# Simple harmonic motion on an incline

1. Feb 4, 2013

### PhizKid

1. The problem statement, all variables and given/known data

2. Relevant equations
F = -dU/dx

3. The attempt at a solution
$U = \frac{1}{2}kx^2 + mgxsin\theta \\\\ F = -(kx + mgsin\theta) \\\\ F = -kx - mgsin\theta \\\\$

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

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

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?

2. Feb 4, 2013

### SammyS

Staff Emeritus
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).