Question
(a) A progressive wave travels through a medium. The wave causes a particle of the medium to vibrate along a line P. The energy of the wave propagates along a line Q.
Compare the directions of lines P and Q if the wave is:
(i) a transverse wave. (1 mark)
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(ii) a longitudinal wave. (1 mark)
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(b) A tube is closed at one end. A loudspeaker is placed near the other end of the tube, as shown in Fig. 5.1.
The loudspeaker emits sound of frequency \(1.7\,\mathrm{kHz}\). The speed of sound in the air in the tube is \(340\,\mathrm{m\,s^{-1}}\). A stationary wave is formed with an antinode A at the open end of the tube.
There is only one other antinode A inside the tube, as shown in Fig. 5.1.
Determine:
(i) the wavelength of the sound. (2 marks)
wavelength = ……………………………………………… \(\mathrm{m}\)
(ii) the length \(L\) of the tube. (1 mark)
\(L=\) ……………………………………………… \(\mathrm{m}\)
(iii) the maximum wavelength of the sound from the loudspeaker that can produce a stationary wave in the tube. (1 mark)
maximum wavelength = ……………………………………………… \(\mathrm{m}\)
(c) Two polarising filters are arranged so that their planes are vertical and parallel. The first filter has its transmission axis at an angle of \(35^\circ\) to the vertical and the second filter has its transmission axis at angle \(\alpha\) to the vertical, as shown in Fig. 5.2.
Angle \(\alpha\) is greater than \(35^\circ\) and less than \(90^\circ\). A beam of vertically polarised light of intensity \(8.5\,\mathrm{W\,m^{-2}}\) is incident normally on the first filter.
(i) Show that the intensity of the light transmitted by the first filter is \(5.7\,\mathrm{W\,m^{-2}}\). (1 mark)
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(ii) The intensity of the light transmitted by the second filter is \(5.2\,\mathrm{W\,m^{-2}}\).
Calculate angle \(\alpha\). (2 marks)
\(\alpha=\) ………………………………………………..\(^\circ\)
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 7.1: Progressive waves — part (b)(i)
• 8.1: Stationary waves — parts (b)(ii) and (b)(iii)
• 7.5: Polarisation — parts (c)(i) and (c)(ii)
▶️ Answer/Explanation
(a)(i) Transverse wave [1 mark]
For a transverse wave, the particles of the medium vibrate perpendicular to the direction in which the energy propagates.
Answer: \(\boxed{\text{P is perpendicular to Q}}\)
(a)(ii) Longitudinal wave [1 mark]
For a longitudinal wave, the particles of the medium vibrate parallel to the direction in which the energy propagates.
Answer: \(\boxed{\text{P is parallel to Q}}\)
(b)(i) Wavelength of the sound [2 marks]
For a progressive wave,
\(v=f\lambda\)
Therefore,
\(\lambda=\frac{v}{f}\)
\(f=1.7\,\mathrm{kHz}=1.7\times10^3\,\mathrm{Hz}\)
\(\lambda=\frac{340}{1.7\times10^3}\)
\(\lambda=0.20\,\mathrm{m}\)
Answer: \(\boxed{0.20\,\mathrm{m}}\)
(b)(ii) Length of the tube [1 mark]
The tube is closed at one end and open at the other. Therefore, there is a node at the closed end and an antinode at the open end.
With one other antinode inside the tube, the length corresponds to \(\frac{3}{4}\lambda\).
\(L=\frac{3}{4}\lambda\)
\(L=\frac{3}{4}(0.20)\)
\(L=0.15\,\mathrm{m}\)
Answer: \(\boxed{0.15\,\mathrm{m}}\)
(b)(iii) Maximum wavelength [1 mark]
For a tube closed at one end, the fundamental stationary wave has
\(L=\frac{\lambda_{\max}}{4}\)
Therefore,
\(\lambda_{\max}=4L\)
\(\lambda_{\max}=4(0.15)\)
\(\lambda_{\max}=0.60\,\mathrm{m}\)
Answer: \(\boxed{0.60\,\mathrm{m}}\)
(c)(i) Intensity after the first polarising filter [1 mark]
Using Malus’ law,
\(I=I_0\cos^2\theta\)
Here, \(I_0=8.5\,\mathrm{W\,m^{-2}}\) and \(\theta=35^\circ\).
\(I=8.5\cos^2 35^\circ\)
\(I=5.7\,\mathrm{W\,m^{-2}}\)
Answer: \(\boxed{5.7\,\mathrm{W\,m^{-2}}}\)
(c)(ii) Angle \(\alpha\) [2 marks]
The angle between the transmission axes of the two filters is
\(\theta=\alpha-35^\circ\)
Using Malus’ law again,
\(5.2=5.7\cos^2\theta\)
\(\cos^2\theta=\frac{5.2}{5.7}\)
\(\theta\approx17^\circ\)
Therefore,
\(\alpha=35^\circ+17^\circ\)
\(\alpha=52^\circ\)
Answer: \(\boxed{52^\circ}\)