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Place the massless junction at $x=0$, with common tension $T$ on both sides. Using a common time factor $e^{-i\omega t}$, write the left-hand wave as $A_i{e}^{i{k}_{1}x}+A_r{e}^{-i{k}_{1}x}$ and the right-hand wave as $A_t{e}^{i{k}_{2}x}$. A stationary junction does not change the frequency, while $k_j=\omega \sqrt{{\rho }_j/T}$.

displacement continuity requires $A_i+A_r=A_t$. Transverse force balance on a massless junction requires equal tension-weighted slopes, giving $k_1(A_i-A_r)=k_2{A}_t$. Hence the signed displacement coefficients areHere ${\rho }_2=4{\rho }_1$, so $k_2=2k_1$ and $\boxed{A_r=-A_i/3,A_t=2A_i/3}$. The negative reflected coefficient signifies inversion. If “reflection and transmission coefficients” instead denotes power fractions, their values are $\mathcal{R}=1/9$ and $\mathcal{T}=8/9$, as derived separately below; power and displacement coefficients must not be conflated.

The reflected wave remains on the incident string, so its wavenumber magnitude is still $k_1$. Its outward power magnitude isThus $|P_r|+P_t=(1/9+8/9)P_i=P_i$, verifying energy conservation. With signed positive-$x$ fluxes, $P_r=-P_i/9$ and the same statement is $P_i+P_r=P_t$. The minus sign of the reflected displacement is a phase reversal, whereas the minus sign of its flux records travel in the opposite direction; these are different uses of a sign.

The transmitted and incident waves have the same angular frequency and tension, but the transmitted wavenumber is twice as large. The time-average string-power formula therefore givesThe extra factor $k_2/k_1$ is essential. Squaring the displacement transmission coefficient alone would give $4/9$, which is not the transmitted power fraction in strings with different linear mass densities.

If ${\rho }_1\gg {\rho }_2$, then $k_1\gg k_2$ and the junction behaves approximately as a free end for the incident string. The limits areThe reflected displacement is not inverted, and at the junction it adds in phase to the incident displacement, producing nearly twice its amplitude. The transmitted displacement is also in phase. A transmission amplitude near two does not create power: the lighter string has much smaller impedance, and $\mathcal{T}=(k_2/k_1)t^2\simeq 4k_2/k_1\to 0$. These are limiting statements for positive densities, not an ordinary propagating-wave model for an exactly massless second string.

If ${\rho }_1\ll {\rho }_2$, then $k_1\ll k_2$ and the wave meets a much larger string impedance. The displacement and power coefficients approachThe junction behaves approximately as a fixed end. The reflected displacement reverses sign, corresponding to a phase shift of $\pi$; the small transmitted displacement is not inverted. For a large but finite density ratio, $t\simeq 2k_1/k_2$ and the transmitted power fraction is about $4k_1/k_2$, tending to zero rather than vanishing identically at every finite ratio.

For equal linear mass densities and common tension, $k_1=k_2$. The junction relations then giveThere is no mechanical-impedance discontinuity, so no reflected wave is needed to satisfy displacement and force continuity. The transmitted displacement is in phase with the incident wave and has the same amplitude; all incident power crosses the junction.

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.

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 $M=2m$, 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.

Let $\delta =x_2-x_1$. The inextensible-string constraint gives the upward displacement $y_2=l-\sqrt{l^2-{\delta }^2}={\delta }^2/(2l)+O({\delta }^4/l^3)$. Thus the quadratic potential energy is $U=k{x}_1^2/2+mg(x_2-x_1{)}^2/(2l)$. The vertical velocity of the second mass is of second order in the small motion, so the quadratic kinetic energy is $m({\dot{x}}_1^2+{\dot{x}}_2^2)/2$.

The Euler-Lagrange equations for this quadratic Lagrangian giveSmall displacements are used both in the quadratic approximation to the gravitational potential energy and in neglecting the higher-order vertical kinetic energy. These are the spring-supported pendulum normal modes at linear order. The corresponding force picture sets the string tension to $mg$ at leading order and its horizontal angle to $\delta /l$.

Define $D=\sqrt{{\kappa }^2/4+q^2}$. Normalizing the first displacement amplitude to one, either row of the eigenvector equation givesThe two normal modes are $x(t)=A_{\pm }{v}_{\pm }\cos ({\omega }_{\pm }t+{\varphi }_{\pm })$, and arbitrary small motion is their superposition. The lower-frequency mode has $r_{-}>0$, so the masses move in phase; the upper-frequency mode has $r_{+}<0$, so they move in opposite phase. The identity $r_{+}{r}_{-}=-1$ verifies orthogonality of these eigenvectors for equal masses.

The product-to-sum identity gives $\sin \Omega t\sin \omega t=[\cos (\Omega -\omega )t-\cos (\Omega +\omega )t]/2$. Thus the first mass must have equal and opposite cosine-mode amplitudes and no sine-mode amplitudes. Equivalently, release the second mass from a displacement $X$ while the first is at equilibrium, both initially at rest:The two normal mode coefficients in $x_1$ are $qX/(2D)$ and $-qX/(2D)$. Therefore, withthe two displacements areThese initial conditions are necessary for a nontrivial product with exactly the stated time origin, as well as sufficient: $x_1(0)={\dot{x}}_1(0)=0$ follows from the product, and its zero third derivative at zero, together with the first coupled equation, forces ${\dot{x}}_2(0)=0$. Since both normal frequencies are positive, $\Omega >\omega >0$. This is the product form used to describe beating, although a well-separated slow envelope requires the two normal frequencies to be close. No such extra near-degeneracy is assumed in the exact formula.

