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SHM - increase in mass?

  1. Jun 15, 2006 #1
    Hi everyone,

    I'm given a graph of the object's displacement and frequency and the object is executing SHM. The question asks what would the graph look like when the mass of the object is increased to lower the amplitude of the oscillation.

    Personally I think that the natural frequency would be reduced besides the decrease in amplitude because of additional damping?

    Why would the increase in mass decrease the amplitude?

    Thanks!
     
  2. jcsd
  3. Jun 15, 2006 #2
    An increase in mass during oscillation would increase the inertia and hence decrease the amplitude. Do not confuse it with damping in the literal sense.

    Mathematically the total energy of oscillation would decrease when you add mass impulsively (assuming of course that the mass increase is done in a manner not to impart momentum along the direction of motion). Whether the amplitude gets affected or not depends on when you add the additional mass (if it is add in discrete steps): at zero displacement or maximum displacement.
     
  4. Jun 15, 2006 #3
    Thanks Maver. Is it to say then, assuming driving force to be the same, for the curve given (displacement against Freq) for an object of mass 2m executing SHM would be of the same as one that has mass m, except that it would have a lower amplitude?

    How would the natural frequency of mass 2m comapre with m? Thanks a lot. The equations I have here doesn't mention anything about mass.

    Thanks again!
     
  5. Jun 15, 2006 #4

    HallsofIvy

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    The differential equation governing "Simple Harmonic Motion" is "mass times acceleration equals force":
    [tex]m \frac{d^2x}{dt^2}= -kx[\tex]
    The simplest example is a spring with spring constant k, no damping, and "natural length" 0. The general solution to that is, of course,
    [tex]x(t)= C cos(\sqrt{\frac{k}{m}}t)+ D sin(\sqrt{\frac{k}{m}}t)[/tex]
    The period is
    [tex]T= 2\pi\sqrt{\frac{m}{k}}[/tex]
    and the frequency is the reciprocal of that
    [tex]\lambda= \frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]
    which certainly does depend on mass, m.

    The simplest situation is to "stretch" the spring a fixed amount and release it from rest: the initial values x(0)= x0, x'(0)= 0. But then it is easily seen that the amplitude (what you seem to be calling "displacement") is just that initial displacement x0.

    Instead, consider the situation in which the mass is intially at x= 0 but has speed v0: the initial conditions x(0)= 0, x'(0)= v0.
    In that case we have x(0)= C= 0 and x'(0)= D[itex]\sqrt{\frac{k}{m} }= v_0[/itex] so [itex]D= v_0\sqrt{\frac{m}{k}}[/itex]. Now
    [tex] x(t)= v_0\sqrt{\frac{m}{k}}sin(\sqrt{\frac{k}{m}}t)[/tex]
    The frequency is still
    [tex]\lambda= \frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]
    and the amplitude is
    [tex]v_0\sqrt{\frac{m}{k}}[/tex]
    You will notice that the frequency is inversely proportional to the mass: if the mass doubles, then the frequency is divided by [itex]\sqrt{m}[/itex], the object moves more slowly.
    However, amplitude is directly proportional to the mass: if the mass doubles, then the amplitude is multiplided by [itex]\sqrt{m}[/itex]- the heavier object will go farther against force.
     
  6. Jun 16, 2006 #5
    Thanks a lot for the help here. Just want to make sure, then for the curve shape of the mass 2m in the graph of amplitude against driving frequency would be of a higher amplitude than the curve of mass m, and natural frequency of mass 2m is less than of m?

    Is this correct?
     
    Last edited: Jun 16, 2006
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