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Simple Harmonic Motion - Mass on a Spring

by MohammedRady97
Tags: harmonic, mass, motion, simple, spring
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MohammedRady97
#1
Feb6-14, 01:17 PM
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For a mass on a spring (vertical set up) undergoing SHM, we equate the restoring force, -kx, to -ω^2 x, coming to a conclusion that ω = [itex]\sqrt{\frac{k}{m}}[/itex]. My question is, is the restoring force |mg - T| Where T is the tension in the spring? Because this seems to be the net force. I am used to equating tension to kx, not the net force.
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nasu
#2
Feb6-14, 02:10 PM
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The oscillation is around the equilibrium position.
In the equilibrium position the spring is stretched by an amount x_o, just enough for the elastic force to be equal to the weight.
So we have
kx_o=mg.

Now if we move it a little from this equilibrium position, let say by pulling it down, the spring will be stretched by an extra amount, x. x is measured from the equilibrium position.
So the elastic force will be
F=k(x_o+x)
and the net force will be
F_net=F-mg= k(x_o+x) -mg= kx_o+kx-mg = kx.
So the net force depends only on the displacement from the equilibrium position. And this is the restoring force.
sophiecentaur
#3
Feb6-14, 06:22 PM
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Quote Quote by MohammedRady97 View Post
For a mass on a spring (vertical set up) undergoing SHM, we equate the restoring force, -kx, to -ω^2 x, coming to a conclusion that ω = [itex]\sqrt{\frac{k}{m}}[/itex]. My question is, is the restoring force |mg - T| Where T is the tension in the spring? Because this seems to be the net force. I am used to equating tension to kx, not the net force.
That looks to be along the right lines to me, except why do you have a Modulus sign there? The restoring force needs to be given a sign to tell you which way it acts. At equilibrium, T = mg. and the equilibrium position is not directly related to the actual spring length. Oscillation is about this position and the tension will increase of decrease according to the 'restoring force'.
I can see your confusion but you just need to relate the physical situation to the maths describing it.


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