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

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SUMMARY

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, is calculated using energy conservation principles. The relevant equation is (1/2) k x² = (1/2) m v², where k is the spring constant, x is the displacement from equilibrium, m is the mass, and v is the velocity. At equilibrium, the maximum velocity occurs, confirming that the equation applies when referencing displacement from the equilibrium position. The correct answer to the problem is 4 m/s (option B).

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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
 
Last edited:
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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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