Write the initial triangular pulse as the function $h$, extended by zero to negative arguments:At $t=0$, the initial displacement and zero velocity give $f+g=h$ and $-c{f}^{\prime }+c{g}^{\prime }=0$ on the positive half-line. Fix the harmless opposite additive constants by taking both travelling waves to vanish at positive infinity. ThenThe second formula follows from the fixed-end boundary condition $f(-ct)+g(ct)=0$. Thus $f$ is one-half of the odd extension $H$ of the initial displacement, andThis is the odd-extension solution for a semi-infinite fixed string: the reflected displacement reverses sign. Evaluating the travelling profiles givesThe first profile is a positive triangle supported on $[L,3L]$, with peak $a/2$ at $2L$. The second has a negative triangle on $[0,2L]$ with minimum $-a/2$ at $L$, and a positive triangle on $[2L,4L]$ with maximum $a/2$ at $3L$.

Let $x_1,x_2,x_3$ be longitudinal displacements from equilibrium, with positive direction to the right. The fixed walls have zero displacement. The first mass feels forces $-k{x}_1$ and $k(x_2-x_1)$; the central mass feels $k(x_1-x_2)$ and $k(x_3-x_2)$; the last mass feels $k(x_2-x_3)$ and $-k{x}_3$. Hooke's law and Newton's second law therefore giveEquivalently,All four springs contribute: the wall springs supply one of the two diagonal terms in the first and third rows. This is a fixed-end chain, not a freely ended chain with a zero-frequency translation mode.

To specify “initial motion” unambiguously, choose each mode to pass through equilibrium at $t=0$ with $\dot{\mathbf{x}}(0)$ proportional to its listed mode shape. The arrows in the figure then describe relative initial velocities:

Reversing every arrow changes only the common phase. A quarter-cycle later these same ratios describe the extremal displacements, with zero instantaneous velocity.

Now replace the central mass by $M$, leaving the outer masses equal to $m$. The antisymmetric mode shape $q=(1,0,-1{)}^T$ gives zero net force on the central mass, since $k{x}_1+k{x}_3=0$. Each outer mass still obeys $m\ddot{x}=-2kx$. ConsequentlyThis middle-mass-independent mode of a fixed three-mass spring chain is exact, not merely a large-$M$ approximation. The other two modes move the central mass and generally change. For example, their symmetric-sector frequencies satisfywhich explicitly depends on $M$.

Work in the small-slope model of a transverse wave on a taut string. Let $T$ be the constant tension, $\rho$ the linear mass density, and take a right-moving sinusoidal displacement $y=A\cos (\kappa x-\omega t)$, with wave number $\kappa =\omega /v$. The energy flux of a transverse string wave through a cut is the transverse force times transverse velocity:A cycle average gives $⟨{\sin }^2⟩=1/2$. Using $T=\rho v^2$, obtained from the string wave equation, yieldsThe sign in the flux formula makes the power positive for propagation in the positive $x$ direction.

The energy per unit equilibrium length is the sum of the kinetic and elastic terms:For a segment of mass $M=\rho \ell$ containing many wavelengths, the spatial average agrees with the cycle average up to a small endpoint correction. Thus $E\simeq \frac{1}{2}M{\omega }^2{A}^2$. Convert the given quantities to SI units: $M=0.080\mathrm{kg}$, $A=1.6\times 10^{-4}\mathrm{m}$, and $\omega =2\pi (120){\mathrm{s}}^{-1}$. ThenThe many-wavelength condition explains why no phase of the finite segment is needed. The result is exact as a cycle average, or for a segment containing an integer number of wavelengths.

The wave equation on a string gives $c_j=\sqrt{T/{\rho }_j}$ and $\omega =c_j{k}_j$, henceThe tension is common across the join, and the frequency is unchanged by reflection and transmission at the stationary boundary.

Use the complex time convention $e^{i\omega t}$. The physical displacement is the real part ofThe first term is incident from the left, the second reflected to the left, and the right-side term transmitted to the right. Continuity of displacement gives $A+B=C$. At a massless join, continuity of the transverse force $T{y}_x$ gives $k_1(A-B)=k_{2}C$. Solving these two equations yieldsThese are displacement-amplitude coefficients for reflection at a density step on a string, not power fractions. Since the mean incident power is $T\omega k_1|A{|}^2/2$, the reflected and transmitted power fractions are $|r{|}^2$ and $(k_2/k_1)|t{|}^2$. They sum to one, as expected for the lossless join.

Convert the fixed position to $x=0.20\mathrm{m}$. The element's displacement is then $y(t)=0.27\sin (2.4-500t)\mathrm{mm}$. Taking time derivatives at fixed position gives its transverse velocity and acceleration:Equivalently the prefactors in SI units are $-0.135\mathrm{m}{\mathrm{s}}^{-1}$ and $-67.5\mathrm{m}{\mathrm{s}}^{-2}$. The phase is measured in radians. At $t=4\mathrm{s}$,These are the transverse motions of the material element, not the propagation speed of the wave along the string.

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.