Simple harmonic motion diff. equation

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SUMMARY

The discussion revolves around solving the differential equation for simple harmonic motion involving a mass of 1 slug and a spring constant of 9 lb/ft. The initial conditions are set with the mass released from 1 foot above the equilibrium position with an upward velocity of \(\sqrt{3}\) ft/s. The solution derived is \(x(t) = \frac{2\sqrt{3}}{3} \sin(3t + \frac{4\pi}{3})\), leading to the velocity equation \(x'(t) = 6\sqrt{3} \cos(3t + \frac{4\pi}{3})\). The user encounters a negative time value when solving for when the mass is moving downward at 3 ft/s, indicating a need to find the first positive time value instead.

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Homework Statement


A mass of 1 slug is suspended from a spring whose spring constant is 9 lb/ft. The mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of \sqrt{3} ft/s. Find the times at which the mass is heading downward at a velocity of 3 ft/s.

Homework Equations


x(double prime) + k/m x=0
(k/m)^(1/2)=w

The Attempt at a Solution


So first, m=1, k=9 and the initial conditions: x(0)=-1 and x(prime)(0)=-\sqrt{3}
x(double prime)+9x=0 which has roots (plus or minus)3i. The solution then looks like

x(t)=C1cos(3t)+C2sin(3t) and then plugging in initial conditions I get C1=-1, C2=-\sqrt{3}/3

Then converting it to another form where A=\sqrt{C1^2+C2^2} and phi=arctan(C1/C2)
x=Asin(wt+phi)
x=2\sqrt{3}/3 sin(3t+4pi/3)
Taking the derivative to get the velocity:
x(prime)=6\sqrt{3}cos(3t+4pi/3)
setting x(prime)=3 and solving for t yields a negative value for t, t=-.97

I know I didn't really show a lot of work in a form that is easy to read or understand, but where am I going wrong here? Thanks for any help you can provide!
 
Last edited:
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Cosine function gets a given value in more than one point; maybe you should solve the first positive value of t where it holds.
 

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