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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 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.

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.

Consider a string element of equilibrium length $dx$. Its mass is $\rho dx$, where $\rho$ is mass per unit length. Under the small-slope approximation, the tension is approximately uniform and its transverse component at an endpoint is $T\partial y/\partial x$. The two endpoint forces have net transverse componentNewton's second law gives $\rho dx{y}_{tt}=Tdx{y}_{xx}$. Cancelling the element length yieldsThis is the one-dimensional wave equation for transverse displacement. Its derivation neglects gravity, bending stiffness and nonlinear changes in tension and segment length; the given small slope makes those omissions consistent to first order in the displacement.

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.

At the same tension, $k\propto \sqrt{\rho }$, so $k_2/k_1=3$, $t=1/2$ and $r=-1/2$. The transmitted power fraction is not $t^2$, because the two strings have different wave impedances. Using the energy flux of a transverse string wave,The reflected fraction is $\mathcal{R}=|r{|}^2=1/4$, so $\mathcal{R}+\mathcal{T}=1$. The transmitted displacement amplitude is $2.5\mathrm{mm}$ and its wavelength is $0.5/3\mathrm{m}$. The negative $r$ indicates a reflected phase reversal; it does not mean negative reflected energy.

The three nearly horizontal segments have lengths $2a,a,a$. To first order in small transverse slopes, the upward force on the first point mass is $T[(y_2-y_1)/a-y_1/(2a)]$. The force on the second is $T[(y_1-y_2)/a-y_2/a]$. Applying Newton's second law to masses $m,2m$ therefore givesThe approximation treats the string as massless, retains the imposed equilibrium tension and neglects gravity, as stipulated.

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.

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 $\delta /l$.

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.

At distance $x$ below the support, the cable must carry the weight of the remaining length $L-x$. Thus its equilibrium tension is $T(x)=\mu g(L-x)$, and the local transverse wave speed isThe linear mass density cancels. More precisely the small-displacement equation with variable tension is $\mu y_{tt}={\partial }_x[T(x)y_x]$; its local characteristic speed is the expression above, even though a first-spatial-derivative term is also present.

The wave travel time on a hanging uniform cable is obtained from $dt=dx/c(x)$:Although the local speed tends to zero at the free end, the endpoint singularity in the travel-time integral is integrable. This finite result belongs to the ideal flexible-cable model, with bending stiffness neglected.

Assume small transverse slopes, a flexible string, constant tension $T$ and uniform linear mass density $\rho$. For an element of equilibrium length $dx$, the vertical force is the difference of the vertical tension components at its ends. To first order in slope,Its mass is $\rho dx$ and its transverse acceleration is $y_{tt}$. Newton's second law therefore gives $\rho dx{y}_{tt}=Tdx{y}_{xx}$, orThis is the one-dimensional wave equation for small-amplitude transverse waves. Changes in string length and horizontal acceleration are second order in slope and are neglected in this linear approximation.

Take an ideal flexible mechanical string with constant tension $T>0$ and uniform linear mass density $\mu >0$. Let $x$ measure distance along its straight equilibrium position and let $y(x,t)$ be its small transverse displacement. Neglect damping, bending stiffness and external distributed forces, and assume $|y_x|\ll 1$.

A segment of reference length $\Delta x$ has mass $\mu \Delta x$. The endpoint tension forces have transverse resultantsince $\sin \theta \simeq \tan \theta =y_x$ for small slopes. Newton's second law, divided by $\Delta x$ and taken to the limit, givesHere $t$ is time, the subscripts denote partial derivatives, and $c$ is the wave speed. The transverse acceleration is $y_{tt}$; $y_{xx}$ describes the small-slope curvature responsible for the net transverse force.

Put $\xi =x+vt$ and $F(\xi )=A{e}^{-{\xi }^2/x_0^2}$, with $x_0\ne 0$. The chain rule givesFor a nonzero pulse the wave equation on a string therefore requiresIt is not a solution for arbitrary $v$; the zero pulse $A=0$ is the trivial exception. Here $\mu$ is linear mass density, and $T>0$ is tension. With $v>0$, the maximum lies at $x=-vt$, so the Gaussian function translates towards decreasing $x$ by $v(t_2-t_1)$ between the two times. Its amplitude and width stay constant.

