Equation of Motion Problem involving a spring?

Question

a 16-pound weight  is attached to a spring whose constant is k=4.5 lb/ft. Beginning at t=0 seconds, a force equal to f(t)=4sin(3t)+2cos(3t) acts on the system. Assume no damping forces are present.

QUESTION: Find the equation of motion if the weight is released from rest from the equilibrium position.

I can understand the equation of simple harmonic motion and te equation of damped motion, but I am having having trouble with the equation of Forced Motion, which is what this problem is. 


the equation of forced motion is : [f(t)/m] = x" + (Beta/m)x' + (k/m)x, where Beta = damping motion = 0, k is the constant, m is the mass, x is the displacement, and x' is the instantaneous velocity and x" is the acceleration of the mass and f(t) is the external force acting on the vibrating mass.


Any insight on how to approach this equation will  be helpful.

Dan B · Accepted Answer

In the equation of forced motion replace f(t) with 4sin(3t)+2cos(3t) , m with 16, Beta with zero, and k with 4.5 to get

[4sin(3t)+2cos(3t)]/16 = x" + (4.5/16)x

multiply by 16 to get [4sin(3t)+2cos(3t)] = 16x" + 4.5x

Anonymous · Answer

partly A all you need to do is understand the conservation of ability. you be attentive to that the preliminary spring potenital is P=kx^2 and because you have the amplitude and the spring consistent you may calculate that. The preliminary spring ability comes out to .03 joules. next you placed you preliminary ability on your very final ability. Your equation now's P=P+ok (p being ability and ok being kinetic ability). utilising substitution with with kinetic ability equations and ability ability you presently have kx^2=kx^2+.5mv^2. you may enter all your givens and then it comes out to look like 2.8(.104)^2=2.8(.052)^2+.5(m)(.266)^2. After some Algebra, your mass will come out too .17kg. desire that enables you. stable success with something.

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Equation of Motion Problem involving a spring?

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