Q.(i) For the wave on a string described in Exercise 14.11, do all the points on the string oscillate with the same

Concept: The Exercise 14.11 wave is a standing wave, $y=0.06\sin(2\pi x/3)\cos(120\pi t)$ m, giving $f=60\ \text{Hz}$, $\lambda=3\ \text{m}$ ($v=180\ \text{m/s}$). Because the equation is a product of a purely $x$-dependent factor and a purely $t$-dependent factor $\cos(120\pi t)$, every point on the string shares the exact same time-dependence, only scaled (and possibly sign-flipped) differently.

(i)(a) Frequency: Every point's displacement is proportional to the same function $\cos(120\pi t)$ (only the multiplying coefficient $0.06\sin(2\pi x/3)$ changes with $x$), so every point completes a full cycle in the same time -- the frequency is the same everywhere, $f = 60\ \text{Hz}$.

(i)(b) Phase: Nodes occur where $\sin(2\pi x/3)=0$, i.e. every $\lambda/2 = 1.5\ \text{m}$. Between two adjacent nodes (one "loop"), $\sin(2\pi x/3)$ keeps a single sign, so every point in that loop reaches its maximum displacement at the same instant -- they are all in phase with each other. Crossing a node, $\sin(2\pi x/3)$ changes sign, which flips $y \to -y$, i.e. adds a phase of $\pi$ -- so points in adjacent loops are exactly out of phase. …

Recognising a standing wave

Exercise 14.11 gives a wave function that separates into a pure function of $x$ times a pure function of $t$:

This product form $f(x)g(t)$ is the signature of a standing (stationary) wave. From the equation:

• Angular frequency: $\omega = 120\pi\ \text{rad/s} \Rightarrow f=\dfrac{\omega}{2\pi}=60\ \text{Hz}$
• Wave number: $k = \dfrac{2\pi}{3}\ \text{rad/m} \Rightarrow \lambda = 3\ \text{m}$
• Wave speed: $v=f\lambda = 60\times3 = 180\ \text{m/s}$

(i) Do all points share the same frequency, phase, amplitude?

1. Frequency -- YES, the same for all points. The time-dependence $\cos(120\pi t)$ is identical for every $x$.
2. Phase -- YES, within each loop. Because spatial and temporal factors are separated, every point between two consecutive nodes reaches maximum, zero, and minimum displacement at the same instants. (Points on opposite sides of a node differ by $\pi$ -- but that's across a node, not within one loop.)
3. Amplitude -- NO, it varies with position. $A(x)=0.06\left|\sin\!\left(\dfrac{2\pi}{3}x\right)\right|$ depends on $x$: zero at nodes, maximum ($0.06\ \text{m}$) at antinodes. …

Step 1: From Exercise 14.11, the wave is standing with $k=2\pi/3\ \text{rad/m}$, $\omega=120\pi\ \text{rad/s}$, so $f=60\ \text{Hz}$, $\lambda=3\ \text{m}$.

Step 2 (frequency): the time factor $\cos(120\pi t)$ is identical for every $x$, so every point oscillates at the same 60 Hz.

Step 3 (phase): because the equation separates as $f(x)g(t)$, all points between two consecutive nodes hit their extreme/zero displacement at the same instant — same phase within a loop; points on either side of a node are exactly out of phase (differ by $\pi$).

Step 4 (amplitude): the coefficient multiplying $\cos(\omega t)$ at each $x$ is $A(x)=0.06|\sin(2\pi x/3)|$ — this depends on $x$, so amplitude is not the same everywhere: zero at nodes, $0.06\ \text{m}$ at antinodes. …