Exercises — Wave equation — derivation for string

WHAT: We want $v$. WHY this formula: speed on a string is fixed by the medium alone (tension and density), so we use $v=\sqrt{T/\mu }$ — no wave shape needed. $v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{64}{0.04}}=\sqrt{1600}=40\text{m/s}.$ Units check: $\sqrt{\text{N}÷(\text{kg/m})}=\sqrt{({\text{kg⋅m/s}}^2)\cdot (\text{m/kg})}=\sqrt{{\text{m}}^2/{\text{s}}^2}=\text{m/s}.$ ✓

(a), (b), (d) are always solutions. The wave equation is linear, so a right-mover plus a left-mover is still a solution — that is (d), the general solution (see Travelling wave function y = A sin(kx - omega t)). (c) is generally NOT a solution: an arbitrary product $f(x)g(t)$ only works for special separated forms (standing waves like $\sin (kx)\cos (\omega t)$), not any product. See Standing waves on a string. WHY: we verified in the parent note that $u=x\pm vt$ makes $y_{tt}=v^2{y}_{xx}$ automatically; a free product has no such built-in $v$.

$y$ is a length (m), $x$ is a length (m), $t$ is a time (s). $\left[\frac{{\partial }^{2}y}{\partial x^2}\right]=\frac{\text{m}}{{\text{m}}^2}={\text{m}}^{-1},\left[\frac{{\partial }^{2}y}{\partial t^2}\right]=\frac{\text{m}}{{\text{s}}^2}.$ WHY it matters: the equation $y_{tt}=v^2{y}_{xx}$ then needs $[v^2]=\frac{{\text{m/s}}^2}{{\text{m}}^{-1}}={\text{m}}^2/{\text{s}}^2$ — a speed squared. The units force $v$ to be a speed.

WHAT: solve $v=\sqrt{T/\mu }$ for $T$. WHY: we know two of the three quantities; algebra gives the third. Square both sides to undo the square root: $v^2=\frac{T}{\mu }\Rightarrow T=\mu v^2=0.02\times 120^2=0.02\times 14400=288\text{N}.$

Same material + same thickness-per-material means $\mu \propto$ area, so ${\mu }_B=2{\mu }_A$. Tension equal. $\frac{v_B}{v_A}=\frac{\sqrt{T/{\mu }_B}}{\sqrt{T/{\mu }_A}}=\sqrt{\frac{{\mu }_A}{{\mu }_B}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}\approx \mathrm{0.707.}$ WHY: heavier per length ($\mu \uparrow$) → more inertia to shake → slower wave. The thick string runs at about $70.7\%$ of the thin one's speed.

Step 1 — speed from the wave. For a sine on a non-dispersive string the Dispersion relation omega = vk gives $v=\frac{\omega }{k}=\frac{200}{5}=40\text{m/s}.$ WHY: in Example 3 of the parent note, plugging the sine into the wave equation forced ${\omega }^2=v^2{k}^2$, i.e. $v=\omega /k$. Step 2 — tension from the medium. Now use $v=\sqrt{T/\mu }$ backwards: $T=\mu v^2=0.05\times 40^2=0.05\times 1600=80\text{N}.$

Look at the figure below.

WHAT the shape tells us: at a crest the string bends like a hill — it curves downward, so ${\partial }^{2}y/\partial x^2<0$ (the slope goes from positive on the left to negative on the right, i.e. it is decreasing). WHY that gives a force: the net vertical force from the parent note is $F_y=T\frac{{\partial }^{2}y}{\partial x^2}dx.$ With ${\partial }^{2}y/\partial x^2<0$, the force $F_y<0$ — downward. A crest is always pulled back toward the axis. WHAT IT LOOKS LIKE: the two tension arrows at the ends of the little hill both point slightly down-and-outward; their leftover vertical parts add to a downward tug (red arrow in the figure).

