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An ideal uniform, flexible string under constant tension $T>0$ has transverse displacement $y(x,t)$ satisfyingwhere $\mu >0$ is linear mass density, meaning mass per unit length of the straight reference string. The approximation assumes small slope $|y_x|\ll 1$, transverse motion, negligible bending stiffness, negligible damping, and no distributed external force. Constant tension and uniform linear mass density make $c$ constant. Some texts use $\rho$ for this same linear mass density, giving $c=\sqrt{T/\rho }$; here $\rho$ must have units $\mathrm{kg}{\mathrm{m}}^{-1}$, rather than units of volume mass density. ^[ref]

The units provide a useful check: $T/\mu$ has units $(\mathrm{kg}\mathrm{m}{\mathrm{s}}^{-2})/(\mathrm{kg}{\mathrm{m}}^{-1})={\mathrm{m}}^2{\mathrm{s}}^{-2}$. For $T=100\mathrm{N}$ and $\mu =0.010\mathrm{kg}{\mathrm{m}}^{-1}$, the wave speed is $100\mathrm{m}{\mathrm{s}}^{-1}$, so a pulse crosses a one-metre string in $0.010\mathrm{s}$. Quadrupling tension doubles wave speed; quadrupling linear mass density halves it. These square-root laws explain how material and loading set the time scale of the vibration.

For a length-$L$ string with fixed endpoints, the homogeneous Dirichlet boundary conditions are $y(0,t)=y(L,t)=0$. Its nonzero normal modes haveThus $cn\pi /L$ is an angular frequency, in radians per second, whereas $cn/(2L)$ is the frequency in hertz. The wavelength is $2L/n$, and each node of a standing wave lies at $x=jL/n$ for some $j=0,\dots ,n$. For the numerical string above with $L=1\mathrm{m}$, the first three frequencies are $50$, $100$ and $150\mathrm{Hz}$, with angular frequencies $100\pi$, $200\pi$ and $300\pi {\mathrm{s}}^{-1}$. ^[ref]

Each normal mode is a standing wave, but it also has a travelling wave description:At a fixed endpoint an incident displacement pulse reflects with reversed sign, allowing incident and reflected displacements to cancel at the endpoint. Periodic reflection for a fixed-end string implements both endpoints through odd, $2L$-periodic extensions of the initial data. This is the bridge between moving pulses and the stationary Fourier sine basis. ^[ref]

A short derivation comes from Newton's second law. The transverse force on a small segment is approximately $T[y_x(x+\delta x,t)-y_x(x,t)]$, and its mass is $\mu \delta x$. Dividing the force balance by $\delta x$ and taking the limit gives $\mu y_{tt}=T{y}_{xx}$. Separation of variables then gives ${\varphi }^{\prime \prime }+k^2\varphi =0$; the fixed-end boundary conditions select $k=n\pi /L$, while the temporal equation is a harmonic oscillator equation with $\omega =ck$.

For a point mass of constant mass $m$ observed in an inertial frame, the net work along its actual path equals its change in kinetic energy:The middle expression is a line integral; the dot product selects the component of force along the displacement. Positive net work increases speed, negative net work decreases it, and zero net work leaves the endpoint speeds equal. Equal endpoint speeds do not imply equal velocities: a perpendicular force can continually change direction while doing zero work. ^[ref]

The bridge to Newton's second law is short: take the dot product of $m\dot{\mathbf{v}}={\mathbf{F}}_{\mathrm{net}}$ with velocity. The chain rule gives $d(m|\mathbf{v}{|}^2/2)/dt={\mathbf{F}}_{\mathrm{net}}\cdot \mathbf{v}$, equating the rate of change of kinetic energy to instantaneous power. Integrating gives the theorem. No conservative force assumption entered this argument; friction also contributes to the net work.

For the harmonic oscillator with spring stiffness $k=2N/m$, release a mass $m=0.5\mathrm{kg}$ from rest at $x=0.3\mathrm{m}$. The spring exerts $F=-kx$. Its work before the first passage through $x=0$ isThus $m{v}^2/2=0.09\mathrm{J}$ gives speed $0.6m/s$ there. The velocity is negative on this passage because the position is decreasing. The energy calculation alone determines its magnitude, so the direction still requires information about the motion.

For a time-independent potential energy $V$, the conservative force contributes work $-\Delta V$. Separating all other work as $W_{\mathrm{other}}$ gives $\Delta (T+V)=W_{\mathrm{other}}$. Consequently conservation of mechanical energy requires zero net other work; the work-energy theorem itself does not require that conservation. With damping, mechanical energy decreases even though the theorem remains valid. ^[ref]