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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.

Measure $x_1,x_2$ as horizontal displacements from the respective downward equilibrium positions; the absolute horizontal coordinates are $x_1$ and $L+x_2$. Each vertical displacement is of second order in its small angle. Thus the spring is horizontal to first order, its extension is $x_2-x_1$, and Hooke's law gives horizontal forces $k(x_2-x_1)$ and $-k(x_2-x_1)$. Combining these with the linearized simple pendulum restoring force yieldsFor the normal modes of two spring-coupled equal pendulums, add and subtract these ordinary differential equations. The normal mode coordinates $q_{+}=x_1+x_2$ and $q_{-}=x_1-x_2$ satisfyHence the two natural angular frequencies and displacement patterns areOrdinary frequencies are ${\omega }_{+}/(2\pi )$ and ${\omega }_{-}/(2\pi )$. In the first normal mode, both particles swing together with equal amplitudes and the spring is unextended to first order. In the second normal mode, the particles swing in opposite directions with equal amplitudes, so the spring adds a restoring force and increases the angular frequency.

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.

For polar coordinates, the rotating unit vectors satisfy ${\dot{\mathbf{e}}}_r=\dot{\theta }{\mathbf{e}}_{\theta }$ and ${\dot{\mathbf{e}}}_{\theta }=-\dot{\theta }{\mathbf{e}}_r$. Applying the product rule twice to $\mathbf{r}=r{\mathbf{e}}_r$ gives the velocity and accelerationNewton's second law for a central force now givesMultiplying the azimuthal force equation by $r/m$ yieldsThis is conservation of specific angular momentum. Here $F(r)$ is the signed radial component of the force; a negative value points inward.

When $h\ne 0$, $\theta$ is a valid local parameter along the trajectory. Put $u=1/r$ and let primes denote derivatives with respect to $\theta$. The chain rule yieldsSubstituting into the radial Newton's second law equation and dividing by $-m{h}^2{u}^2$ proves the Binet equation:

The free-body diagram shows a meridional section of the right circular cone. On the first particle, gravity acts vertically downwards, tension points along the straight line to the apex, and the normal reaction is perpendicular to the surface. The smooth mechanical constraint supplies no tangential friction. The second particle has only its downward gravitational weight and the upward tension.

Adding the forces on each particle and applying Newton's second law givesAt $z_1=-r_1$, a normal vector is ${\mathbf{e}}_r+{\mathbf{e}}_z$. Its sign depends on the supporting face of the cone; either sign has zero dot product with every permitted velocity. The diagram chooses the reaction towards the upper side of the surface.

The massless inextensible string has a common tension magnitude $T$. Since ${\mathbf{r}}_1=r_1({\mathbf{e}}_r-{\mathbf{e}}_z)$, the unit vector from the first particle towards the apex is $(-{\mathbf{e}}_r+{\mathbf{e}}_z)/\sqrt{2}$. The second particle lies below the apex, soThe string length constraint isThe velocity of the first particle in cylindrical coordinates is ${\dot{r}}_1{\mathbf{e}}_r+r_1\dot{\theta }{\mathbf{e}}_{\theta }+{\dot{z}}_1{\mathbf{e}}_z$. Therefore the powers delivered by the two tension forces areThusThe tension can transfer energy between the particles, but its total power on the pair is zero.

All forces on the first particle lie in the plane spanned by ${\mathbf{e}}_r$ and ${\mathbf{e}}_z$: this holds for gravity, tension, and the normal reaction. Hence the azimuthal component of Newton's second law vanishes:Using the product rule givesStrictly, $h$ is the axial specific angular momentum; the actual axial angular momentum is $m_{1}h$, which is conserved as well.

Differentiate the stated total energy using the chain rule and then substitute the two Newton's second law equations:The gravitational potential energy terms have canceled the power of gravity. Also $\mathbf{N}$ is proportional to ${\mathbf{e}}_r+{\mathbf{e}}_z$, whose dot product with ${\dot{\mathbf{r}}}_1$ is ${\dot{r}}_1+{\dot{z}}_1=0$. Thus the stationary smooth mechanical constraint does no work. The remaining tension powers cancel by part (b), proving $\dot{E}=0$ and conservation of total energy while the massless inextensible string remains taut and contact is maintained.

