaded (easy to arrange by giving it a good stretching!) Then imagine the spring and mass are laid on a frictionless horizontal surface ( dry, clean ice). The rest position of the mass will now be different, with respect to the fixing of the spring but the restoring force per cm of displacement will still be the same. So the period will be exactly the same as when the mass is hanging down. The amplitude will still be the maximum displacement relative to the rest position.
     
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