Simple harmonic oscillation

OK, I was never very good at math but can someone explain step by step how:

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becomes:

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set k=1
set m=1
F=-x can be thought of as being like a spring. the force is proportional to displacement.
set A=1
set f=1/2π
set φ=0
therefore x(t)=sin(t)
F=ma
a=d^2x/dt^2
the second derivative of x (which equals sin(t)) is -sin(t)
therefore F=-sin(t)=-x
the force is proportional to the displacement

 

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It's easier to work backwards; to show that if:
\(x(t) = A sin(2\pi ft + \phi\)
then:
\(m \frac{d^2x}{dt^2}=-kx\)

...where:
\(k = m(2\pi f)^2\)

 

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In general, ignoring power series expansions, sinusoidal and exponential
functions are the most common functions which have derivatives of the same
form as the original function. This is why they are so often the solution to
equations of time or motion; differential equations.

For the simple harmonic motion problem you offer, the second derivative
is proportional to the original function.

 

For any ODE of the form \(A y'' + B y' + c y = 0\) where A,B,C are constants you should consider the trial solution \(y = e^{\lambda x}\) and see what equation \(\lambda\) must obey. For SHM you'll find \(\lambda\) is complex and then you use \(e^{ikx} = cos(kx) + i sin(kx)\) to obtain the general solutions in terms of sin and cos.

If you've got no idea what I'm talking about, just say and I'll be more explicit.

 

Just to add to this: Assuming a solution of the form \(y=e^{\lambda x}\) yields \(y' = \lambda e^{\lambda x} = \lambda y\) and \(y''=\lambda^2 y\). Plugging these into the ODE yields \((A\lambda^2+B\lambda+C)y = 0\). This means either \(y=0\) or \(A\lambda^2+B\lambda+C = 0\). The former is the trivial solution, which is not of interest. The latter is a quadratic equation in \(\lambda\) with solutions

\(\lambda = \frac{-B\pm\sqrt{B^2-4AC}}{2A}\)

When \(B^2 \ne 4AC\) there will be two distinct roots, call them \(\lambda_1\) and \(\lambda_2\) and the solutions will be of the form

\(y=c_1e^{\lambda_1 x} + c_2e^{\lambda_2 x}\)

When \(B^2 = 4AC\) there will be a double root \(\lambda=-\,\frac 1 2 \, \frac B A\). In this case, the solutions are of the form

\(y=(c_1+c_2x)e^{\lambda x}\)

 

Thanks James,

I guess the only step that I didn't fully understand here is the above. Can you explain this in a bit more detail?

What is a "trial solution" and how did you come up with that structure?

 

(d^2x/dt^2)=-(k/m) x
d/dx [(1/2)(dt/dx)^-2] =-(k/m) x # A trick for ODE in this form x''=f(x)
(dt/dx)^-2 = 2 ∫ -(k/m) x dx = A - (k/m) x^2
± dx/dr = 1/√(A - (k/m) x^2)
± ∫ dt = ∫ dx/√(A - (k/m) x^2)
±(t-t_0) = √(m/k) arctan( √(m/k) x / √(A - (k/m) x^2) ) = √(m/k) arcsin( √(k/m) x/ √A)
x = Amplitude sin(√(k/m) t + phase)

So technically, a guess isn't necessary if you have a repertoire of useful methods to rely upon.

 

As an example, would it be possible to give me another solution for that differential equation that has a completely different structure to it?

 

x''=6x^2+1

Solution: x(t) = ℘(t-t_0,-2, C) where d = ℘(a,b,c) solves the integral a = Integral from infinity up to d of 1/sqrt(4f^3 - bf - c) df
(x'(t))^2 = (4x^3 + 2x - C) = (4℘(t-t_0,-2, C)^3 + 2 ℘(t-t_0,-2, C) - C)

http://mathworld.wolfram.com/WeierstrassEllipticFunction.html

x''=x

Solution: x(t) = Ae^t + Be^-t, x'(t) = Ae^t - Be^-t, x''(t) = Ae^t + Be^-t

x''= 2 x x'

Solution: x(t) = A tan( A t + B ), x'(t) = (A / cos(A t + B )) ^2, x''(t) = 2 A^3 tan(A t + B ) / (cos(A t + B ))^2

 

wolframalpha.com

this site is good!

 

I remember calling this case "degenerate" and note that the term with the x multiplying has opposite parity than the one without the x factor.

Is this why there are two components to the solution still? I.e. if Y(x) satisfies the original DE, then xY(x) will too.

Perhaps this speculation of mine is false. Are these two different terms related to parity or not?

 

What is parity?

A derivation of that formula would be to write the equation in first order matrix-vector form (introducing new variables and arranging them as components of a vector) and then factorizing the matrix into its Jordan normal form.

 

Perhaps not spelled correctly? but I was refering to symetric or anti-symetric function wrt the origin. For example Cos(x) has "even parity" and Sin(x) has "odd parity." (or is "anti- symetric.")

 

You mean the coefficient in front of the exponential? Otherwise the whole function (including the exp) does not have that kind of symmetry.

 

I think the whole function (the complete solution) does not have a pure parity, but is a mix of both. I am guessing that it can be separated into the sum of a symetric and an anti-symetric function (and assuming that if separated, each will still be a solution to the DE, but not the most general solution - their sum is that.)

I may be all wrong about this - I was just struck by the possible link to parity and wondered if it was real.

 

I completely agree and thank you for making the parity concept more clear (I did not as thought it was well known.) In my last post I explicitly stated that the complete solution was NOT of either parity; and speculated that it could be separated into two parts, one of even and one of odd parity. The main basis of this speculation was that most functions can be written as Fourier expansions. I.e. as weighted summation of sin and cos terms. For those that can be written as fourier expansions, the separation into two pure parity funcitions is obviously possible. (Collect all the sin terms of the Fourier expansion to make the odd parity function and conversely all the cos terms make the even parity function.)

What I have doubt about is whether or not these seaparated "pure parity" function each individually are a solution to the original DE. I tend to think they are not, except in special cases.

For example, D.H.'s post 7 solution for the degenerate case has (c1 + xc2) times an exponential, but that exponential can be written as sin and cos terms (one is imaginary, but not sure now which is - its been a long time.) Thus, even the general solution should have even parity made of the [(c1)cos + (xc2)sin] terms and the odd parity made by [(c1)sin + (xc2)cos] terms. But again, I tend to doubt either alone is a solultion to the general DE.

As I vaguely recall when the DE is descibing an electron via QM, then there may be two solutions but the "real one" is anti- symteric (odd parity, if you prefer) and this is some how related to the Pauli exclusion principle (Fermion nature of the electron). It has been more than 40 years so I am some what vague on all this now, but hope someone will straighten out what I am trying o remember and develop this thought further, if there is anything to it.