Net force $\propto$ curvature $y_{xx}$. It is zero where the curvature changes sign — the inflection points, where the bump switches from curving down (near the top) to curving up (on the tails). Differentiate twice: $y=h(1+s^2{)}^{-1},s=x/a.$ $\frac{dy}{ds}=-2hs(1+s^2{)}^{-2},\frac{d^{2}y}{d{s}^2}=h\frac{6s^2-2}{(1+s^2{)}^3}.$ Set numerator to zero: $6s^2-2=0\Rightarrow s^2=\frac{1}{3}\Rightarrow s=\pm \frac{1}{\sqrt{3}}$, i.e. $x=\pm \frac{a}{\sqrt{3}}\approx \pm 0.577a.$ WHY: at those two points the string is momentarily straight (curvature $=0$), so the tension pulls cancel exactly — no transverse force there, even though the string is moving.

The step: we replaced $\sin \theta$ by $\tan \theta =\partial y/\partial x$ and $\cos \theta$ by $1$. Why linear: $\partial y/\partial x$ is a linear function of $y$ (double $y$, double the slope). Feeding that straight into $F=ma$ keeps every term proportional to $y$ or its derivatives — no $y^2$, no products. A linear equation obeys superposition: two solutions add to give a solution (that is why Problem 1.2(d) works). What breaks: the exact vertical force is $T\sin \theta$, and $\sin \theta =\frac{y_x}{\sqrt{1+y_x^2}}$. That $\sqrt{1+y_x^2}$ is nonlinear in $y_x$. For large slopes it can no longer be dropped; superposition fails and pulses steepen/distort. So the small-angle step is a genuine physical assumption (small amplitude), not a shortcut.

Speeds (same $T$, different $\mu$): $v_1=\sqrt{\frac{90}{0.01}}=\sqrt{9000}\approx 94.87\text{m/s},v_2=\sqrt{\frac{90}{0.04}}=\sqrt{2250}\approx 47.43\text{m/s}.$ Ratio: $\frac{v_1}{v_2}=\sqrt{\frac{{\mu }_2}{{\mu }_1}}=\sqrt{\frac{0.04}{0.01}}=\sqrt{4}=2.$ The light half carries the wave exactly twice as fast. Wave type: the string element moves perpendicular to the string ($y$-direction) while the wave runs along $x$ — this is a transverse wave. (In Transverse vs Longitudinal Waves, transverse = displacement ⟂ propagation.)

Chain the relations. From Dispersion relation omega = vk, $v=\omega /k$. Substitute the definitions: $v=\frac{\omega }{k}=\frac{2\pi f}{2\pi /\lambda }=f\lambda .$ And $v=\sqrt{T/\mu }$, so $\boxed{f\lambda =\sqrt{\frac{T}{\mu }}}.$ Numbers: first $v=\sqrt{100/0.01}=\sqrt{10000}=100\text{m/s}$. Then $f=\frac{v}{\lambda }=\frac{100}{0.5}=200\text{Hz}.$ WHY this is powerful: it welds a shape property ($\lambda$), a timing property ($f$), and a medium property ($T,\mu$) into one line. Change any one and the others adjust so the product stays $=v$.

During overlap the displacement is the sum $y=f(x-vt)+f(x+vt)$ — the pulses pass through each other, adding at each point (superposition). This building block is exactly how Standing waves on a string arise (equal opposite travelling waves). Confirm it solves the equation: the wave equation is linear, and we already verified each piece separately: $\frac{{\partial }^2}{\partial t^2}[f(x-vt)]=v^2\frac{{\partial }^2}{\partial x^2}[f(x-vt)],$ and identically for $f(x+vt)$. Adding two true equations gives a true equation, so the sum satisfies $y_{tt}=v^2{y}_{xx}$. WHY it holds: no step introduced a product like $y^2$, so nothing spoils the additivity.