Put $D=m{R}^2+{\mathcal{I}}_{33}=m(R^2+a^2/2)$. The angular acceleration is constant. The initial rest conditions and $x(0)=\varphi (0)=0$ giveFor the upward rightward pull in the diagram, $0\le \alpha \le \pi /2$, the sign of $x(t)$ is the sign of $R\cos \alpha -a$. Consequently it rolls towards the horizontal pull when $0\le \alpha <\arccos (a/R)$. At $\alpha =\arccos (a/R)$ it remains at rest, with balanced forces and torques. For $\arccos (a/R)<\alpha \le \pi /2$, it rolls to the left. In particular a horizontal pull, $\alpha =0$, makes it roll to the right, since $R>a$. If other pulling angles are allowed, the general condition for motion in the same horizontal direction as the pull is $\cos \alpha (R\cos \alpha -a)>0$.

The required static friction at each rail follows from the translational equation:For $0\le \alpha \le \pi /2$ it is positive, agreeing with the leftward force convention. The assumed rolling without slipping is physically possible only if the rails can supply this force while maintaining contact. With coefficient of static friction ${\mu }_s$, the necessary contact inequalities areFor the usual pulling range, these becomeIf either fails, the proposed motion requires lift-off or more static friction than the rails provide. Equality $T\sin \alpha =mg$ is also incompatible here: it gives $N=0$ but a nonzero required friction. The question supplies no coefficient of static friction, so it determines the required friction exactly but gives no numerical slip threshold. If $F$ is regarded as a prescribed force rather than an available contact response, it must equal $F_{\mathrm{req}}$ for the rolling constraint to hold.

The initial transverse velocity gives $h=a(L/a)=L$. From $\dot{r}=-h{u}^{\prime }$, the initial conditions are $u(0)=1/a$ and $u^{\prime }(0)=1/a$. DefineThe Binet equation becomes the constant-coefficient linear ordinary differential equationIts particular solution is $({\nu }^2-1)/(a{\nu }^2)$. Applying the two initial conditions to the complementary solution givesThe physical branch is the connected range containing $\theta =0$ on which the right-hand side is positive. A zero corresponds to escape to infinite radius, rather than continuation to negative $r$.

The source calls $F$ an inward force magnitude, but here $F(r)=(r-a)/(a{r}^3)$ changes sign at $r=a$. The stated initial condition has inward radial velocity, so the particle immediately enters $r<a$, where the force is outward. The consistent interpretation is that $F$ is the signed inward radial component of the force per unit mass.

The uniform rod has linear mass density $m/\ell$. Integrating squared distance from the pivot gives its moment of inertia about the axis perpendicular to the plane of motion:Measure $\theta$ from the downward vertical. The center of mass is at distance $\ell /2$ from the pivot, so gravity produces the restoring torque $-mg(\ell /2)\sin \theta$. The fixed-axis rotational equation of motion is $I_O\ddot{\theta }={\tau }_O$. Therefore the physical pendulum obeysThe pivot force has zero torque about the pivot.

The lowest positive root belongs to the symmetric family and lies in $(0,\pi /2)$; all antisymmetric roots are at least $\pi$. Let $\alpha =m/(2\rho \ell )\gg 1$. The equation $\cot z=\alpha z$ forces its first root to tend to zero: for any fixed $\epsilon >0$, the right side eventually exceeds $\cot \epsilon$ at $z=\epsilon$, so the decreasing-minus-increasing difference has its zero below $\epsilon$.

The Taylor series $\cot z=z^{-1}-z/3+O(z^3)$ then givesSince $\omega =cz/\ell$ and $c=\sqrt{T/\rho }$, the lowest angular frequency isPhysically, at such a low frequency each half-string is nearly straight. If the mass moves up by $Y$, the two slopes are approximately $Y/\ell$ and $-Y/\ell$, so the restoring force is $-2TY/\ell$. The resulting harmonic oscillator equation, $m\ddot{Y}+(2T/\ell )Y=0$, has exactly the leading angular frequency above. The next correction from the same expansion is $\omega /{\omega }_0=1-\rho \ell /(3m)+O((\rho \ell /m{)}^2)$.

