Simple harmonic motion on an incline

AI Thread Summary
The discussion centers on the calculation of equilibrium position for a block on an incline using simple harmonic motion principles. The user derived the equilibrium position as 0.075 m but questioned the solution's interpretation that this distance is from an initial position of 0.450 m. Clarification was provided that the 0.075 m represents the stretch of the spring from its unstretched length, not the distance from the incline's top. The importance of correctly applying Hooke's Law was emphasized to avoid confusion regarding displacement and equilibrium. Understanding these concepts is crucial for accurately solving problems involving forces and motion on an incline.
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Homework Statement


WpQ7NqB.png



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?
 
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PhizKid said:

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).
 
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