Simple Harmonic Oscillator Help

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SUMMARY

The discussion focuses on solving a simple harmonic oscillator problem involving a particle oscillating between x = 40mm and x = 160mm, with an acceleration defined by a = k(100-x). The key points include determining the constant k and the velocity at x = 120mm. The approach involves manipulating the second-order differential equation derived from the acceleration equation. A more efficient method suggested involves substituting y = x - 100 to simplify the equation to y'' = -ky, allowing for easier resolution without needing to solve the entire differential equation.

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Homework Statement



A particle oscillates between the points x = 40mm and x = 160mm with an acceleration a = k(100-x) where k is a constant. The velocity of the particle is 18mm/s when x=100 and zero at x = 40mm and x = 160mm. Determine a) the value of hte constant k, b) the velocity when x = 120mm

Homework Equations



[tex]a = k(100-x)[/tex]

The Attempt at a Solution



This looked like a simple harmonic oscillator to me.

So I went:

[tex]a = 100k - kx[/tex]

[tex]\frac{d^2x}{dt^2} = 100k - kx[/tex]
Define:
[tex]\dot x = \frac{\mathrm{d}x}{\mathrm{d}t}[/tex]
Then Observe:
[tex]\frac{\mathrm{d}^2 x}{\mathrm{d} t^2} = \ddot x = \frac{\mathrm{d}\dot {x}}{\mathrm{d}t}\frac{\mathrm{d}x}{\mathrm{d}x}=\frac{\mathrm{d}\dot {x}}{\mathrm{d}x}\frac{\mathrm{d}x}{\mathrm{d}t}=\frac{\mathrm{d}\dot{x}}{\mathrm{d}x}\dot {x}[/tex]
Then substitute:
[tex]\frac{d\dot x}{dx}\dot x = 100k-kx[/tex]

[tex]d\dot x = (100k-kx)dx[/tex]

[tex]\int \dot x d\dot x = \int (100k-kx)dx[/tex]

[tex]\dot x^2 = 50kx - kx^2 + c[/tex]

I got that far in the manipulation, then I got stuck. Where do i go from here or what have I done wrong? My current approach is to solve for the differential then differentiate to get an equation for the velocity. Is there a better approach?
 
Last edited:
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Welcome to PF!

noodle_snacks said:
This looked like a simple harmonic oscillator to me.

So I went:

[tex]a = 100k - kx[/tex]

[tex]\frac{d^2x}{dt^2} = 100k - kx[/tex]
Define:
[tex]\dot x = \frac{\mathrm{d}x}{\mathrm{d}t}[/tex]
Then Observe:
[tex]\frac{\mathrm{d}^2 x}{\mathrm{d} t^2} = \ddot x = \frac{\mathrm{d}\dot {x}}{\mathrm{d}t}\frac{\mathrm{d}x}{\mathrm{d}x}=\frac{\mathrm{d}\dot {x}}{\mathrm{d}x}\frac{\mathrm{d}x}{\mathrm{d}t}=\frac{\mathrm{d}\dot{x}}{\mathrm{d}x}\dot {x}[/tex]
Then substitute:
[tex]\frac{d\dot x}{dx}\dot x = 100k-kx[/tex]

[tex]d\dot x = (100k-kx)dx[/tex]

[tex]\int \dot x d\dot x = \int (100k-kx)dx[/tex]

[tex]\dot x^2 = 50kx - kx^2 + c[/tex]

I got that far in the manipulation, then I got stuck. Where do i go from here or what have I done wrong? My current approach is to solve for the differential then differentiate to get an equation for the velocity. Is there a better approach?

Hi noodle_snacks! Welcome to PF! :smile:

hmm … a bit long-winded …

I'd have started by saying "Let y = x - 100"

Then that gives you y'' = -ky, which you may be able to solve on sight.

If not, then continue y''y' = -kyy', and so on.

It isn't any better … but it is easier! :biggrin:

You got stuck at:
[tex]\dot x^2 = 50kx - kx^2 + c[/tex]

So square-root it, and you get dx/√(...) = constant, and you can use trigonometric substitution to solve that. :smile:
 
Last edited:


tiny-tim said:
So square-root it, and you get dx/√(...) = constant, and you can use trigonometric substitution to solve that. :smile:
Just to point out here that there is no need to actually solve your final differential equation. The question only asks you to determine the value of k and the value of the velocity for a given displacement, both of which can be done by just plugging numbers into the ODE without actually solving it.
 

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