Equation of motion and normal modes of a coupled oscillator

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SUMMARY

The discussion centers on the equation of motion for a simple oscillator, represented by the differential equation mx''(t) = -kx(t) + mg. The frequency is determined as ω = √(k/m), with the gravitational term mg being constant. The participant expresses confusion regarding the amplitude of the motion, having identified the equation of motion as x(t) = Acos(ωt + φ). Clarification is provided that the equation represents Newton's second law rather than a direct expression for displacement over time.

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  • Familiarity with differential equations
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  • Basic concepts of amplitude and frequency in physics
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Students of physics, particularly those studying mechanics and oscillatory systems, as well as educators seeking to clarify concepts related to coupled oscillators and their equations of motion.

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Homework Statement
Find the equation of motion of a mass on a spring
Relevant Equations
x(t) = Acos(ωt + φ)
This is a question from an exercise I don't have the answers to.

I have been trying to figure this out for a long time and don't know what to do after writing

mx''¨(t)=−kx(t)+mg
I figure that the frequency ω=√(k/m) since the mg term is constant and the kx term is the only term that changes.

I have the equation of motion written as x(t) = Acos(ωt + φ) but do not know how to go about finding the amplitude either.

Any help would be great thanks!

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It looks like this is a simple (not coupled) oscillator.

It also looks like you have this backwards, in a sense. The equation of motion is not the expression for ##x## as a function of time, but rather it is Newton's second law, involving gravity and the restoring force that the spring exerts on the mass. In other words: It is a differential equation.
 

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