For a nonzero separated solution $U=X(x)T(t)$, separation of variables gives $X^{\prime \prime }/X=T^{\prime }/T=-\lambda$. The zero Dirichlet boundary conditions force $X=\sin nx$, $\lambda =n^2$, with integers $n\ge 1$; the time factors are $e^{-n^{2}t}$. Expanding the initial profile in this Fourier sine series requires only the two already displayed spatial modes. ThusBoth terms individually solve the heat equation, vanish at the endpoints and give the required initial profile at $t=0$. The initial function is also $4{\sin }^{3}x\ge 0$ on the bar; the later solution stays nonnegative and tends to zero as heat leaves through the fixed-temperature ends.

Parametrize the line by $\mathbf{r}(t)=(t,t,t)$, $0\le t\le 1$. Its tangent vector is $(1,1,1)$. On this line the three force components sum to $\mu t+(1-\lambda )t+(\lambda -1-\mu )t=0$. Therefore the line integral is $\boxed{\int \mathbf{F}\cdot d\mathbf{r}=0}$ for every parameter choice.

Taking the negative gradient of the potential energy gives the force

Set $dF=0$ in the relative total differential and use the two equal $1\%$ mass increases:Thus the separation must increase by $1\%$. In this special case the result is also exact: the mass product increases by $1.01^2$, and preserving the inverse-square force requires $r_{\mathrm{new}}^2=1.01^2{r}^2$, hence $r_{\mathrm{new}}=1.01r$.

For the force magnitude in Newton's law of universal gravitation, the total differential isDividing by the positive value $F$ gives the particularly useful relative formFor finite small changes this gives the first-order approximation to the fractional change; higher-order products of increments are omitted.

A rigid rotation of a regular tetrahedron permutes its four vertices, giving an injective group homomorphism from its rotation group to $S_4$: fixing every vertex forces the rotation to be the identity. Its seven nontrivial rotation axes are four lines through a vertex and the opposite face centre, and three lines through midpoints of opposite edges.

Each vertex-face axis admits rotations by $\pm 2\pi /3$, giving eight three-cycles on the vertices. Each opposite-edge axis admits a half-turn, giving three double transpositions. Along with the identity there are twelve rotations. All their induced permutations are even, so the injective image is a twelve-element subgroup of $A_4$, hence the whole group. ThereforeIts action on ${\mathbb{R}}^3$ supplies a three-dimensional representation. To prove irreducibility even after complexification, choose verticesThe group contains the half-turns $D_x=\mathrm{diag}(1,-1,-1)$ and their cyclic variants, as well as the coordinate cycle $(x,y,z)\mapsto (y,z,x)$. If a nonzero invariant subspace over the complex numbers contains a vector with nonzero $x$ component, applying $(I+D_x)/2$ shows it contains the $x$ axis. If another component is nonzero, use the corresponding projection instead. The coordinate cycle then supplies the other two axes. Every nonzero invariant subspace is therefore the full space, provingThis also gives an elementary proof without presupposing the character table.

The endpoint constraints are $y(-a)=y(a)=0$. The constants in the three costs contribute only $2aK$, $2a{K}^{\prime }$ and $2a{K}^{\prime \prime }$ respectively, so they do not affect the minimizing shapes. For companies B and C, the fixed arc length isThe original PDF has $(y^{\prime }{)}^2$ as company A's first variable term. The local TeX transcription has lost that derivative dot and incorrectly repeats $y^2$; the calculation below follows the PDF.

Company A. The relevant functional is $J_A=\int (y^{\prime 2}+y^2-2xy)dx$. The Euler-Lagrange equation givesIts general solution is $y=x+c_1\cosh x+c_2\sinh x$. Adding and subtracting the endpoint equations gives $c_1=0$ and $c_2=-a/\sinh a$. ThusThis curve is odd, negative on the left half and positive on the right half; company A has no lower-half-plane restriction. It is the unique global minimum. Indeed, for any endpoint-zero variation $\eta$, integration by parts and the differential equation cancel all linear terms, leavingunless $\eta =0$.