A reference string segment $dx$ has mass $\mu dx$ and transverse velocity $y_t$. Its kinetic energy is $dK=\mu y_t^{2}dx/2$. In the small-slope approximation its extra arc length isThe work against constant tension $T$ supplies the corresponding elastic potential energy $dU=T(d\ell -dx)$. To quadratic order, the energy density per unit reference length is thereforeFor the travelling wave $y=A\cos (kx-\omega t+\varphi )$, the wave equation implies ${\omega }^2=(T/\mu )k^2$. Its kinetic energy density and elastic energy density are equal, soThe energy per wavelength of a transverse string wave comes from integration over any full wavelength $\lambda =2\pi /|k|$:The printed expression for energy per wavelength is missing a factor $\lambda$. Its right-hand side is the mean energy density, with units of energy per length. For a concrete counterexample within the small-slope regime, take $\mu =1\mathrm{kg}{\mathrm{m}}^{-1}$, $T=1\mathrm{N}$, $A=0.01\mathrm{m}$ and $\omega =\pi {\mathrm{s}}^{-1}$. Then $\lambda =2\mathrm{m}$ and direct integration gives $E_{\lambda }=10^{-4}{\pi }^2\mathrm{J}$, whereas the printed right-hand side is $5\times 10^{-5}{\pi }^2\mathrm{J}{\mathrm{m}}^{-1}$. It cannot equal an energy.

Assume $\rho$ is mass per unit equilibrium length, the tension has constant magnitude $T$, the string has negligible bending stiffness, and there is no damping or external transverse force. Assume small slopes $|y_x|\ll 1$ and purely transverse motion. Then a short segment has mass $\rho \Delta x$ to first order. If $\alpha$ is the angle of its tangent, the small-slope approximation gives $\sin \alpha \simeq \tan \alpha =y_x$.

The vertical components of tension at the two ends give net transverse force $T[y_x(x+\Delta x,t)-y_x(x,t)]$. Newton's second law therefore givesto first order. Dividing by $\Delta x$ and taking its limit to zero yieldsThus $c$ is the propagation speed of small transverse waves.

Take the pivot as origin, with $x$ horizontal and $y$ vertically upward. For the simple pendulum, putThe velocity and acceleration areThe forces are the tension $-T{\mathbf{e}}_r$ and gravity $(0,-mg)$. Taking the dot product of Newton's second law with ${\mathbf{e}}_{\theta }$ eliminates the tension and givesNear the downward stable equilibrium $\theta =0$, the small-angle approximation $\sin \theta \simeq \theta$ gives $\ddot{\theta }=-(g/\ell )\theta$ to first order. Since $x=\ell \sin \theta \simeq \ell \theta$, its linearization isThis is simple harmonic motion with angular frequency $\sqrt{g/\ell }$; the restoring sign confirms the stable equilibrium.

Use $g$ as the gravitational acceleration, so the weights are $mg$ and $\sqrt{2}mg$, consistently with the requested equations. The printed introductory phrase calling $g$ a force magnitude omits the factor $m$.

For the particle on a cone coupled to a hanging mass, write $z=-r$ and use the cylindrical coordinates unit vectors ${\mathbf{e}}_r,{\mathbf{e}}_{\theta },{\mathbf{e}}_z$. The ideal string segment from the cone particle to the hole has length $\sqrt{2}r$ and direction $(-{\mathbf{e}}_r+{\mathbf{e}}_z)/\sqrt{2}$. If $h$ is the downward length of the hanging segment and $s_0$ the fixed total length, thenThe light ideal string and smooth hole give the same tension $T$ on both sides. Applying the vertical component of Newton's second law to the hanging particle givesWrite the normal reaction on the cone particle as $N({\mathbf{e}}_r+{\mathbf{e}}_z)/\sqrt{2}$, allowing $N$ to be signed according to the side of the ideal constraining surface. The acceleration in cylindrical coordinates and Newton's second law giveThere is no azimuthal force. Multiplying the middle equation by $r/m$ proveswhere the positive angular direction is chosen along the initial velocity. This is conservation of angular momentum about the vertical axis.

Subtract the radial equation from the vertical equation to eliminate the normal reaction. Since $\ddot{r}=-\ddot{z}$, this givesSubstitute the tension and the conserved angular momentum, using $r=-z$:Multiplication by $\dot{z}$ now gives an exact time derivative:Initially $z=-a$ and $\dot{z}=0$, because the initial velocity is horizontal. ThereforeThis is also total mechanical energy divided by $m$, with an irrelevant constant removed: the two particles contribute $(1+\sqrt{2})m{\dot{z}}^2$ to the radial and vertical kinetic energy, while their combined gravitational potential energy is $-mgz$ plus a constant.

Let the uniform string have constant tension $\tau >0$ and linear mass density $\mu >0$. Assume small transverse motion with $|u_x|\ll 1$, negligible longitudinal motion, and no external transverse force. If the string makes angle $\beta$ with the horizontal, then $\tan \beta =u_x$, and the small-angle approximation gives $\sin \beta \simeq u_x$ and $ds\simeq dx$ to first order.

The net upward force on the segment from $x$ to $x+\Delta x$ is thereforeBy Newton's second law, its mass $\mu \Delta x$ times its transverse acceleration equals this force, to the order retained in the small displacement model. Divide by $\Delta x$ and let $\Delta x\to 0$ to obtainThus the wave equation on a string iswhere $c$ is the wave speed. The horizontal force balance makes the horizontal component of tension constant; to first order it equals $\tau$.