The vibratring string

Simple harmonic motion

Let $y(t)$ denote the displacement of the mass at time $t$.

Apply Hooke’s Law:

$k(k>0)$ denotes the spring constant.

$x$ denotes the displacement.

$F$ denotes the force applied by the spring.

$$\begin{array}{rl}F& =-kx\\ F& =-ky(t)\end{array}$$

Apply Newton’s Law:

$m$ denotes the mass.

$a$ denotes the acceleration.

$F$ denotes the force.

$$\begin{array}{rl}F& =ma\\ -ky(t)& =ma\\ x& =y(t)\\ v& =y^{{}^{\prime }}(t)\\ a& =y^{{}^{″}}(t)\\ -ky(t)& =m{y}^{{}^{″}}(t)\end{array}$$

Let $c=\sqrt{k/m}$ to simplify this equation.

$$\begin{array}{rl}-ky(t)& =m{y}^{{}^{″}}(t)\\ 0& =m{y}^{{}^{″}}(t)+ky(t)\\ m{y}^{{}^{″}}(t)+ky(t)& =0\\ y^{{}^{″}}(t)+\frac{k}{m}y(t)& =0\\ c& =\sqrt{k/m}\\ c^2& =k/m\\ y^{{}^{″}}(t)+\frac{k}{m}y(t)& =0\\ y^{{}^{″}}(t)+c^{2}y(t)& =0\end{array}$$

The general solution of that equation is given by:

$$y(t)=a\cos ct+b\sin ct$$

Given $y(0)$,

$$\begin{array}{rl}y(0)& =a\cos (c\times 0)+b\sin (c\times 0)\\ y(0)& =a\cos 0+b\sin 0\\ y(0)& =a\times 1+b\times 0\\ y(0)& =a\\ a& =y(0)\end{array}$$

Given $y^{{}^{\prime }}(0)$,

$$\begin{array}{rl}y(t)& =a\cos ct+b\sin ct\\ y^{{}^{\prime }}(t)& =-ac\sin ct+bc\cos ct\\ y^{{}^{\prime }}(0)& =-ac\sin (c\times 0)+bc\cos (c\times 0)\\ y^{{}^{\prime }}(0)& =-ac\sin 0+bc\cos 0\\ y^{{}^{\prime }}(0)& =-ac\times 0+bc\times 1\\ y^{{}^{\prime }}(0)& =bc\\ b& =\frac{y^{{}^{\prime }}(0)}{c}\end{array}$$

Since $a=y(0)$ and $b=\frac{y^{{}^{\prime }}(0)}{c}$:

$$\begin{array}{rl}y(t)& =a\cos ct+b\sin ct\\ y(t)& =y(0)\cos ct+\frac{y^{{}^{\prime }}(0)}{c}\sin ct\end{array}$$

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Given Angle Subtraction Theorem:

$\alpha ,\beta \in \mathbb{R}$

$$\begin{array}{rl}\cos (\alpha -\beta )& =\cos \alpha \cos \beta +\sin \alpha \sin \beta \end{array}$$

Let $\alpha =ct$, $\beta =\phi$:

$$\begin{array}{rl}\cos (ct-\phi )& =\cos ct\cos \phi +\sin ct\sin \phi \end{array}$$

Let $A>0$:

$$\begin{array}{rl}A\cos (ct-\phi )& =A\cos ct\cos \phi +A\sin ct\sin \phi \\ A\cos (ct-\phi )& =(A\cos \phi )\cos ct+(A\sin \phi )\sin ct\end{array}$$

Given:

$$\begin{array}{rl}y(t)& =a\cos ct+b\sin ct\\ A\cos (ct-\phi )& =(A\cos \phi )\cos ct+(A\sin \phi )\sin ct\end{array}$$

Thus:

$$\begin{array}{rl}a& =A\cos \phi \\ b& =A\sin \phi \\ y(t)& =A\cos (ct-\phi )\\ A\cos (ct-\phi )& =a\cos ct+b\sin ct\end{array}$$

Given the Pythagorean Identity:

$${\sin }^2\theta +{\cos }^2\theta =1$$

Let $\theta =\phi$:

$$\begin{array}{rl}{\sin }^2\phi +{\cos }^2\phi & =1\\ (\frac{b}{A}{)}^2+(\frac{a}{A}{)}^2& =1\\ b^2+a^2& =A^2\\ A& =\sqrt{a^2+b^2}\end{array}$$

For equation:

$$A\cos (ct-\phi )=a\cos ct+b\sin ct$$

$A$ denotes the amplitude.

$c$ denotes natural frequency.

$\phi$ denotes phase.

$2\pi /c$ denotes the "period" of the motion.

Derivation of the wave equation

$$\frac{1}{c^2}\frac{{\partial }^{2}u}{\partial t^2}=\frac{{\partial }^{2}u}{\partial x^2}$$

Let $x=aX$, $t=bT$, $U(X,T)=u(x,t)$ ($a$ and $b$ are positive constants):

$$\begin{array}{rl}\frac{{\partial }^{2}U}{\partial X^2}& =a^2\frac{{\partial }^{2}u}{\partial x^2}\\ \frac{{\partial }^{2}U}{\partial X^2{a}^2}& =\frac{{\partial }^{2}u}{\partial x^2}\\ \\ \frac{{\partial }^{2}U}{\partial T^2}& =b^2\frac{{\partial }^{2}u}{\partial t^2}\\ \frac{{\partial }^{2}U}{\partial T^2{b}^2}& =\frac{{\partial }^{2}u}{\partial t^2}\end{array}$$

Replace original equation:

$$\begin{array}{rl}\frac{1}{c^2}\frac{{\partial }^{2}u}{{\partial }^{2}t}& =\frac{{\partial }^{2}u}{{\partial }^{2}x}\\ \frac{1}{c^2}\frac{{\partial }^{2}U}{\partial T^2{b}^2}& =\frac{{\partial }^{2}U}{\partial X^2{a}^2}\\ \frac{a^2}{c^2{b}^2}\frac{{\partial }^{2}U}{\partial T^2}& =\frac{{\partial }^{2}U}{\partial X^2}\end{array}$$

If we choose $a,b$ properly, we can let $\frac{a^2}{c^2{b}^2}=1$, and get:

$$\frac{{\partial }^{2}U}{\partial T^2}=\frac{{\partial }^{2}U}{\partial X^2}$$