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Subsection 7.1.1 Solution of Wave Equation for Vibrating String

Let us consider an elastic string of length \(l\) tightly stretched between points O and P, as shown in figure Figure 7.1.2. If y be the displacement of string at point A(x,y) at any time t, then the wave equation for a vibrating string is given by
\begin{equation} \frac{\partial^{2} y}{\partial x^{2}} = \frac{1}{v^{2}}\frac{\partial^{2} y}{\partial t^{2}}\tag{7.1.6} \end{equation}
where \(v\) is a constant. Equation (7.1.6) shows that the displacement y is a function of x and t, i.e., \(y=f(x,t)\) hence from the method of separation of variables, the solution for equation (7.1.6) is
\begin{equation} y(x,t) = X(x)T(t)\tag{7.1.7} \end{equation}
where \(X\) is a function of \(x\) only and \(T\) is a function of \(t\) only. Now
\begin{equation*} \frac{\partial y}{\partial t}=X\frac{\partial T}{\partial t}\hspace{1cm} \text{or,}\quad \frac{\partial^{2} y}{\partial t^{2}}=X\frac{\partial^{2} T}{\partial t^{2}} \end{equation*}
and
\begin{equation*} \frac{\partial y}{\partial x}=T\frac{\partial X}{\partial x} \end{equation*}
or,
\begin{equation*} \frac{\partial^{2} y}{\partial x^{2}}=T\frac{\partial^{2} X}{\partial x^{2}} \end{equation*}
substituting these values in equation (7.1.6), we get -
\begin{equation*} T\frac{\partial^{2} X}{\partial x^{2}}=\frac{1}{v^{2}}X\frac{\partial^{2} T}{\partial t^{2}} \end{equation*}
or,
\begin{equation} \frac{1}{X}\frac{\partial^{2} X}{\partial x^{2}}=\frac{1}{v^{2}T}\frac{\partial^{2} T}{\partial t^{2}}=\lambda \quad \text{(say)}.\tag{7.1.8} \end{equation}
This is because each side of equation (7.1.8) contains only one variable which is possible if each side must be seperately equal to some constant, say, \(\lambda\text{.}\) Now, according to the value of \(\lambda\) there arises three cases.
Figure 7.1.2.
Case I. When \(\lambda=0\text{,}\) we have from equation (7.1.8),
\begin{equation} \begin{cases} \frac{\partial^{2} X}{\partial x^{2}} = 0 & \Rightarrow X=c_{1}x+c_{2}\\ \frac{\partial^{2} T}{\partial t^{2}} = 0 & \Rightarrow T=c_{3}t+c_{4} \end{cases}\tag{7.1.9} \end{equation}
Case II. when \(\lambda \gt 0\text{,}\) for simplicity assume \(\lambda=k^{2}\text{.}\)
\begin{equation} \begin{cases} \frac{\partial^{2} X}{\partial x^{2}} -k^{2}X = 0 & \Rightarrow X=c_{1}e^{kx}+c_{2}e^{-kx}\\ \frac{\partial^{2} T}{\partial t^{2}}-k^{2}v^{2}T = 0 & \Rightarrow T=c_{3}e^{kvt}+c_{4}e^{-kvt} \end{cases}\tag{7.1.10} \end{equation}
Case III. when \(\lambda \lt 0\text{,}\) i.e., \(\lambda=-k^{2}\)(say),
\begin{equation} \begin{cases} \frac{\partial^{2} X}{\partial x^{2}} +k^{2}X = 0 & \Rightarrow X=c_{1}\cos kx+c_{2}\sin kx\\ \frac{\partial^{2} T}{\partial t^{2}}+k^{2}v^{2}T = 0 & \Rightarrow T=c_{3}\cos kvt+c_{4}\sin kvt \end{cases}\tag{7.1.11} \end{equation}
The boundary conditions are
\begin{equation} y(0,t) = 0 = y(l,t),\tag{7.1.12} \end{equation}
the initial conditions are
\begin{equation} y(x,0) = f(x) \quad \text{and} \quad \left(\frac{\partial y}{\partial t}\right)_{t=0}=f'(x)\tag{7.1.13} \end{equation}
Imposing the boundary conditions on case I and II, we obtain \(X(x)=0\) for \(y(x,t) =0\text{.}\) So equations (7.1.9) and (7.1.10) fail to give the solution of equation (7.1.6) and hence an equation (7.1.11) is only capable to provide the complete solution of equation (7.1.6) as it is periodic in \(x\) and \(t\text{.}\) Therefore with the case III the equation (7.1.7) becomes -
\begin{equation} y(x,t) = (c_{1}\cos kx+c_{2}\sin kx)(c_{3}\cos kvt+c_{4}\sin kvt)\tag{7.1.14} \end{equation}
or,
\begin{equation*} 0=(c_{1}(c_{3}\cos kvt+c_{4}\sin kvt) \end{equation*}
from equation (7.1.12), \(y(x,t)=0 \) at \(x=0\) \(\Rightarrow (c_{1} =0)\) for any arbitrary time.