Exact vertical force on the element (net of the two ends): $F_y=T\sin \theta {|}_{x+dx}-T\sin \theta {|}_x=T\frac{\partial }{\partial x}\left(\sin \theta \right)dx.$ Now $\tan \theta =y_x$, so $\sin \theta =\frac{y_x}{\sqrt{1+y_x^2}}$. This is the exact geometry. Newton's second law (Newton's Second Law) on mass $dm=\mu dx$: $T\frac{\partial }{\partial x}\left(\frac{y_x}{\sqrt{1+y_x^2}}\right)dx=\mu dx\frac{{\partial }^{2}y}{\partial t^2}.$ The dropped term: for small slopes $y_x\ll 1$, expand $\frac{1}{\sqrt{1+y_x^2}}\approx 1-\frac{1}{2}{y}_x^2+\cdots$. The correction $-\frac{1}{2}{y}_x^2$ is second order in the slope. Dropping it leaves $\sin \theta \approx y_x$, so $T\frac{\partial }{\partial x}(y_x)=\mu y_{tt}\Rightarrow T{y}_{xx}=\mu y_{tt}\Rightarrow \frac{{\partial }^{2}y}{\partial t^2}=\frac{T}{\mu }\frac{{\partial }^{2}y}{\partial x^2}.$ Recovered with $v=\sqrt{T/\mu }$. ✓ The whole approximation lives in one place: throwing away the $y_x^2$ correction.

(a) $\mu \to 0$: $v=\sqrt{T/\mu }\to \infty$. A massless string has no inertia, so a disturbance appears everywhere instantly — an idealisation, never literally realised. (b) $T\to 0$: $v=\sqrt{T/\mu }\to 0$. With no tension there is no restoring pull, so transverse disturbances don't propagate — the string just flops. This is the degenerate "no wave" limit. (c) $y_{xx}=0$ everywhere: then $F_y=T{y}_{xx}dx=0$, so $y_{tt}=0$. No net force, no acceleration. A straight (possibly tilted) string stays as it is or drifts at constant transverse velocity — consistent with Newton's first law. No curvature ⇒ no wave force. (d) Sudden kink: at a kink the slope $y_x$ is discontinuous, so $y_{xx}$ is infinite (a spike) there. The idealised equation predicts an infinite force at a point — physically the kink is instantly smoothed and splits into two travelling pulses ($f(x-vt)$ and $g(x+vt)$). Real strings have finite bending stiffness, which caps this. The clean wave equation assumes smooth slopes, so a true kink sits just outside its strict validity.

Fundamental frequency on a fixed-length string is $f=\frac{v}{2L}=\frac{1}{2L}\sqrt{\frac{T}{\mu }}$ (the length sets $\lambda =2L$; see Standing waves on a string). (i) Change $T$ only, keep $L,\mu$: $f\propto \sqrt{T}$. To double $f$: $2=\sqrt{\frac{T_{new}}{T_{old}}}\Rightarrow \frac{T_{new}}{T_{old}}=4.$ Tension must quadruple (×4) to double the pitch — which is why over-tightening snaps strings. (ii) Halve density, keep $T,L$: $f\propto 1/\sqrt{\mu }$, so $\frac{f_{new}}{f_{old}}=\sqrt{\frac{{\mu }_{old}}{{\mu }_{new}}}=\sqrt{2}\approx \mathrm{1.414.}$ The pitch rises by a factor $\sqrt{2}$ (about a musical tritone), not by 2. Doubling pitch this way would need $\mu$ cut to one quarter.

• L1: units of curvature $y_{xx}$? ::: ${\text{m}}^{-1}$
• L2: $v=120$, $\mu =0.02$ ⇒ $T$? ::: $288\text{N}$
• L3: sign of $y_{xx}$ at a crest, and force direction? ::: negative; force downward
• L4: state $f\lambda =?$ ::: $\sqrt{T/\mu }$
• L5: factor on $T$ to double pitch? ::: ×4