Codex @codex  0

Put $k_1=k_3=k$ and $m_1=m_2=m$. The mass-normalized stiffness matrix isIts symmetric eigenvector $(1,1)$ has eigenvalue $k/m$, while its antisymmetric eigenvector $(1,-1)$ has eigenvalue $(k+2k_2)/m$. ThusThe symmetric frequency does not depend on the middle spring constant because that spring has no extension in an in-phase normal mode. The antisymmetric motion stretches the middle spring and has the higher frequency for $k_2>0$.

Let $x_1,x_2$ be displacements from equilibrium, with positive directions chosen alike. Hooke's law and Newton's second law giveFor a normal mode $x_j=a_j\cos \Omega t$, the determinant condition isSolving the quadratic for ${\Omega }^2$ givesThe angular frequencies are the positive square roots. Positive masses and spring constants make the stiffness quadratic form $k_1{x}_1^2+k_2(x_2-x_1{)}^2+k_3{x}_2^2$ positive definite, ensuring both squared frequencies are positive. Distinct spring constants do not require equal masses or any symmetry approximation.

Seek a separated normal mode $y(x,t)=X(x)Q(t)$. Substitution in the wave equation gives $X{Q}^{\prime \prime }=c^2{X}^{\prime \prime }Q$. For an oscillatory mode choose the separation constant $-{\omega }^2$:The fixed endpoints impose $X(0)=X(l)=0$. In $X=A\sin kx+B\cos kx$, the first condition gives $B=0$, and the second requires $\sin kl=0$. Nonzero displacement therefore requiresEach spatial shape $\sin (n\pi x/l)$ has fixed nodes and an oscillating overall amplitude: a standing wave. A general lossless motion is a superposition of these standing normal modes, not necessarily one single-frequency standing wave. The $n=0$ solution is trivial under both fixed-end conditions and does not supply an additional vibrational mode.

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.

The original diagram has three horizontal spans $L$, $2L$ and $L$. In the small-slope approximation the massless string segments are straight and their transverse force components are tension times slope. The horizontal tension components cancel to first order.

For the first bead, the left segment contributes $-T{y}_1/L$ and the middle segment contributes $T(y_2-y_1)/(2L)$. For the second bead, the middle contribution is $T(y_1-y_2)/(2L)$ and the driven right segment contributes $T(h-y_2)/L$. Newton's second law therefore givesWith ${\omega }_0^2=T/(2ML)$ and $h=h_0\cos \omega t$, the coupled equations of motion areThe factor two in the driving term follows from the right span having length $L$, while the middle span has length $2L$. Treating all three spans as equal would change both the coupling and the normal mode frequencies.

The natural normal modes belong to the homogeneous system with both end displacements fixed at zero. Seek $\mathbf{y}=\mathbf{a}\cos \Omega t$, soThe characteristic polynomial is $(3{\omega }_0^2-{\Omega }^2{)}^2-{\omega }_0^4=0$. The eigenvectors $(1,1)$ and $(1,-1)$ therefore giveThe symmetric mode leaves the middle segment horizontal, while the antisymmetric mode bends it strongly and has the higher restoring stiffness. Defining $q_{+}=(y_1+y_2)/2$ and $q_{-}=(y_1-y_2)/2$ diagonalizes the driven equations too:This shows that the boundary drive excites both natural modes; neither natural resonance is removed by a zero drive projection.

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.

For either normal mode, let $k_n=n\pi /L$ and write its contribution as $B_n\sin (k_{n}x)\sin ({\omega }_{n}t)$. The product-to-sum identity givesThis standing wave is a superposition of two travelling waves of equal amplitude $|B_n|/2$, moving oppositely with phase velocity ${\omega }_n/k_n=c$. Their cancellation at $x=0,L$ enforces the fixed-end boundary conditions at every time.

