Thursday, 30 November 2017

homework and exercises - Damped Simple Harmonic Motion Proof?



I was reading about damped simple harmonic motion but then I saw this equation:


bvkx=ma


b is the damping constant. Then it said by substituting dx/dt for v and d2x/dt2 for a we will have:



md2xdt2+bdxdt+kx=0


Then it says the solution of the equation is: (this is my problem)


x(t)=xmebt/2mcos(ωt+ϕ)


I don't understand the last part. How can we reach the x(t)? I know very little of calculus can you please explain how to solve this?



Answer



The differential equation you quote is fairly standard in university physics/engineering course but definitely requires some calculus to solve. As a first step, if you know how to differentiate products and chains, you can substitute the given solution into the differential equation and verify that it is indeed a solution. It contains two arbitrary constants (here xm and ϕ) as you would expect of a second order differential equation (DE).


If you wanted to solve it, you still need some kind of guess as to what the function might look like; here the trial function would be:


x(t)=Aeλt


and then substitute this parametrised solution into the original DE to obtain a quadratic in λ.


x(t)=Aeλtdxdt=λAeλt=λx(t)

and d2xdt2=λ2Aeλt=λ2x(t)



so that


mλ2+bλ+k=0.


as x(t),A0


Depending on the relative values of m, k and b, you will get a quadratic with two, one or no real (i.e. one or two complex) solutions.


Given that you have two solutions λ1 and λ2, the intermediate result for x(t) will be then


x(t)=Aeλ1t+Beλ2t

For the solution you have been given, the corresponding quadratic in λ will have no real solutions, and so the λs will be complex, and the real part of the solution will give you the damped exponential at the front of the solution, and the imaginary parts will give you a wave-like term. xm and ϕ will be related to A and B and are determined by boundary conditions.


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