Solving the Simple Harmonic Oscillator Equation of Motion: Tips and Tricks

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The discussion focuses on solving the equation of motion for a physical system represented by md2x/dt2 + c(dx/dt) - kx = 0. Participants suggest using the damped harmonic oscillator as a reference for finding solutions, indicating that the expected solutions will resemble those of damped systems. The conversation highlights the need to identify distinct types of solutions, including underdamping, overdamping, and critical damping. A hint is provided to substitute x = e^rt into the differential equation to progress further. The thread emphasizes the importance of understanding the behavior of the system based on these solutions.
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Homework Statement



A physical system is designed having the following equation of motion

md2x/dt2 + c(dx/dt) - kx = 0.

(a) From the corresponding subsidiary equation, find the solution to this equation of motion. (HINT: use the solution of the damped harmonic oscillator as a guide).
(b) How many distinct types of solution and hence physical behaviour does it exhibit ( does it have solutions that correspond to underdamping, overdamping, critical damping?)

Homework Equations





The Attempt at a Solution



From the hint, I expect the solutions to the system to be similar to the damped solution.

So the damped solution was x = -β +- √ (β2 - ω2 )

β = c/m and ω2 = k/m

So now I am stuck! Anyone care to point me in the right direction?
Thanks!
 
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Hint: x=e^rt is a solution of that equation (so substitute it into the DE and see what you get).
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

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