Exercises — Mechanical waves — transverse and longitudinal

WHAT to check: the angle between particle motion and wave travel. Particles move perpendicular (up/down) to travel (rightward) → transverse. Features: crests (highest points) and troughs (lowest points). Why: transverse ⇔ perpendicular ⇔ crest/trough language; longitudinal ⇔ parallel ⇔ compression/rarefaction language.

No. The curve plots pressure (or displacement) versus position, not the physical sideways path of particles. Sound is longitudinal: air particles oscillate along the travel direction (back and forth), forming compressions and rarefactions. Air has no shear rigidity, so it cannot carry transverse mechanical waves in its bulk. See Sound waves.

Compare term-by-term with $y=A\sin (\omega t-kx)$:

• $A=0.01\text{m}$ — the multiplier out front is the maximum displacement.
• $\omega =200\text{rad/s}$ — the number multiplying $t$.
• $k=4\text{rad/m}$ — the number multiplying $x$. Why so easy: the standard form is the dictionary — matching positions decodes the symbols.

Tool: $v=f\lambda$ (in one period the wave advances one wavelength). Solve for $\lambda$: $\lambda =\frac{v}{f}=\frac{343}{256}=1.340\text{m}$ Why $v$ on top: $v$ is fixed by air, the fork fixes $f$, so $\lambda$ must adjust to $v/f$.

Tool: transverse string speed $v=\sqrt{T/\mu }$ = √(restoring ÷ inertia). $v=\sqrt{\frac{50}{2.0\times 10^{-3}}}=\sqrt{25000}=158.1\text{m/s}$ Why the square root: it's a tug-of-war between the tension pulling a bent element straight (restoring) and its mass resisting acceleration (inertia).

Read $\omega =628\text{rad/s}$, $k=2\text{rad/m}$.

• $f=\frac{\omega }{2\pi }=\frac{628}{2\pi }=99.95\text{Hz}\approx 100\text{Hz}$.
• $\lambda =\frac{2\pi }{k}=\frac{2\pi }{2}=3.1416\text{m}$.
• $v=\frac{\omega }{k}=\frac{628}{2}=314\text{m/s}$ (matches $f\lambda =100\times 3.1416$). Why two routes agree: $v=\omega /k=f\lambda$ are the same relation dressed differently.

$\omega =400\text{rad/s}$, $k=5\text{rad/m}$, $A=0.05\text{m}$. (a) $v=\frac{\omega }{k}=\frac{400}{5}=80\text{m/s}$ — speed of the pattern. (b) Particle velocity $v_p=\frac{\partial y}{\partial t}=A\omega \cos (400t-5x)$; its maximum magnitude is $v_{p,\max }=A\omega =0.05\times 400=20\text{m/s}.$ Here $v=80>v_{p,\max }=20$. No contradiction: they measure different things — how fast the shape travels sideways vs how fast a single particle wiggles up/down. They needn't be equal or ordered any particular way.

Tool: the slope–velocity link $v_p=-v\frac{\partial y}{\partial x}$ (both come from differentiating the same $y$; see figure). $v_p=-(60)(+0.03)=-1.8\text{m/s}.$ Negative → the particle moves downward. Why the minus sign: for a wave moving right, a particle takes on what its left neighbour just had. Look at the figure: on a positive (upward) slope, the point just to the left is lower, so a moment later this particle will be lower too → it is heading down.

Frequency is set by the source and does not change on crossing a boundary (the boundary particle is driven at the source's rate) → $f=500\text{Hz}$ in both.

• Air: ${\lambda }_1=\frac{v_1}{f}=\frac{340}{500}=0.680\text{m}$.
• Water: ${\lambda }_2=\frac{v_2}{f}=\frac{1480}{500}=2.960\text{m}$. Why $\lambda$ grows in water: faster medium, same $f$, so $\lambda =v/f$ stretches.

(a) $\mu =\frac{m}{L}=\frac{9.6\times 10^{-3}}{1.2}=8.0\times 10^{-3}\text{kg/m}$. (b) $v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{120}{8.0\times 10^{-3}}}=\sqrt{15000}=122.5\text{m/s}$. (c) $\lambda =\frac{v}{f}=\frac{122.5}{210}=0.583\text{m}$. Why this order: mass/length feeds the speed; the source's $f$ then sets $\lambda$ via $v=f\lambda$.

