23.8: Appendix 23A- Solution to Simple Harmonic Oscillator Equation

In our analysis of the solution of the simple harmonic oscillator equation of motion, Equation (23.2.1),

\[-k x=m \frac{d^{2} x}{d t^{2}} \nonumber \]

we assumed that the solution was a linear combination of sinusoidal functions,

\[x(t)=A \cos \left(\omega_{0} t\right)+B \sin \left(\omega_{0} t\right) \nonumber \]

where \(\omega_{0}=\sqrt{k / m}\). 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 \(t=0 \text { by } x(t=0) \equiv x_{0}\). Let \(v_{x}=d x / d t\) denote the x -component of the velocity at time t and denote the x -component of the velocity at time \(t=0 \text { by } v_{x}(t=0) \equiv v_{x, 0}\). The constancy of the mechanical energy is then expressed as

\[E=K+U=\frac{1}{2} k x^{2}+\frac{1}{2} m v^{2} \nonumber \]

We can solve Equation (23.A.3) for the square of the x -component of the velocity,

\[v_{x}^{2}=\frac{2 E}{m}-\frac{k}{m} x^{2}=\frac{2 E}{m}\left(1-\frac{k}{2 E} x^{2}\right) \nonumber \]

Taking square roots, we have

\[\frac{d x}{d t}=\sqrt{\frac{2 E}{m}} \sqrt{1-\frac{k}{2 E} x^{2}} \nonumber \]

(why we take the positive square root will be explained below).

Let \(a_{1} \equiv \sqrt{2 E / m}\) and \(a_{2} \equiv k / 2 E\). It’s worth noting that \(a_{1}\) has dimensions of velocity and w has dimensions of [length] to the power −2 . Equation (23.A.5) is separable,

\[\begin{array}{l}
\frac{d x}{d t}=a_{1} \sqrt{1-a_{2} x^{2}} \\
\frac{d x}{\sqrt{1-a_{2} x^{2}}}=a_{1} d t
\end{array} \nonumber \]

We now integrate Equation (23.A.6),

\[\int \frac{d x}{\sqrt{1-a_{1} x^{2}}}=\int a_{1} d t \nonumber \]

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 \(\cos \theta=\sqrt{a_{2}} x\) with the differentials \(d \theta\) and \(d x\) related by \(-\sin \theta d \theta=\sqrt{a_{2}} d x\). The integration variable is

\[\theta=\cos ^{-1}(\sqrt{a_{2}} x) \nonumber \]

Equation (23.A.7) then becomes

\[\int \frac{-\sin \theta d \theta}{\sqrt{1-\cos ^{2} \theta}}=\int \sqrt{a_{2}} a_{1} d t \nonumber \]

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 \(\sqrt{a_{2}} a_{1}\) on the right has dimensions of inverse time, \([\text { length }]^{-1}\left[\text {length } \cdot \operatorname{time}^{-1}\right]=\left[\text {time }^{-1}\right]\) so \(\sqrt{a_{2}} a_{1}\) dt is dimensionless. Using the trigonometric identity \(\sqrt{1-\cos ^{2} \theta}=\sin \theta\), Equation (23.A.9) reduces to

\[\int d \theta=-\int \sqrt{a_{2}} a_{1} d t \nonumber \]

Although at this point in the derivation we don’t know that \(\sqrt{a_{2}} a_{1}\), which has dimensions of frequency, is the angular frequency of oscillation, we’ll use some foresight and make the identification

\[\omega_{0} \equiv \sqrt{a_{2}} a_{1}=\sqrt{\frac{k}{2 E}} \sqrt{\frac{2 E}{m}}=\sqrt{\frac{k}{m}} \nonumber \]

and Equation (23.A.10) becomes

\[\int_{\theta=\theta_{0}}^{\theta} d \theta=-\int_{t=0}^{t} \omega_{0} d t \nonumber \]

After integration we have

\[\theta-\theta_{0}=-\omega_{0} t \nonumber \]

where \(\theta_{0} \equiv-\phi\) is the constant of integration. Because \(\theta=\cos ^{-1}(\sqrt{a_{2}} x(t))\) Equation (23.A.13) becomes

\[\cos ^{-1}(\sqrt{a_{2}} x(t))=-\left(\omega_{0} t+\phi\right) \nonumber \]

Take the cosine of each side of Equation (23.A.14), yielding

\[x(t)=\frac{1}{\sqrt{a_{2}}} \cos \left(-\left(\omega_{0} t+\phi\right)\right)=\sqrt{\frac{2 E}{k}} \cos \left(\omega_{0} t+\phi\right) \nonumber \]

At t = 0

\[x_{0} \equiv x(t=0)=\sqrt{\frac{2 E}{k}} \cos \phi \nonumber \]

The x -component of the velocity as a function of time is then

\[v_{x}(t)=\frac{d x(t)}{d t}=-\omega_{0} \sqrt{\frac{2 E}{k}} \sin \left(\omega_{0} t+\phi\right) \nonumber \]

At t = 0,

\[v_{x, 0} \equiv v_{x}(t=0)=-\omega_{0} \sqrt{\frac{2 E}{k}} \sin \phi \nonumber \]

We can determine the constant \(\phi\) by dividing the expressions in Equations (23.A.18) and (23.A.16),

\[-\frac{v_{x, 0}}{\omega_{0} x_{0}}=\tan \phi \nonumber \]

Thus the constant \(\phi\) can be determined by the initial conditions and the angular frequency of oscillation,

\[\phi=\tan ^{-1}\left(-\frac{v_{x, 0}}{\omega_{0} x_{0}}\right) \nonumber \]

Use the identity

\[\cos \left(\omega_{0} t+\phi\right)=\cos \left(\omega_{0} t\right) \cos (\phi)-\sin \left(\omega_{0} t\right) \sin (\phi) \nonumber \]

to expand Equation (23.A.15) yielding

\[x(t)=\sqrt{\frac{2 E}{k}} \cos \left(\omega_{0} t\right) \cos (\phi)-\sqrt{\frac{2 E}{k}} \sin \left(\omega_{0} t\right) \sin (\phi) \nonumber \]

and substituting Equations (23.A.16) and (23.A.18) into Equation (23.A.22) yields

\[x(t)=x_{0} \cos \omega_{0} t+\frac{v_{x, 0}}{\omega_{0}} \sin \omega_{0} t \nonumber \]

agreeing with Equation (23.2.21).

So, what about the missing \(\pm\) 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 \(\phi\) by \(\pi-\phi\) (or \(\phi-\pi\)) when the block’s velocity changes direction. Changing from the positive square root to the negative and changing \(\phi\) to \(\pi-\phi\) have the collective action of reproducing Equation (23.A.23).