Simple harmonic motion on an incline

[PhizKid]

Homework Statement

Homework Equations

F = -dU/dx

The Attempt at a Solution

$U = \frac{1}{2}kx^2 + mgxsin\theta \\\\<br /> F = -(kx + mgsin\theta) \\\\<br /> F = -kx - mgsin\theta \\\\$

We want to set the force = 0 because that's when the block is in equilibrium with no forces acting on it.

$0 = -kx - mgsin\theta \\\\<br /> x = -\frac{mgsin\theta}{k} \\\\<br /> x = -\frac{\frac{14.0}{g}gsin\theta}{k} \\\\<br /> x = -\frac{14.0sin(40 deg)}{120} \\\\<br /> x = -0.075$

So since 0.075 m is the equilibrium position, it is the distance from the top of the incline to the equilibrium position, but the solution says that 0.075 m is the displacement from the position 0.450 m to equilibrium. I didn't even include 0.450 m in the equation, how can I assume that x is the distance from 0.450 to 0.075? Why is it not some arbitrary position to 0.075?

 

[SammyS]

Homework Statement

[ img]http://i.imgur.com/WpQ7NqB.png[/PLAIN]

Homework Equations

F = -dU/dx

The Attempt at a Solution

$U = \frac{1}{2}kx^2 + mgxsin\theta \\\\<br /> F = -(kx + mgsin\theta) \\\\<br /> F = -kx - mgsin\theta \\\\$

We want to set the force = 0 because that's when the block is in equilibrium with no forces acting on it.

$0 = -kx - mgsin\theta \\\\<br /> x = -\frac{mgsin\theta}{k} \\\\<br /> x = -\frac{\frac{14.0}{g}gsin\theta}{k} \\\\<br /> x = -\frac{14.0sin(40 deg)}{120} \\\\<br /> x = -0.075$

So since 0.075 m is the equilibrium position, it is the distance from the top of the incline to the equilibrium position, but the solution says that 0.075 m is the displacement from the position 0.450 m to equilibrium. I didn't even include 0.450 m in the equation, how can I assume that x is the distance from 0.450 to 0.075? Why is it not some arbitrary position to 0.075?

The solution you refer to is correct. The 0.075 m is the amount the spring is stretched from its unstretched length.

Check Hooke's Law again.

It may say $\ \ F_\text{Spring}=-k(x - x_0)\,, \$ where x is the length of the spring, and x₀ is the unstretched length.

Or it may say $\ \ F_\text{Spring}=-k\,x\,, \$ where x is the amount the spring is stretched (from its unstretched length).