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Use $\sin \theta \cos \theta =\frac{1}{2}\sin 2\theta$ with $\theta =\pi x/l$. The initial shape becomesThe sine functions are orthogonal on $[0,l]$, so this decomposition directly identifies the two nonzero spatial normal mode coefficients. Since the string is released from rest, every time-dependent sine coefficient vanishes and the subsequent motion isThus the signed mode amplitudes are $A_1=A$, $A_2=A/2$ and $A_n=0$ for $n\ge 3$; their magnitude amplitudes are $|A|$ and $|A|/2$. The factor of two in the second angular frequency is fixed by the same boundary conditions as its spatial wavelength.

For a single normal mode $y_n=A_n\sin (k_{n}x)\cos ({\omega }_{n}t+{\varphi }_n)$, the kinetic energy is the integral of $\frac{1}{2}\rho y_t^2$. Using ${\int }_0^l{\sin }^2(k_{n}x)dx=l/2$ givesThe extra string length is ${\int }_0^l[\sqrt{1+y_x^2}-1]dx\simeq \frac{1}{2}{\int }_0^l{y}_x^{2}dx$. Multiplying by tension gives the quadratic potential energy, so ${\int }_0^l{\cos }^2(k_{n}x)dx=l/2$ impliesSince $\rho {\omega }_n^2=T{k}_n^2$, the sum is time independent:For the released shape, $A_1=A$ and $A_2=A/2$, soThe second mode's smaller displacement is offset by its doubled spatial wavenumber and angular frequency. orthogonality of distinct sine and cosine spatial modes eliminates cross terms in the integrated energies, so the total is their sum. Initially the kinetic energy is zero and all the energy is stored in the string's deformation; during the motion the total remains constant.

At $\omega =\sqrt{3}{\omega }_0$, $s=3$, so the periodic amplitudes becomeThe second bead is stationary in this particular motion, although the endpoint oscillates. The first bead moves at twice the endpoint amplitude and in opposite phase. This is an antiresonance of the second-bead response, not a natural resonance of the complete two-bead system.

The force balance makes the cancellation explicit. At $y_2=0$, the middle segment supplies transverse force $T{y}_1/(2L)=-Th/L$, while the end segment supplies $Th/L$; these cancel on the second bead. The first bead then oscillates between two effectively fixed neighbours with restoring coefficient $3T/(2L)$, whose single-coordinate natural angular frequency is $\sqrt{3}{\omega }_0$.

As elsewhere in the undamped problem, this exact nodal motion presumes initial conditions selecting the periodic particular response. Additional free normal modes would move the second bead as well; they do not decay automatically.

The fixed-end boundary conditions require $f(0)=f(L)=0$. For $\alpha =0$, the spatial solution is linear and both endpoints force it to vanish; for $\alpha >0$, the exponential solutions likewise cannot satisfy both conditions nontrivially. Write $\alpha =-k^2$ with $k>0$. Then $f=A\sin (kx)$ and $\sin (kL)=0$, givingTheir angular frequencies are ${\omega }_n=cn\pi /L$. Unlike the freely translating chain, this fixed string has no translational zero normal mode.

The zero initial displacement selects sine time factors. Put ${\omega }_1=\pi c/L$. The Fourier sine series of the initial velocity has coefficients $V_1=V_0/2$, $V_2=-3V_0/4$, with all other coefficients zero. Since differentiation in time multiplies a sine by its angular frequency,The normal mode coordinates are $q_1(t)=V_0\sin ({\omega }_{1}t)/(2{\omega }_1)$ and $q_2(t)=-3V_0\sin (2{\omega }_{1}t)/(8{\omega }_1)$. Their peak displacement amplitudes are $|V_0|/(2{\omega }_1)$ and $3|V_0|/(8{\omega }_1)$. Both the initial displacement and the initial velocity can be checked directly from this expression.

Write the displacement as $\mathbf{x}=\mathbf{a}\cos (\omega t+\varphi )$. The generalized eigenvalue problem isThe symmetry of the two end atoms suggests an end-antisymmetric shape $(1,0,-1)$ and two end-symmetric shapes. Direct substitution gives the normal modes of three masses connected by two equal springs:These three independent eigenvectors exhaust the three-dimensional system. The ordinary frequencies are $\omega /(2\pi )$; the values $\omega$ above are angular frequencies.

