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{aligned} F & = - k x \\ F & = - k y (t) \end{aligned}

Apply Newton’s Law:

$m$ denotes the mass.

$a$ denotes the acceleration.

$F$ denotes the force.

\begin{aligned} F & = m a \\ - k y (t) & = m a \\ x & = y (t) \\ v & = y^{^{'}} (t) \\ a & = y^{^{″}} (t) \\ - k y (t) & = m y^{^{″}} (t) \end{aligned}

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

\begin{aligned} - k y (t) & = m y^{^{″}} (t) \\ 0 & = m y^{^{″}} (t) + k y (t) \\ m y^{^{″}} (t) + k y (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{aligned}

The general solution of that equation is given by:

y (t) = a \cos c t + b \sin c t

Given $y (0)$ ,

\begin{aligned} 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{aligned}

Given $y^{^{'}} (0)$ ,

\begin{aligned} y (t) & = a \cos c t + b \sin c t \\ y^{^{'}} (t) & = - a c \sin c t + b c \cos c t \\ y^{^{'}} (0) & = - a c \sin (c \times 0) + b c \cos (c \times 0) \\ y^{^{'}} (0) & = - a c \sin 0 + b c \cos 0 \\ y^{^{'}} (0) & = - a c \times 0 + b c \times 1 \\ y^{^{'}} (0) & = b c \\ b & = \frac{y^{^{'}} (0)}{c} \end{aligned}

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

\begin{aligned} y (t) & = a \cos c t + b \sin c t \\ y (t) & = y (0) \cos c t + \frac{y^{^{'}} (0)}{c} \sin c t \end{aligned}

Given Angle Subtraction Theorem:

$α, β \in R$

\begin{aligned} \cos (α - β) & = \cos α \cos β + \sin α \sin β \end{aligned}

Let $α = c t$ , $β = φ$ :

\begin{aligned} \cos (c t - φ) & = \cos c t \cos φ + \sin c t \sin φ \end{aligned}

Let $A > 0$ :

\begin{aligned} A \cos (c t - φ) & = A \cos c t \cos φ + A \sin c t \sin φ \\ A \cos (c t - φ) & = (A \cos φ) \cos c t + (A \sin φ) \sin c t \end{aligned}

Given:

\begin{aligned} y (t) & = a \cos c t + b \sin c t \\ A \cos (c t - φ) & = (A \cos φ) \cos c t + (A \sin φ) \sin c t \end{aligned}

Thus:

\begin{aligned} a & = A \cos φ \\ b & = A \sin φ \\ y (t) & = A \cos (c t - φ) \\ A \cos (c t - φ) & = a \cos c t + b \sin c t \end{aligned}

Given the Pythagorean Identity:

\sin^{2} θ + \cos^{2} θ = 1

Let $θ = φ$ :

\begin{aligned} \sin^{2} φ + \cos^{2} φ & = 1 \\ (\frac{b}{A})^{2} + (\frac{a}{A})^{2} & = 1 \\ b^{2} + a^{2} & = A^{2} \\ A & = \sqrt{a^{2} + b^{2}} \end{aligned}

For equation:

A \cos (c t - φ) = a \cos c t + b \sin c t

$A$ denotes the amplitude.

$c$ denotes natural frequency.

$φ$ denotes phase.

$2 π / c$ denotes the "period" of the motion.

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

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

\begin{aligned} \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{aligned}

Replace original equation:

\begin{aligned} \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{aligned}

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}}