What Is the Speed of a Mass at Equilibrium in Simple Harmonic Motion?

AI Thread Summary
The discussion revolves around calculating the speed of a 1.0 kg mass at the equilibrium position in simple harmonic motion, attached to a vertical spring with a force constant of 400 N/m and an amplitude of 10 cm. Participants clarify that at equilibrium, the acceleration is zero and the velocity is at its maximum. The conservation of energy equation is applied, where the initial energy equals the final energy, specifically referencing the position relative to the equilibrium point. It is confirmed that the equation is valid for vertical springs as long as the position is measured from the equilibrium position. The conversation concludes with an understanding that the principles of energy conservation apply in this context.
Soniteflash
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Homework Statement


A 1.0 Kg mass is attached to the end of a vertical ideal spring with a force constant of 400 N/m. The mass is set in simple harmonic motion with an amplitude of 10 cm. The speed of the 1.0 kg mass at the equilibrium position is.

A. 2 m/s
B. 4 m/s
C. 20 m/s
D. 40 m/s
E. 200 m/s

Homework Equations


EI= EF

The Attempt at a Solution


I know that at equilibrium the acceleration is zero and that velocity is max.
So using conservation of energy I thought this:
EI= EF
(1/2) k x2 = (1/2) m v2

I have a feeling that this equation does not apply because the spring is vertical
For vertical springs all I know is that the equilibrium position at which the mass simply hangs and does not move.

So at equilibrium the the amount of force required to keep it in equilibrium is
k yeq = mg
 
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The equation does apply as long as you are referencing x to the equilibrium position.
 
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paisiello2 said:
The equation does apply as long as you are referencing x to the equilibrium position.
Ah, Ok that makes sense then.
 
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