For $M=2m$, take the mode shapes $v_0=(1,1,1)$, $v_a=(1,0,-1)$ and $v_s=(1,-1,1)$. Their angular frequencies are $0$, $\omega =\sqrt{k/m}$ and $\sqrt{2}\omega$. The initial displacement decomposes asAll initial velocities vanish, so the zero-mode coefficient stays constant and the two oscillatory coefficients use cosines. ThusThe conserved centre of mass is $(x_1+2x_2+x_3)/4=d/4$. Omitting the zero mode would incorrectly force this to zero. The two vibrational frequencies are in the ratio $\sqrt{2}$, so there is no common nonzero period.

Write the complex displacements as $y_1=(A{e}^{i{k}_{1}x}+B{e}^{-i{k}_{1}x})e^{-i\omega t}$ and $y_2=C{e}^{i{k}_{2}x-i\omega t}$. At a massless join, continuity of displacement and transverse force givesEliminating $B$ gives the displacement transmission coefficient and reflection coefficient,The angular frequency is shared across a stationary interface, whereas the wavelength changes. The equal-tension assumption is used in the slope condition.

Let $\kappa =T/(2ma)$ and seek a normal mode $\mathbf{y}=\mathbf{a}{e}^{i\omega t}$. Its eigenvalue equation isor ${\lambda }^2-5\lambda +4=(\lambda -1)(\lambda -4)=0$, where $\lambda ={\omega }^2/\kappa$. ThusThese are angular frequencies; the corresponding ordinary frequencies are $f_{\pm }={\omega }_{\pm }/(2\pi )$.

Put $\kappa =k/m$, $q=g/l$. The positive stiffness-per-mass matrix isSubstitution of a normal mode $x=v\cos \omega t$ requires $Kv={\omega }^{2}v$. Its characteristic equation is ${\lambda }^2-(\kappa +2q)\lambda +\kappa q=0$. ThereforeFor $k,m,g,l>0$, their sum and product are positive, so both squared angular frequencies are positive and ${\omega }_{+}>{\omega }_{-}>0$.

For a plane wave, ${\partial }_t$ acts as $-i\omega$ and ${\partial }_x$ as $ik$. Substitution givesso, for $\omega \ne {\omega }_d$, the dispersion relation isAssume ${\omega }_n>{\omega }_d>0$ and real angular frequency. The ratio is negative exactly on ${\omega }_d<\omega <{\omega }_n$, giving imaginary wavenumbers and an evanescent wave. Thus the propagating frequency gap is $({\omega }_d,{\omega }_n)$. At ${\omega }_d$ there is no finite-$k$ solution, while ${\omega }_n$ has $k=0$; neither endpoint is an ordinary finite-wavelength propagating mode.

Choose the branch $k=(\omega /c)\sqrt{(\omega -{\omega }_n)/(\omega -{\omega }_d)}$, so the phase velocity is positive in the allowed bands. Direct differentiation gives the phase velocity and group velocityThe opposite spatial branch reverses both signs. On the chosen branch, the requested limits areThe band-edge limits are taken from their propagating sides. More precisely, with $ϵ=\omega -{\omega }_n\downarrow 0$ and $\delta ={\omega }_d-\omega \downarrow 0$,

For the final case, set $K=k_{0}c>0$. Substitution of $T(t)e^{i{k}_{0}x}$ at ${\omega }_d=0$ gives $T^{(3)}+(3iK/2)T^{\prime \prime }+K^2{T}^{\prime }=0$. Its characteristic polynomial factors as $r(r+2iK)(r-iK/2)$, henceThe initial data give $A+B+C=1$, $-2B+C/2=0$ and $-4B-C/4=-5$. Thus $B=1$, $C=4$, $A=-4$, andThe time-independent contribution is essential: static modes lost by cancelling frequency would be missed if one divided the ${\omega }_d=0$ dispersion relation by $\omega$ and retained only its two nonzero-frequency roots.