Build each symbol:

• $A=0.030\text{m}$.
• $\omega =2\pi f=2\pi (25)=157.08\text{rad/s}$.
• $\lambda =\frac{v}{f}=\frac{50}{25}=2.0\text{m}\Rightarrow k=\frac{2\pi }{\lambda }=\pi =3.1416\text{rad/m}$ (or $k=\omega /v=157.08/50=3.1416$, same). Assemble in the $+x$ form $y=A\sin (\omega t-kx)$: $\boxed{y(x,t)=0.030\sin (157.08t-3.1416x)\text{(SI)}}$ Why the sign: $-kx$ makes far-away points lag, which is exactly a wave moving in $+x$. For $-x$ travel you'd use $+kx$.

Same $T$, so $v=\sqrt{T/\mu }\Rightarrow v\propto 1/\sqrt{\mu }$. (a) $\frac{v_2}{v_1}=\sqrt{\frac{{\mu }_1}{{\mu }_2}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$ → speed halves. (b) $f$ is fixed by the source and unchanged across the join. With $\lambda =v/f$: $\frac{{\lambda }_2}{{\lambda }_1}=\frac{v_2}{v_1}=\frac{1}{2}$ → wavelength halves. Why $f$ is the invariant: the junction particle is shared; it can only oscillate at one rate — the driving rate.

$A=0.02\text{m}$, $\omega =300\text{rad/s}$, $k=15\text{rad/m}$. Particle: $v_p=\partial y/\partial t=A\omega \cos (\varphi )$, acceleration $a_p={\partial }^{2}y/\partial t^2=-A{\omega }^2\sin (\varphi )=-{\omega }^{2}y$, where $\varphi =300t-15x$. (a) $a_{p,\max }=A{\omega }^2=0.02\times 300^2=1800{\text{m/s}}^2$. (b) At a crest $y=+A$, $\sin \varphi =1$ so $\cos \varphi =0$: $v_p=A\omega \cos \varphi =0$ (momentarily at rest), and $a_p=-{\omega }^{2}A=-1800{\text{m/s}}^2$ (max magnitude, pointing back toward equilibrium, i.e. downward). (c) At $y=0$, $\sin \varphi =0$ so $\cos \varphi =\pm 1$: $v_p=\pm A\omega =\pm 6\text{m/s}$ (maximum speed), and $a_p=-{\omega }^2(0)=0$. Why this is just SHM: $a_p=-{\omega }^{2}y$ is the SHM signature — extreme displacement ⇔ max acceleration & zero speed; equilibrium ⇔ zero acceleration & max speed. See Simple Harmonic Motion.

(a) Substitute $B=\gamma P$ into $v=\sqrt{B/\rho }$: $v=\sqrt{\gamma P/\rho }$. (b) $v=\sqrt{\gamma P/\rho }$. If $P\to 2P$ and $\rho \to 2\rho$, then $P/\rho$ is unchanged, so $v$ is unchanged. (This is why sound speed depends on temperature, not on how much you squeeze at fixed $T$.) (c) As $\rho \to 0$ at fixed $P$, $P/\rho \to \infty$, so $v\to \infty$ (unbounded). Physically the ideal-gas model breaks down long before this — very rarefied gas can't relay the disturbance particle-to-particle. This is the degenerate limit: no medium ($\rho \to 0$) means the "wave" concept itself fails.

Tool: $v_p=-v\cdot (\partial y/\partial x)$, wave moving right so $v>0$.

• Point $P$ (rising front): here the snapshot slope $\partial y/\partial x>0$ (curve going up as $x$ increases). Then $v_p=-v(\text{positive})<0$ → $P$ moves down? — careful: check the figure. On the leading (right) edge of a rightward bump the slope is negative (curve coming down toward the flat string ahead). So $v_p=-v(\text{negative})>0$ → up: the front rises as the pulse arrives. ✓
• Point $Q$ (back/left edge): slope is positive there (curve climbing up into the bump), so $v_p=-v(\text{positive})<0$ → down: the trailing edge falls as the pulse leaves. ✓
• Zero particle speed: at the peak of the pulse the slope $=0$, so $v_p=0$ — that particle is momentarily at its highest point, instantaneously at rest. Why the edges behave oppositely: the leading edge is being lifted as the shape advances into it; the trailing edge is being lowered as the shape moves off it. The peak is the turning point. This matches the crest/equilibrium logic of L5.1.

L1 ::: Recognise wave type and read standard-form symbols. L2 ::: Single-formula application: $v=f\lambda$, $v=\sqrt{T/\mu }$, decode $y(x,t)$. L3 ::: Particle vs wave velocity; slope–velocity link; $f$ fixed across media. L4 ::: Chain relations to build $v$, $\lambda$, and full $y(x,t)$. L5 ::: SHM acceleration link, $v=\sqrt{\gamma P/\rho }$ limits, pulse geometry & degenerate cases.