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bcjochim07
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Homework Statement
A particle of mass m is confined to a horizontal plane. It is elastically bound to its equilibrium position by an isotropic elastic force,
F=mω^2r
Where r is the displacement of the particle from equilibrium and ω is a real constant parameter.
From Newton's law, we obtain the equation of motion:
r'' + ω^2r = 0.
In rect. coordinates r(t) = ix(t) + jy(t).
Obtain expressions for x(t) and y(t).
Homework Equations
The Attempt at a Solution
The x-component of force :
Fx = -x(t)mω^2 = mx''(t)
x''(t) + x(t)ω^2 = 0
The solution to this DE is x(t) = c1cos(ωt) + c2sin(ωt)
Then I'll apply the initial conditions given in the problem to obtain the constants.
Here's the part I'm unsure about
When I solve the differential equation for the y component I get
y(t) = a1cos(ωt) + a2sin(ωt). I'm thinking I have to multiply this by t so that the solutions are linearly independent:
y(t) = a1*tcos(ωt) + a2*tsin(ωt). Is this correct?