3.3 Waves — coverage pack

Answer

• $3.2\times10^{2}\,\text{m s}^{-1}$

Method: The period is $T=2.5\times10^{-3}\,\text{s}$, so $f=1/T=400\,\text{Hz}$. Hence $v=f\lambda=400\times0.80=3.2\times10^{2}\,\text{m s}^{-1}$.

Answer

• $135^{\circ}$, or $2.36\,\text{rad}$

Method: The separation is $x/\lambda=0.24/0.64=0.375$ cycles. Therefore $\Delta\phi=360^{\circ}\times0.375=135^{\circ}$. In radians, $\Delta\phi=2\pi\times0.375=0.75\pi=2.36\,\text{rad}$.

Answer

• $0.84\,\text{m s}^{-1}$; $3.95\,\text{rad}$ (or $226^{\circ}$); $0.26\,\text{s}$

Method: First, $v=f\lambda=2.4\times0.35=0.84\,\text{m s}^{-1}$. The phase change is $\Delta\phi=2\pi x/\lambda=2\pi(0.22/0.35)=3.95\,\text{rad}$, which is $226^{\circ}$. The propagation delay is $t=x/v=0.22/0.84=0.262\,\text{s}$, so to two significant figures it is $0.26\,\text{s}$.

Answer

• Longitudinal; the molecules oscillate parallel to the direction of energy propagation.

Method: Sound in air consists of compressions and rarefactions, so it is longitudinal. The molecules move back and forth parallel to the direction in which the wave transfers energy.

Answer

• The electric field oscillates vertically, so it drives charges most effectively in a vertical aerial; rotation reduces the field component along the aerial and a horizontal aerial receives a minimum signal.

Method: Vertical polarisation means that the electric field oscillates vertically. The field component parallel to the conducting aerial drives its charges and induces the received signal. Rotating the aerial reduces this parallel component, which becomes zero in the ideal horizontal orientation.

Answer

• $3.53\,\text{s}$; polarisation selects one direction of field oscillation, which is possible only for a transverse wave.

Method: For the radio pulse, $t_{\rm radio}=1200/(3.00\times10^8)=4.00\times10^{-6}\,\text{s}$. For sound, $t_{\rm sound}=1200/340=3.529\,\text{s}$. The difference is $3.529-0.000004=3.53\,\text{s}$. A polariser transmits a selected direction of oscillation; a longitudinal wave has oscillations only along its travel direction and cannot show this effect, so the observation supports a transverse model.

Answer

• A node has zero amplitude, whereas an antinode has maximum amplitude.

Method: At a node, the two component waves always cancel. At an antinode, their displacements reinforce to give the largest oscillation amplitude.

Answer

• $1.1\times10^2\,\text{Hz}$

Method: Use $f=(2l)^{-1}\sqrt{T/\mu}$. Thus $f=[2(0.75)]^{-1}\sqrt{45/(1.8\times10^{-3})}=(1/1.50)\sqrt{25000}=105\,\text{Hz}$, which is $1.1\times10^2\,\text{Hz}$ to two significant figures.

Answer

• $13\,\text{N}$; $4$ nodes and $3$ antinodes

Method: For harmonic number $n$, $f_n=n(2l)^{-1}\sqrt{T/\mu}$. Rearranging gives $T=\mu(2lf_n/n)^2$. Hence $T=2.5\times10^{-3}[2(0.90)(120)/3]^2=2.5\times10^{-3}(72)^2=12.96\,\text{N}$, so $T=13\,\text{N}$. The third harmonic has three loops, so it has three antinodes and four nodes including the fixed ends.

Answer

• They have the same frequency and a constant phase difference.

Method: Coherence requires equal frequency so the relative phase does not drift, and a constant phase difference so fixed maxima and minima can form.

Answer

• $3.6\,\text{mm}$

Method: Convert to SI units: $\lambda=6.00\times10^{-7}\,\text{m}$ and $s=4.0\times10^{-4}\,\text{m}$. Then $w=\lambda D/s=(6.00\times10^{-7})(2.4)/(4.0\times10^{-4})=3.6\times10^{-3}\,\text{m}=3.6\,\text{mm}$.

Answer

• $0.260\,\text{mm}$; a white central fringe with coloured fringes on either side, violet nearer the centre and red farther out.

Method: From the first to the ninth bright fringe there are eight spacings, so $w=28.8/8=3.60\,\text{mm}=3.60\times10^{-3}\,\text{m}$. Rearranging $w=\lambda D/s$ gives $s=\lambda D/w=(520\times10^{-9})(1.80)/(3.60\times10^{-3})=2.60\times10^{-4}\,\text{m}=0.260\,\text{mm}$. At zero path difference every visible wavelength reinforces, giving a white central fringe. Since $w\propto\lambda$, violet fringes are closer to the centre and red fringes are farther away.

Answer

• The central maximum becomes wider.

Method: A smaller slit-width-to-wavelength ratio produces more diffraction, so the angular spread and hence the central maximum width increase.

Answer

• $13.6^{\circ}$

Method: The line density is $400\times10^3\,\text{m}^{-1}$, so $d=1/(400\times10^3)=2.50\times10^{-6}\,\text{m}$. For $n=1$, $\sin\theta=n\lambda/d=(589\times10^{-9})/(2.50\times10^{-6})=0.2356$. Therefore $\theta=13.6^{\circ}$.

Answer

• Third order at $59.8^{\circ}$

Method: The spacing is $d=1/(600\times10^3)=1.667\times10^{-6}\,\text{m}$. Since $n\lambda\le d$, $n\le d/\lambda=(1.667\times10^{-6})/(480\times10^{-9})=3.47$, so the largest integer order is $n=3$. Then $\sin\theta=3(480\times10^{-9})/(1.667\times10^{-6})=0.864$, giving $\theta=59.8^{\circ}$. For $n=4$, $\sin\theta=1.152>1$, so no real diffraction angle exists.

Answer

• $1.50$

Method: Use $n=c/v=(3.00\times10^8)/(2.00\times10^8)=1.50$. Refractive index has no unit.

Answer

• Yes; the critical angle is $41.1^{\circ}$.

Method: For glass to air, $\sin c=n_{\rm air}/n_{\rm glass}=1/1.52$. Hence $c=\sin^{-1}(1/1.52)=41.1^{\circ}$. The ray travels from higher to lower refractive index and $44.0^{\circ}>41.1^{\circ}$, so total internal reflection occurs.

Answer

• The critical angle is $78.5^{\circ}$, so the ray undergoes total internal reflection; wavelength-dependent speeds and different ray path lengths produce pulse broadening.

Method: At the core-cladding boundary, $\sin c=n_{\rm clad}/n_{\rm core}=1.47/1.50=0.980$, so $c=78.5^{\circ}$. The ray is travelling from higher to lower refractive index and its incidence angle is $80.0^{\circ}>c$, so total internal reflection guides it. Material dispersion occurs because different wavelengths have different refractive indices and therefore different speeds. Modal dispersion occurs because rays at different angles travel different path lengths. Their arrival times spread out, broadening the received pulse and limiting the data rate.