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Simple Harmonic Motion  Mass on a Spring 
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#1
Feb614, 01:17 PM

P: 101

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.



#2
Feb614, 02:10 PM

P: 1,970

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=Fmg= k(x_o+x) mg= kx_o+kxmg = kx. So the net force depends only on the displacement from the equilibrium position. And this is the restoring force. 


#3
Feb614, 06:22 PM

Sci Advisor
Thanks
PF Gold
P: 12,170

I can see your confusion but you just need to relate the physical situation to the maths describing it. 


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