26 Jun 2025
Vibrational Modes of a Stretched String and Rectangular Membrane
Separated standing-wave solutions and normal-mode frequencies for fixed string and rectangular membrane boundaries.
The syllabus asks for solutions of the wave equation, not its derivation. We therefore take
\[\boxed{u_{tt}=c^2u_{xx}}\]as given for a string, where $[c]={\rm length}/{\rm time}$.
String fixed at both ends
Let $0<x<L$ and impose
\[u(0,t)=u(L,t)=0.\]Set $u(x,t)=X(x)T(t)$. Then
\[XT''=c^2X''T, \qquad \frac{T''}{c^2T}=\frac{X''}{X}=-k^2.\]The separated equations are
\[X''+k^2X=0, \qquad T''+c^2k^2T=0.\]The spatial solution is $X=A\cos kx+B\sin kx$. The condition $X(0)=0$ gives $A=0$, and $X(L)=0$ gives
\[\sin(kL)=0, \qquad k_n=\frac{n\pi}{L}.\]Thus
\[\boxed{ u(x,t)=\sum_{n=1}^{\infty} \left[A_n\cos(\omega_nt)+B_n\sin(\omega_nt)\right] \sin\frac{n\pi x}{L}},\]with
\[\boxed{\omega_n=ck_n=\frac{n\pi c}{L}}, \qquad f_n=\frac{\omega_n}{2\pi}=\frac{nc}{2L}.\]The coefficients follow from the initial displacement and velocity by sine-series orthogonality.
Rectangular membrane
For $0<x<a$, $0<y<b$, take the given two-dimensional wave equation
\[u_{tt}=c^2(u_{xx}+u_{yy})\]with all four edges fixed. Put $u=XYT$. Division by $c^2XYT$ gives
\[\frac{T''}{c^2T}=\frac{X''}{X}+\frac{Y''}{Y}.\]Choose
\[\frac{X''}{X}=-k_x^2, \qquad \frac{Y''}{Y}=-k_y^2.\]Fixed edges require
\[X_m=\sin\frac{m\pi x}{a}, \qquad Y_n=\sin\frac{n\pi y}{b}.\]The time equation is
\[T''+\omega_{mn}^2T=0,\]where
\[\boxed{ \omega_{mn}=c\pi \sqrt{\left(\frac ma\right)^2+\left(\frac nb\right)^2}}.\]One normal mode is therefore
\[\boxed{ u_{mn}=A_{mn} \sin\frac{m\pi x}{a} \sin\frac{n\pi y}{b} \cos(\omega_{mn}t+\delta_{mn})}.\]
The dimensions are consistent: each term under the square root has dimension ${\rm length}^{-2}$, so $[\omega_{mn}]={\rm time}^{-1}$.
The separated string and membrane modes are substituted into their given wave equations in the Unit II Maxima worksheet.
Discussion