- #1

Jalhalla

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## Homework Statement

An object of mass

*m*= 50 kg is placed on top of 10 vertical springs. After this, the springs come to a new equilibrium position, 0.05 m below the original one, so

*x*eq = –0.05 m. The springs are stretched upward 0.15 m from the equilibrium position. The spring is released after being stretched, and simple harmonic motion takes place, meaning that, during each "bounce"—up and down—the mass returns to the same maximum height

*x*=

*x*eq +

*s*= –0.05 m + 0.15 m = 0.1 m, and the same minimum height

*x*´. Determine the force constant

*k*of each spring, and then determine the minimum height

*x*´ of the mass in order for simple harmonic motion to occur.

## Homework Equations

*F*x =

*kx*

*F*g =

*mg*

*F*x´ =

*kx*´

## The Attempt at a Solution

At the top of its motion,

*F*x =

*F*g

10

*kx*=

*mg*

*k*= (

*mg*)/10

*x*= [(50 kg)(9.8 N/kg)]/[(10)(0.1 m)] = 490 N/m

[Did I disclude any negatives that should have been present?]

At the bottom of its motion,

*F*x´ = –

*F*g

10

*kx*´ = –mg

*x*` = (–

*mg*)/(10

*k*) = [–(50 kg)(9.8 N/kg)]/[10(490 N/m)] = –0.1 m

So,

*x*= –

*x*´—did I do this correctly?

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