Question
(a) A stationary wave is formed on a string \(XY\) that has a length of \(0.48\,\mathrm{m}\). Fig. 5.1 shows the string at one instant in time.
The speed of the wave on the string is \(1400\,\mathrm{m\,s^{-1}}\).
(i) On Fig. 5.1, draw a cross (×) at one position that is a node and another cross at one position that is an antinode. Label the node N and the antinode A. [1 mark]
______________________________________________________________________
(ii) Show that the wavelength of the wave produced is \(0.32\,\mathrm{m}\). Explain your reasoning. [1 mark]
______________________________________________________________________
(iii) Calculate the frequency of the wave. [2 marks]
frequency = ______________________________ \(\mathrm{Hz}\)
(b) A source of sound waves of frequency \(780\,\mathrm{Hz}\) is on a rotating platform. The speed of the source is \(39\,\mathrm{m\,s^{-1}}\). The sound is detected by an observer that is a large distance from the rotating platform, as shown in Fig. 5.2.
(i) The speed of sound in air is \(320\,\mathrm{m\,s^{-1}}\).
Calculate the maximum frequency of the sound detected by the observer. [2 marks]
maximum frequency = ______________________________ \(\mathrm{Hz}\)
(ii) At time \(t=0\), the observer detects the sound emitted by the source when it was in the position shown in Fig. 5.2.
On Fig. 5.3, sketch the variation with \(t\) of the frequency of the sound detected by the observer for one complete rotation of the platform. Calculations are not required. [2 marks]
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 7.1: Progressive waves — part (a)(iii), using \(v=f\lambda\)
• 7.3: Doppler effect for sound waves — parts (b)(i) and (b)(ii)
▶️ Answer/Explanation
(a)(i) [1 mark]
A node is a point that remains stationary, so it occurs where the solid and dashed wave profiles intersect. An antinode occurs at a peak or trough where the amplitude is maximum.
A suitable node is at \(X\) or \(Y\), or at an intersection of the solid and dashed lines. A suitable antinode is at any peak or trough.
Answer: \( \boxed{\text{N at a node and A at a peak or trough}} \)
(a)(ii) [1 mark]
The string contains \(1.5\) wavelengths over its length of \(0.48\,\mathrm{m}\).
Therefore,
\(0.48=1.5\lambda\)
\(\lambda=\frac{0.48}{1.5}\)
\(\lambda=0.32\,\mathrm{m}\)
Answer: \( \boxed{0.32\,\mathrm{m}} \)
(a)(iii) [2 marks]
Use the wave equation
\(v=f\lambda\)
Rearranging,
\(f=\frac{v}{\lambda}\)
\(f=\frac{1400}{0.32}\)
\(f=4375\,\mathrm{Hz}\approx4.4\times10^3\,\mathrm{Hz}\)
Answer: \( \boxed{4.4\times10^3\,\mathrm{Hz}} \)
(b)(i) [2 marks]
The maximum frequency occurs when the source is moving directly towards the stationary observer.
For a moving source,
\(f_{\mathrm{o}}=\frac{f_{\mathrm{s}}v}{v-v_{\mathrm{s}}}\)
Substituting the values,
\(f_{\mathrm{o}}=\frac{(780)(320)}{320-39}\)
\(f_{\mathrm{o}}\approx888\,\mathrm{Hz}\)
\(f_{\mathrm{o}}\approx890\,\mathrm{Hz}\)
Answer: \( \boxed{890\,\mathrm{Hz}} \)
(b)(ii) [2 marks]
As the source rotates, its velocity component towards the observer varies continuously from maximum towards the observer, through zero, to maximum away from the observer.
Hence the observed frequency varies smoothly above and below the source frequency of \(780\,\mathrm{Hz}\).
At \(t=0\), the source is moving perpendicular to the direction of the observer, so its velocity component towards the observer is zero. Therefore the graph starts at the mean frequency \(780\,\mathrm{Hz}\), falls to a minimum, rises to a maximum, and returns to \(780\,\mathrm{Hz}\) after one complete rotation.
Answer: \( \boxed{\text{a smooth periodic curve about }780\,\mathrm{Hz}\text{, starting at the mean value}} \)