Oscillations of a Block Mass attached to a Spring on an Incline

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Homework Statement

If the block is pulled slightly down the incline and released, what is the period of the resulting oscillations?

Relevant Equations

See images.

Why doesn't the incline angle play a role in changing the $m$ component of this equation?
$T = 2π\sqrt{\frac{m}{k}}$

FOR QUESTION 25, PART B:

ANSWER:

 

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Is it because even the spring is dealing with gravity on an incline?
Both, the mass and the spring are dealing with the same gravity/g, so, the equation/ratio remains the same.

 

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Compare a horizontal spring to a vertical spring with mass. If there is no difference here, then there will be no difference for intermediate angles.

For a horizontal spring, the equation of motion is given by:

$m\frac{d^2x}{dt^2}=-kx \tag{1}$

Where x is the displacement from the equilibrium position, which is the unloaded spring length, and m is the mass of the object. Now if we have a vertical spring the equation of motion is given by:

$m\frac{d^2y}{dt^2}=-ky - mg \tag{2}$

Where y displacement from the unloaded spring length. The new equilibrium position is found by force balance with zero acceleration:

$0=-ky_{eq} - mg \rightarrow y_{eq} = -mg/k \tag{3}$

So if we plug 3 into the right hand side of 2, we get:

$-ky - mg = -k(y-mg/k) = -k(y-y_{eq}) \tag{4}$

Now using 4 in 2 gives:

$m\frac{d^2y}{dt^2}=-k(y-y_{eq})\tag{5}$

And with a change in variables of $x=y-y_{eq}$ in 5, we have an expression identical to 1.

So, it seems that the only influence the weight of the object has is to change the equilibrium position, but that it doesn't change the equation of motion. A linear spring always oscillates about the equilibrium position, not the natural length of the spring. In the horizontal case, the equilibrium position happens to be the natural length of the spring, but not in the vertical (or inclined) case. Since the equations of motion are the same, the periods are the same.

 

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Brilliant analysis. THANK YOU!