Hence equation (7.1.14) becomes
\begin{equation} y(x,t) = c_{2}\sin kx(c_{3}\cos kvt+c_{4}\sin kvt)\tag{7.1.15} \end{equation}
or,
\begin{equation*} 0= c_{2}\sin kl(c_{3}\cos kvt+c_{4}\sin kvt) \end{equation*}
from equation (7.1.12) \(y(x,t) = 0 \) at \(x=l\) or, \(\sin kl = 0 = \sin n\pi\) \(\because \quad c_{2}\neq 0.\text{.}\) or,
\begin{equation} kl=n\pi \qquad \therefore \quad k=\frac{n\pi}{l}\tag{7.1.16} \end{equation}
where \(n=1,2,3,\cdots\) (\(n \neq 0,\) otherwise \(k=0\text{,}\)the case I.)
Therefore equation (7.1.15) becomes -
\begin{equation*} y(x,t) = c_{2}\sin \frac{n\pi x}{l}\left(c_{3}\cos \frac{n\pi vt}{l}+c_{4}\sin \frac{n\pi vt}{l}\right) \end{equation*}
\begin{equation} = \left(b_{n}\cos \frac{n\pi vt}{l}+d_{n}\sin \frac{n\pi vt}{l}\right)\sin \frac{n\pi x}{l}\tag{7.1.17} \end{equation}
by replacing \(c_{2}c_{3}=b_{n} \) and \(c_{2}c_{4}=d_{n}\text{.}\) Imposing the initial condition \(y(x,0) =f(x)\) from equation (7.1.13) in equation (7.1.17), we get -
\begin{equation} y(x,0) =f(x)=b_{n}\sin \frac{n\pi x}{l} \tag{7.1.18} \end{equation}
Now, differentiating equation (7.1.17) w. r. t. ’t’, we get -
\begin{equation*} \frac{\partial y}{\partial t}=\left(-\frac{n\pi v}{l} b_{n}\sin \frac{n\pi vt}{l} + \frac{n\pi v}{l}d_{n}\cos \frac{n\pi vt}{l}\right) \sin \frac{n\pi x}{l} \end{equation*}
from equation (7.1.13), we have
\begin{equation*} \left(\frac{\partial y}{\partial t}\right)_{t=0}= f'(x) \end{equation*}
\begin{equation} \therefore \quad f'(x) = \left(\frac{\partial y}{\partial t}\right)_{t=0} = \frac{n\pi v}{l}d_{n}\sin \frac{n\pi x}{l}\tag{7.1.19} \end{equation}
The motion of string is harmonic because the damping motion of the string is neglected by assuming the small displacement of string hence there are number of solutions obtained for equation (7.1.6). Therefore the desired solution is given by summing the equation (7.1.17) for \(n=0\) to \(\infty\text{,}\) i.e.,
\begin{equation} y(x,t) = \sum\limits_{n=1}^{\infty} \left[b_{n}\cos \frac{n\pi vt}{l}+d_{n}\sin \frac{n\pi vt}{l}\right]\sin \frac{n\pi x}{l}\tag{7.1.20} \end{equation}
Now, the constants \(b_{n}\) and \(d_{n}\) are determined by multiplying equations (7.1.18) and (7.1.19) on both sides by \(\sin \frac{n\pi x}{l}\) and then integrating from \(x=0\) to \(x=l\text{,}\) from equation (7.1.18),
\begin{equation*} \int\limits_{0}^{l}f(x)\sin \frac{n\pi x}{l}\,dx = \int\limits_{0}^{l}b_{n}\sin^{2} \frac{n\pi x}{l}\,dx \end{equation*}
\begin{equation*} = \frac{b_{n}}{2}\int\limits_{0}^{l}\left(1-\cos \frac{2n\pi x}{l}\right)\,dx = \frac{b_{n}}{2}.l \end{equation*}
\begin{equation} \therefore \quad b_{n}=\frac{2}{l}\int\limits_{0}^{l}f(x)\sin \frac{n\pi x}{l}\,dx\tag{7.1.21} \end{equation}
from equation (7.1.19),
\begin{equation} \frac{n\pi v}{l} d_{n}=\frac{2}{l}\int\limits_{0}^{l}f'(x)\sin \frac{n\pi x}{l}\,dx\tag{7.1.22} \end{equation}
By substituting the value of \(b_{n}\) and \(d_{n}\) in equation (7.1.20), we get the complete solution of equation (7.1.6).
Note: If we assume the initial velocity of the string \(f'(x)=0\text{,}\) then from equation (7.1.19) we have \(d_{n}=0\text{.}\) Thus equation (7.1.20) reduces to
\begin{equation} y(x,t) = \sum\limits_{n=1}^{\infty} b_{n}\cos \frac{n\pi vt}{l}\sin \frac{n\pi x}{l}\tag{7.1.23} \end{equation}
\begin{equation*} =\frac{1}{2}\sum\limits_{n=1}^{\infty} b_{n}\left[\sin \frac{n\pi}{l}(x-vt)+ \sin \frac{n\pi}{l}(x+vt)\right] \end{equation*}
\begin{equation*} = \frac{1}{2}\sum\limits_{n=1}^{\infty} b_{n}\sin \frac{n\pi}{l}(x-vt)+ \frac{1}{2}\sum\limits_{n=1}^{\infty}b_{n}\sin \frac{n\pi}{l}(x+vt) \end{equation*}
\begin{equation} \therefore y(x,t) = \frac{1}{2}[f(x-vt)+f(x+vt)]\tag{7.1.24} \end{equation}
which is the solution of equation (7.1.6), where the function \(f\) is odd and periodic in period \(2l\text{.}\)