Company B. Minimize $\int ydx$ under the length constraint. Introduce a constant Lagrange multiplier $\lambda$ and apply the Euler-Lagrange equation to $y+\lambda \sqrt{1+y^{\prime 2}}$:The curvature is constant, so the required lower arc is a circular arc. Writing its radius as $R\ge a$ and imposing the endpoints givesSince $\sqrt{1+y_B^{\prime 2}}=R/\sqrt{R^2-x^2}$, its length determines the radius:Equivalently, introduce $0<\theta \le \pi /2$ with $a=R\sin \theta$. Then $L/(2a)=\theta /\sin \theta$, which increases strictly from $1$ to $\pi /2$. The stipulated range of $L$ therefore selects one minor arc. At $L=\pi a$ it becomes the lower semicircle, with vertical endpoint tangents; it is still a single-valued finite-length graph.

Global minimality follows from the convexity of the augmented slope cost with $\lambda =R$. For a competitor $y=y_B+\eta$,Consequently the augmented cost difference is at least $\int (\eta +x{\eta }^{\prime })dx=[x\eta {]}_{-a}^a=0$. Competitors have the same length, so the multiplier contribution cancels and the original cost is minimized. This proves that a circular arc maximizes area below a fixed chord: minimizing negative signed height maximizes the area between the fence and the chord.

Company C. The variable cost is $J_C=\int yds=\int y\sqrt{1+y^{\prime 2}}dx$. The augmented integrand is $(y+\lambda )\sqrt{1+y^{\prime 2}}$. Since it has no explicit $x$ dependence, the Beltrami identity givesThe hanging branch with $b>0$ integrates to $y+\lambda =b\cosh [(x-x_0)/b]$. Equal endpoint heights force $x_0=0$, and the endpoint value fixes $\lambda =b\cosh (a/b)$. ThusIts length isWriting $q=a/b$, the equation $L/(2a)=\sinh q/q$ has one positive solution for every $L>2a$: its derivative has numerator $q\cosh q-\sinh q>0$. The curve is a catenary, is even, and lies strictly below the chord except at the endpoints.

A direct argument proves this is a global minimum, rather than merely an extremum of an augmented functional. Parametrize any competitor by arc length $s\in [-L/2,L/2]$, from its left endpoint to its right endpoint. Then $x^{\prime 2}+y^{\prime 2}=1$, $\int x^{\prime }ds=2a$, and integration by parts givesThe Cauchy-Schwarz inequality gives the last step. Equality requires $x^{\prime }=b/\sqrt{b^2+s^2}$ and $y^{\prime }=s/\sqrt{b^2+s^2}$, whose integrals are precisely the displayed catenary with the stated length. Thus it attains a lower bound valid for every competitor. This proves that a catenary minimizes the integral of height along a fixed-length curve.

The sketch uses $a=1$ and $L=2.7$ for the two fixed-length cases. The vertical scale in the company A panel is enlarged to show its small odd deflection. No comparison between the three companies' absolute prices is required or possible from the shapes alone, since their additive cost constants are independent.

For a breathing displacement take ${\mathbf{u}}_i=q(t){\mathbf{r}}_i$. Every spring has equilibrium length $\sqrt{2}$ in the given length units, so its linear extension is $\sqrt{2}q$. At the vertex ${\mathbf{r}}_i$, the four adjacent equilibrium vertices sum to zero. Therefore the spring force there isIt remains radial and proportional to the original displacement at every vertex, proving that this symmetric displacement is a normal mode. Newton's second law givesThe first is angular frequency and the second is cycles per unit time. For $k$ in $\mathrm{N}{\mathrm{m}}^{-1}$ and $m$ in $\mathrm{kg}$, these have units $\mathrm{rad}{\mathrm{s}}^{-1}$ and $\mathrm{Hz}$ respectively. An energy check gives $T=3m{\dot{q}}^2$ and $V=12k{q}^2$, leading to the same ratio ${\omega }_b^2=24k/(6m)$.

The Fourier transform of a derivative is $i\omega$ times the transform under the stated convention. Transforming the linear ordinary differential equation givesThe applied force is $F(t)=H(t)e^{-t}$, with $H$ the Heaviside step function, so $\tilde{F}=(1+i\omega {)}^{-1}$. Partial fractions giveUsing the transform pair $H(t)e^{-ct}\leftrightarrow (c+i\omega {)}^{-1}$ yieldsIts value and first derivative both vanish at $t=0^{+}$, so the joining introduces no spurious Dirac delta terms. Equivalently, this is the convolution of the force with the causal impulse response $H(t)(e^{-2t}-e^{-3t})$.