Put $\mathbf{x}=q{e}^{i\omega t}$ in solution. The normal mode equation is $Aq=\lambda q$, with $\lambda =m{\omega }^2/k$. Its characteristic polynomial isThe three eigenvalues are $2-\sqrt{2}$, $2$, and $2+\sqrt{2}$. Solving the corresponding linear equations yieldsEach mode shape may be multiplied by any nonzero scalar. Their squared Euclidean norms are $4,2,4$, respectively, so unit-normalized eigenvectors are $q_1/2,q_2/\sqrt{2},q_3/2$. They are mutually orthogonal, as expected for the real symmetric stiffness matrix with equal masses.

The entries in the table are angular frequencies; the corresponding ordinary frequencies in cycles per unit time are $f_j={\omega }_j/(2\pi )$. All three are positive because the fixed walls remove uniform translation as a free motion.

A normal mode is a motion of a linear oscillatory system in which every coordinate has the same angular frequency and a fixed relative amplitude and phase. For a conservative small-displacement system it has the formwhere $q$ is the mode shape, $K$ the stiffness matrix, and $M$ the mass matrix. Superposing normal modes gives the general small-oscillation motion when their eigenvectors form a complete basis.

Two common examples are standing transverse oscillations of a string fixed at both ends, and oscillations of coupled masses and springs. The string's mode shapes are sinusoidal along its length; the coupled system's mode shapes are fixed patterns of particle displacements. The following parts determine those patterns for the three-mass system in the source figure.

Assume small planar oscillations about the vertical equilibrium, with the connecting spring relaxed in that equilibrium. Let $x_j=l{\theta }_j$ be the first-order horizontal displacements of the two simple pendulums. The gravitational restoring force on bob $j$ is $-(mg/l)x_j$; the spring extension to first order is $x_2-x_1$. Thus the linearized equations of motion areAdding and subtracting them diagonalizes the system. For $q_{+}=x_1+x_2$ and $q_{-}=x_1-x_2$,The normal modes and angular frequencies are thereforeThe in-phase mode does not stretch the spring, while the opposite-phase mode adds a restoring force of magnitude $2k$ times each bob's displacement. These are the small-oscillation frequencies, not exact amplitude-independent frequencies for arbitrary pendulum angles.

Take constant complex amplitude $A$ and real wavenumber $k$ and angular frequency $\omega$. Substitution of the complex exponential givesThus every nonzero plane wave of the given form solves the equation precisely whenThe wavenumber specifies spatial wave phase change per unit length, with wavelength $\lambda =2\pi /|k|$. The angular frequency specifies temporal wave phase change per unit time, with period $2\pi /|\omega |$ and ordinary frequency $|\omega |/(2\pi )$. The constant $m$ in this equation has dimensions time divided by length squared; without an additional dimensional coefficient, it is not an ordinary physical mass.

For propagating oscillatory solutions assume real $m\ne 0$. If $m=0$, the equation becomes $y_{xx}=0$ and only $k=0$ among these spatial plane waves is possible; the usual dispersion relation branch and velocity discussion would degenerate.

A normal mode of a stable undamped linear coupled oscillator has a single angular frequency and a fixed ratio of component displacements. All components share a sinusoidal time factor, with relative oscillation phases of zero or $\pi$ for real mode shapes. A general free motion is a superposition of the normal modes. For this system the symmetric stiffness matrix gives real eigenvalues and orthogonal eigenvectors; positive spring constants make the squared angular frequencies positive.

Write $m=m_1=m_2>0$ and $\kappa =k_{12}$. Assume ideal massless linear springs, negligible damping and positive $k_1,k_2,\kappa$. By Hooke's law and Newton's second law,The coupling spring extends by $x_2-x_1$, giving equal and opposite coupling forces.

For a normal mode, take $\mathbf{x}=\mathrm{Re}(\mathbf{a}{e}^{i\omega t})$. Nonzero eigenvectors requireSolving this quadratic equation in $m{\omega }^2$ givesA corresponding mode shape has ratioand the general real solution is $\mathbf{x}=C_{-}{\mathbf{a}}_{-}\cos ({\omega }_{-}t+{\varphi }_{-})+C_{+}{\mathbf{a}}_{+}\cos ({\omega }_{+}t+{\varphi }_{+})$. The arbitrary amplitudes and oscillation phases account for four independent initial data.

