Simple Harmonic Oscillator Help

That is, you can solve for k by setting x=100 and a=0 in the ODE, and you can solve for the velocity by setting x=120 and solving for a, then plugging that into the ODE with x=120.
  • #1
noodle_snacks
3
0

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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  • #2
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:
  • #3


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.
 

1. What is a simple harmonic oscillator?

A simple harmonic oscillator is a type of oscillating system that follows the laws of simple harmonic motion, where the restoring force is directly proportional to the displacement from the equilibrium position. Examples of simple harmonic oscillators include a mass-spring system and a pendulum.

2. What is the equation for a simple harmonic oscillator?

The equation for a simple harmonic oscillator is given by x(t) = A*cos(ωt + φ), where x is the displacement from equilibrium, A is the amplitude, ω is the angular frequency, and φ is the phase angle.

3. How is the period of a simple harmonic oscillator calculated?

The period of a simple harmonic oscillator is calculated using the formula T = 2π/ω, where T is the period and ω is the angular frequency. This means that the period is inversely proportional to the angular frequency.

4. What is the relationship between mass and frequency in a simple harmonic oscillator?

The frequency of a simple harmonic oscillator is inversely proportional to the square root of the mass. This means that as the mass increases, the frequency decreases. This relationship is described by the equation f = 1/2π * √(k/m), where f is the frequency, k is the spring constant, and m is the mass.

5. How does damping affect a simple harmonic oscillator?

Damping is the process of reducing the amplitude of an oscillator over time, and it can have different effects on a simple harmonic oscillator. Underdamping occurs when the amplitude decreases gradually, while overdamping occurs when the amplitude decreases more rapidly. Critical damping is when the oscillator returns to equilibrium without oscillating. Damping also affects the frequency of the oscillator, with higher levels of damping resulting in lower frequencies.

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