Simple Harmonic Oscillator Help

[noodle_snacks]

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

a = k(100-x)

The Attempt at a Solution

This looked like a simple harmonic oscillator to me.

So I went:

a = 100k - kx

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

d\dot x = (100k-kx)dx

\int \dot x d\dot x = \int (100k-kx)dx

\dot x^2 = 50kx - kx^2 + c

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?

 

[tiny-tim]

Welcome to PF!

This looked like a simple harmonic oscillator to me.

So I went:

a = 100k - kx

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

d\dot x = (100k-kx)dx

\int \dot x d\dot x = \int (100k-kx)dx

\dot x^2 = 50kx - kx^2 + c

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!

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!

You got stuck at:

\dot x^2 = 50kx - kx^2 + c

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

 

[Hootenanny]

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

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.