For small transverse displacements, the string's kinetic energy and extra elastic potential energy areThe orthogonality of trigonometric functions removes cross terms between the two normal modes. For a mode $B_n\sin (k_{n}x)\sin ({\omega }_{n}t)$, use $T{k}_n^2=\rho {\omega }_n^2$ and ${\int }_0^L{\sin }^2(k_{n}x)dx={\int }_0^L{\cos }^2(k_{n}x)dx=L/2$ to obtainConsequently the modal energy distribution isThe second mode contributes more total energy at every time, despite its smaller displacement amplitude. The total energy is $13\rho L{V}_0^2/64$, with fractions $4/13$ and $9/13$ in the first and second modes. Each mode exchanges kinetic energy and potential energy, while its sum is conserved; the comparison is of these sums, not of instantaneous kinetic energies alone.

A normal mode has a fixed pattern of relative displacements, with every coordinate proportional to one common time-dependent factor. There are three independent normal modes in this three-coordinate model: two internal vibrations and one zero-frequency rigid-body translation. The latter moves all three atoms together without stretching either spring. This count concerns the specified one-dimensional model, rather than a molecule with unrestricted three-dimensional motion.

For the zero normal mode, $x_1=x_2=x_3=A+Bt$. The molecule translates uniformly; a common constant displacement is the special case $B=0$. For the lower vibrational normal mode, put $\omega =\sqrt{k/m}$:The end atoms move oppositely with equal amplitudes, while the central atom remains stationary. In the higher vibrational normal mode, with $\Omega =\sqrt{k/m+2k/M}$,The ends move together and the centre moves in the opposite direction, with central amplitude $2m/M$ times an end amplitude. Both internal modes keep the centre of mass fixed and obey mass-weighted orthogonality. The zero mode instead describes motion of that centre of mass.

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 $f_{\pm }={\omega }_{\pm }/(2\pi )$.

The independent normal modes span the two displacement coordinates. With arbitrary real constants $A,B,C,D$, the general real solution isFour constants are required for two second-order ordinary differential equations, corresponding to two initial displacements and two initial velocities.

Release from rest gives $B=D=0$. At $t=0$, $A+2C=d$ and $A-C=0$, so $A=C=d/3$. ConsequentlyThe frequency ratio is exactly two, so both motions repeat after $2\pi /{\omega }_{-}$. These are commensurate normal modes, rather than a pair of nearly equal frequencies producing slow beats.

A standing wave has a fixed spatial amplitude profile and oscillates in time, with stationary nodes of a standing wave. It arises from equal-amplitude waves of the same frequency travelling in opposite directions. For a uniform string with linear mass density $\mu$, the wave speed is $c=\sqrt{T/\mu }$ and a separated mode has $Y^{\prime \prime }+k^{2}Y=0$.

The fixed boundary requires $Y(0)=0$, while the free boundary requires $T{Y}^{\prime }(L)=0$. Thus $Y=A\sin kx$ and $\cos kL=0$, givingThe fixed end is a node of a standing wave and the free end an antinode. Arbitrary small motions are superpositions of these normal modes; the tension alone determines their frequencies only when $\mu$ is also specified.

In one normal mode, all moving bobs share the same frequency and fixed amplitude ratios. A real mode vector allows their phases to be chosen equal or differing by $\pi$, according to the signs of its components. A component equal to zero means that bob remains at rest in that mode and has no oscillation phase to assign.

With zero initial velocities, choosing the initial displacement proportional to one eigenvector excites only its normal mode. Suitable ratios areThe overall scale must remain small compared with $l$. Release each configuration from rest; its subsequent vector is its initial displacement multiplied by $\cos {\omega }_{j}t$. The middle bob is stationary in the second mode, while it moves oppositely with twice either outer amplitude in the third.

Free ends give $Y^{\prime }(0)=Y^{\prime }(L)=0$, so the oscillatory normal modes areThere is also the zero mode $k_0=0$, with uniform displacement $y_0(x,t)=a+bt$: a free string can translate without a restoring force. It is not a positive-frequency oscillation. The first three oscillatory shapes are $\cos (\pi x/L)$, $\cos (2\pi x/L)$ and $\cos (3\pi x/L)$, with an antinode at each end. The figure includes the uniform mode too, so that the zero-frequency limit is explicit.