Weak coupling alone does not guarantee large beats. A useful representative sketch assumes nearly equal uncoupled angular frequencies. In the resonant case $k_1=k_2=k$ with $\kappa \ll k$, the mode shapes are $(1,1)$ and $(1,-1)$, andFor $x_1(0)=A$, $x_2(0)={\dot{x}}_1(0)={\dot{x}}_2(0)=0$, both normal modes have equal coefficients. With $\bar{\omega }=({\omega }_{+}+{\omega }_{-})/2$ and $\Delta \omega ={\omega }_{+}-{\omega }_{-}$,These rapid oscillations have slowly varying complementary wave envelopes. The initial displacement shifts almost entirely to the other mass at $t=\pi /\Delta \omega$, and returns after $2\pi /\Delta \omega$. Arbitrary initial data give a different superposition; if $|k_1-k_2|\gg \kappa$, the normal modes are localized mainly on individual masses, and strong exchange is suppressed.

Holding one mass fixed leaves the coupling spring attached and stretched by the other mass. Therefore the two constrained angular frequencies areThese differ from the angular frequencies obtained by physically removing the coupling spring. Ordinary frequencies in hertz are each corresponding $\omega /(2\pi )$.

Finally, for $k_1>k_2$ and $\kappa >0$, put $D=\sqrt{(k_1-k_2{)}^2+4{\kappa }^2}>k_1-k_2$. ThenSince ${\omega }_1>{\omega }_2$, taking positive square roots provesAt $\kappa =0$, the outer inequalities become equalities.

To find the normal modes of two arc-spring-coupled beads on a hoop, take $x_i=a_i{e}^{i\Omega t}$ for a normal mode. The resulting generalized eigenvalue problem isIts determinant isThus the angular frequencies areThe ordinary nonzero frequency is $\omega /(2\pi )$. The zero mode has mode shape $(1,1)$ and represents a common rotation with no change in spring length. The oscillatory mode shape is $(1,-1/q)$.

Apply the product-to-sum formula $2\cos u\cos v=\cos (u+v)+\cos (u-v)$ to the amplitude modulation:The amplitude modulation sidebands have angular frequencies $\omega +{\omega }_m$ and $\omega -{\omega }_m$, each with amplitude $b/2$ in this real cosine convention. The carrier wave has angular frequency $\omega$ and amplitude $a$.

This identity is exact for the supplied space-time expression. A physical qualification is needed if every term is meant to be a freely propagating electromagnetic wave with speed of light $c$: the dispersion relation requires $k_{\pm }=(\omega \pm {\omega }_m)/c$, whereas the prescribed modulation at every position leaves all three wavenumbers equal to $k=\omega /c$. A freely propagating modulated wave instead has envelope $a+b\cos ({\omega }_m(t-x/c))$; its analogous expansion has $k_{\pm }$. For example, at $x=t=0$ the prescribed expression has $A_{tt}-c^2{A}_{xx}=-b{\omega }_m^2$, which is nonzero for nontrivial modulation. The given formula can describe modulation imposed throughout space or the temporal signal at one position.

At $\alpha =1$, the supplied and physical systems agree. Their squared angular frequencies are $(g/l)(2\pm \sqrt{2})$. The upper row gives $a_2/a_1=3-\lambda$, soIn the slower normal mode, both masses move horizontally in phase and the lower has the larger amplitude. In the faster normal mode, the lower mass moves in the opposite direction, with magnitude $(\sqrt{2}-1)|a_1|$. These are horizontal displacement ratios. The rod angles instead satisfy ${\theta }_2/{\theta }_1=\pm \sqrt{2}$, because $x_1=l{\theta }_1$, $x_2=l({\theta }_1+{\theta }_2)$; confusing these ratios changes the sketch.

For the supplied equations, ${\lambda }_{+}\to 1$ and $a_2/a_1=\alpha /(\alpha -{\lambda }_{+})\sim -\alpha$. Thus the higher-frequency normal mode becomes an upper-mass oscillation at $\sqrt{g/l}$ with a vanishing, opposite-sign lower horizontal displacement. At $\alpha =0$ the supplied lower equation is ${\ddot{x}}_2=0$ and its other mode has zero frequency.

This limiting picture belongs to the given matrix. For the physical unequal-mass double pendulum, both normal mode angular frequencies tend to $\sqrt{g/l}$ as the lower mass tends to zero; at exactly zero mass its independent equation needs a limiting test-body interpretation. The source's extra $\alpha$ changes precisely this conclusion.