23.8: Appendix 23A- Solution to Simple Harmonic Oscillator Equation
( \newcommand{\kernel}{\mathrm{null}\,}\)
In our analysis of the solution of the simple harmonic oscillator equation of motion, Equation (23.2.1),
−kx=md2xdt2
we assumed that the solution was a linear combination of sinusoidal functions,
x(t)=Acos(ω0t)+Bsin(ω0t)
where ω0=√k/m. We shall now derive Equation (23.A.2).
Assume that the mechanical energy of the spring-object system is given by the constant E . Choose the reference point for potential energy to be the unstretched position of the spring. Let x denote the amount the spring has been compressed (x<0) or stretched (x>0) from equilibrium at time t and denote the amount the spring has been compressed or stretched from equilibrium at time t=0 by x(t=0)≡x0. Let vx=dx/dt denote the x -component of the velocity at time t and denote the x -component of the velocity at time t=0 by vx(t=0)≡vx,0. The constancy of the mechanical energy is then expressed as
E=K+U=12kx2+12mv2
We can solve Equation (23.A.3) for the square of the x -component of the velocity,
v2x=2Em−kmx2=2Em(1−k2Ex2)
Taking square roots, we have
dxdt=√2Em√1−k2Ex2
(why we take the positive square root will be explained below).
Let a1≡√2E/m and a2≡k/2E. It’s worth noting that a1 has dimensions of velocity and w has dimensions of [length] to the power −2 . Equation (23.A.5) is separable,
dxdt=a1√1−a2x2dx√1−a2x2=a1dt
We now integrate Equation (23.A.6),
∫dx√1−a1x2=∫a1dt
The integral on the left in Equation (23.A.7) is well known, and a derivation is presented here. We make a change of variables cosθ=√a2x with the differentials dθ and dx related by −sinθdθ=√a2dx. The integration variable is
θ=cos−1(√a2x)
Equation (23.A.7) then becomes
∫−sinθdθ√1−cos2θ=∫√a2a1dt
This is a good point at which to check the dimensions. The term on the left in Equation (23.A.9) is dimensionless, and the product √a2a1 on the right has dimensions of inverse time, [ length ]−1[length ⋅time−1]=[time −1] so √a2a1 dt is dimensionless. Using the trigonometric identity √1−cos2θ=sinθ, Equation (23.A.9) reduces to
∫dθ=−∫√a2a1dt
Although at this point in the derivation we don’t know that √a2a1, which has dimensions of frequency, is the angular frequency of oscillation, we’ll use some foresight and make the identification
ω0≡√a2a1=√k2E√2Em=√km
and Equation (23.A.10) becomes
∫θθ=θ0dθ=−∫tt=0ω0dt
After integration we have
θ−θ0=−ω0t
where θ0≡−ϕ is the constant of integration. Because θ=cos−1(√a2x(t)) Equation (23.A.13) becomes
cos−1(√a2x(t))=−(ω0t+ϕ)
Take the cosine of each side of Equation (23.A.14), yielding
x(t)=1√a2cos(−(ω0t+ϕ))=√2Ekcos(ω0t+ϕ)
At t = 0
x0≡x(t=0)=√2Ekcosϕ
The x -component of the velocity as a function of time is then
vx(t)=dx(t)dt=−ω0√2Eksin(ω0t+ϕ)
At t = 0,
vx,0≡vx(t=0)=−ω0√2Eksinϕ
We can determine the constant ϕ by dividing the expressions in Equations (23.A.18) and (23.A.16),
−vx,0ω0x0=tanϕ
Thus the constant ϕ can be determined by the initial conditions and the angular frequency of oscillation,
ϕ=tan−1(−vx,0ω0x0)
Use the identity
cos(ω0t+ϕ)=cos(ω0t)cos(ϕ)−sin(ω0t)sin(ϕ)
to expand Equation (23.A.15) yielding
x(t)=√2Ekcos(ω0t)cos(ϕ)−√2Eksin(ω0t)sin(ϕ)
and substituting Equations (23.A.16) and (23.A.18) into Equation (23.A.22) yields
x(t)=x0cosω0t+vx,0ω0sinω0t
agreeing with Equation (23.2.21).
So, what about the missing ± that should have been in Equation (23.A.5)? Strictly speaking, we would need to redo the derivation for the block moving in different directions. Mathematically, this would mean replacing ϕ by π−ϕ (or ϕ−π) when the block’s velocity changes direction. Changing from the positive square root to the negative and changing ϕ to π−ϕ have the collective action of reproducing Equation (23.A.23).