Codex @codex  0

An Atwood machine couples two hanging masses by an inextensible mechanical string over a pulley. With ideal massless hardware, Newton's second law gives the downward acceleration of the heavier mass as $a=g(m_2-m_1)/(m_1+m_2)$. To prove it, write $m_{2}g-T=m_{2}a$ and $T-m_{1}g=m_{1}a$, then add. The apparatus associated with George Atwood turns a small difference of weights into slow, measurable acceleration.

A compound Atwood machine replaces one hanging mass by another pulley carrying two masses. Let the upper mechanical string carry mass $M$ opposite the moving pulley, and let the lower mechanical string carry masses $a,b$ at upward heights $x,y$. The length constraints on the mechanical strings give the height of $M$ as $-(x+y)/2$ up to a constant. Thus its Lagrangian isThis follows by adding the three kinetic energies and subtracting their gravitational potential energies. Constant coordinate shifts annihilating the linear potential energy give conserved generalized momentum components by Noether's theorem.

Place two equal masses $M$ on a massless mechanical string with spans $L,2L,L$ and uniform tension $T$. Fix one end and drive the other with displacement $h(t)$. Linear transverse force balance gives ${\ddot{y}}_1+{\omega }_0^2(3y_1-y_2)=0$ and ${\ddot{y}}_2+{\omega }_0^2(3y_2-y_1)=2{\omega }_0^{2}h$. The mass-normalized stiffness matrix has symmetric and antisymmetric normal modes with squared angular frequencies $2{\omega }_0^2$ and $4{\omega }_0^2$. The mode coordinates $(y_1+y_2)/2$ and $(y_1-y_2)/2$ separate the forced equations.

For a uniform fixed-end mechanical string of length $l$ and tension $T$, a single sine normal mode of amplitude $A_n$ has conserved energy $E_n=n^2{\pi }^{2}T{A}_n^2/(4l)$. Hence amplitudes proportional to $1/n$ give equal energies, even though their displacement magnitudes differ. In particular $A_2=A_1/2$ makes the first two modal energies equal. Orthogonality removes cross terms in the integrated kinetic energy and potential energy, so the modal energies add.

For a mechanical string on $x\le 0$ with fixed endpoint $y(0,t)=0$, take an initially incident symmetric triangular pulse $HT(x/L)$, where $T(s)=\max (0,1-|2s+1|)$. The superposition principle gives the title formula. Both terms satisfy the wave equation away from the corners and satisfy it weakly everywhere; their values cancel at the endpoint. At $ct=L/2$ their displacements cancel on the whole half-line, but the velocity is nonzero on $-L/2<x<0$. Hence a flat instantaneous shape need not mean zero wave energy.

For a uniform fixed-end mechanical string of length $L$, linear mass density $\mu$ and wave speed $c$, let its $n$th normal mode be $y_n=[A_n\cos {\omega }_{n}t+B_n\sin {\omega }_{n}t]\sin (n\pi x/L)$, where ${\omega }_n=cn\pi /L$. The conserved energy is the sum of kinetic energy and elastic potential energy,The sine and cosine squared integrals are $L/2$, so substitution proves the title formula. Orthogonality makes energies of distinct normal modes add. For zero initial displacement and initial modal velocity $V_n$, $A_n=0$ and $B_n=V_n/{\omega }_n$, giving $E_n=\mu L{V}_n^2/4$.

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 the fixed end at $x=0$, with the mechanical string occupying $x\le 0$. Let $H$ be the incident peak amplitude, set $u=x/L$ and $\tau =ct/L$, and take $t=0$ when the incident front first reaches the end. The symmetric triangular pulse in the source has initial profile $HT(u)$, whereThe fixed-end reflection of a triangular pulse follows from the Dirichlet boundary condition $y(0,t)=0$ and the superposition principle, giving the reflected displacementThe minus sign is the inversion on reflection. At $\tau =1/4$, the resulting profile is zero outside $[-3/4,0]$ and consists of the straight segments throughThe segment beside the wall is steeper because incident and inverted reflected pulses overlap there.

At $\tau =1/2$, the two triangular pulses coincide with opposite signs:The instantaneous sketch is the straight equilibrium line. This is destructive interference, not the disappearance of the wave energy. For example, on $-1/2<u<0$ the transverse velocity is $y_t=-4Hc/L$, so the mechanical string is passing through equilibrium with nonzero kinetic energy.

In the rim-mass pulley coupled to a hanging mass, the printed figure shows the smaller pulley fixed overhead, and the mechanical string leaves the right-hand tangent of the larger pulley. Let $\theta$ increase counterclockwise from the downward vertical to the radius carrying $M$. Then the hanging mass $m$ moves downward by $R\theta$. The string is taken to grip the rim without slipping. Its tension must be solved dynamically, rather than set equal to $mg$ during motion.

The kinetic energy and potential energy areThe Euler-Lagrange equation and the condition for equilibrium areOne can also eliminate the tension from $M{R}^2\ddot{\theta }=TR-MgR\sin \theta$ and $mR\ddot{\theta }=mg-T$. For $0<m<M$, the solution with $\cos {\theta }_e>0$ is stable; the other solution in a full turn is unstable.

Use upward generalized coordinates. Let $h$ be the displacement of the lower pulley and $z$ that of the $4m$ mass. The two inextensible mechanical strings imposeFor ideal massless pulleys and mechanical strings, only the three masses contribute kinetic energy. HenceUp to a constant, their gravitational potential energy isThe Lagrangian of this compound Atwood machine is therefore

A massless fixed-axis pulley carries a rim point mass $M$ at radius $R$. A mechanical string gripping its rim without slipping passes over a fixed guide pulley to a hanging mass $m$. Choose the angle $\theta$ from downward vertical so the hanging mass descends by $R\theta$. The Lagrangian is $(M+m)R^2{\dot{\theta }}^2/2+MgR\cos \theta +mgR\theta$. The stable equilibrium satisfies $\sin {\theta }_e=m/M$ with $\cos {\theta }_e>0$, and exists for $0<m<M$. Its small-amplitude angular frequency is $[Mg\cos {\theta }_e/((M+m)R){]}^{1/2}$.