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S.H.M for mass spring system

  • Thread starter repugno
  • Start date
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I've been trying this question for quite some time and have given up.

Prove that a mass,M , suspended from a fixed point by a helical spring, which obeys hooke's law, undergoes simple harmonic motion when it is displaced vertically from its equilibrium position.

Say I displace the mass spring upwards, e metres. Then I can say:

Mg + ke = Ma ?

When the mass is at far bottom then

kx - Mg = -Ma ?

When I equate these two I get kx=-ke, which proves nothing. :yuck:

Thanks
 

HallsofIvy

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What do you have to use? You certainly can't do it by looking at kinetic energy not at a few positions. I would do it by setting up the differential equation governing the motion and showing that the solution is harmonic.
 
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The book gives the answer very vaguely which confuses me more. It does not have any calculus involved..

k(e + x) - Mg = -Ma

I have no Idea how they've derived this

Then they say 'in the equilibrium position, the pull of the spring ke is equal to the weight'

ke = Mg

But when the mass is at the equilibrium position the force on the mass by the spring is 0

Subsittuting above..

Mg + kx - Mg = -Ma
kx=-Ma

a=-(k/M)x

Compare with a=-(omega)^2x

Doing some math it also shows that T = 2(pi)(M/k)^(1/2)

Thanks for your help
 
551
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repugno said:
Then they say 'in the equilibrium position, the pull of the spring ke is equal to the weight'

ke = Mg

But when the mass is at the equilibrium position the force on the mass by the spring is 0
The spring still has a tension, even in equilibrium. It's the sum of the forces on the mass that is 0, which is the same as that first statement.
 
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Nylex said:
The spring still has a tension, even in equilibrium. It's the sum of the forces on the mass that is 0, which is the same as that first statement.
Thanks, it's just occured to me that I've been looking at the situation the wrong way. :zzz: . problem solved
 

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