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).