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For an undamped mass, spring system subject to a harmonic force, the equation is: ma+kx=Fsin(ωt)

where a=d^2/dx^2

After solving the diff eqns, the steady state amplitude is:

X=(F/m)/(ω^2-ω0^2)

where ω is the frequency and ω0 is the natural frequency =sqrt(k/m)

according to the amplitude equation, the amplitude will increase if you increase the stiffness.

I am struggling to understand why this is true, as i understand it, the spring should be resisting the motion of the mass and hence the stiffer the spring the less it moves. Think about if you increase the stiffness to infinity, equivelant to placing the mass on a rigid floor, the displacement of the mass should surely reduce to zero. Am I missing something?

Anyone able to explain?

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# Spring Stiffness and Amplitude, not understanding

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