Solve Modeling Higher-Order Diff Eqn: Free Undamped Motion w/ Spring

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The discussion focuses on solving a higher-order differential equation related to free undamped motion of a mass-spring system. The mass of 1 slug is suspended from a spring with a spring constant of 9 lb/ft, initially released from 1 foot above the equilibrium position with an upward velocity of √3 ft/s. The correct differential equation for the motion is identified as x(t) = -cos(3t) - (√3/3)sin(3t). The goal is to find the times when the mass is moving downward at a velocity of 3 ft/s, leading to a quadratic equation that can be solved for sin(3t) and subsequently for t.

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Matthewmccoy6
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A mass of 1 slug is suspended from a spring whose spring constant is 9lb/ft. The mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of square root 3 ft/s. Find the times at which the mass is heading downward at a velocity of 3 ft/s.

I found the differential equation to be x(t)=Acos(3t) + Bsin(3t).

What I did was I took downward as positive while upward as negative. I'm having trouble finding t.

My equation for the motion was x(t)= -cos 3t - (square root 3/ 3) sin 3t

Please help, thanks!
 
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Matthewmccoy6 said:
A mass of 1 slug is suspended from a spring whose spring constant is 9lb/ft. The mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of square root 3 ft/s. Find the times at which the mass is heading downward at a velocity of 3 ft/s.

I found the differential equation to be x(t)=Acos(3t) + Bsin(3t).
That is NOT a differential equation. I presume you mean you found that as the general solution to the differential equation.

What I did was I took downward as positive while upward as negative. I'm having trouble finding t.

My equation for the motion was x(t)= -cos 3t - (square root 3/ 3) sin 3t

Please help, thanks!
So you need to solve
[tex]-cos(3t)- (\sqrt{3}/3)sin(3t)= 3[/tex].
That is the same as
[tex]cos(3t)= 3- (\sqrt{3}/3)sin(3t)[/tex]

Squaring both sides of that gives
[tex]cos^2(3t)= 9- (2\sqrt{3}/3)sin(3t)+ sin^2(3t)[/tex]

Replace [itex]cos^2(3t)[/itex] with [itex]1- sin^2(3t)[/itex] and you have
[tex]1- sin^2(3t)= 9- (2\sqrt{3}/3)sin(3t)+ sin^2(3t)[/tex]
a quadratic equation you can solve for sin(3t) and then for t.

Be sure to check if an "extraneous" solutions were introduced by the squaring.
 


I know, this is a differential equation problem though and using the equation of simple harmonic motion I found the above equation.
Also, I was hoping there was a simple identity that I could use to find t rather than something really unpleasant. Thanks though.
 

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