IBDP Physics- C.4 Standing waves- IB Style Questions For HL Paper 2 -FA 2025
Question
The diagram shows the direction of a sound wave travelling in a metal sheet.
(a) Particle P in the metal sheet performs simple harmonic oscillations. When the displacement of P is \(3.2\,\mu\text{m}\) the magnitude of its acceleration is \(7.9\,\text{m s}^{-2}\). Calculate the magnitude of the acceleration of P when its displacement is \(2.3\,\mu\text{m}\). [2]
(b) The wave is incident at point Q on the metal–air boundary. The wave makes an angle of \(54^\circ\) with the normal at Q. The speed of sound in the metal is \(6010\,\text{m s}^{-1}\) and the speed of sound in air is \(340\,\text{m s}^{-1}\). Calculate the angle between the normal at Q and the direction of the wave in air. [2]
(c) The frequency of the sound wave in the metal is \(250\,\text{Hz}\). Determine the wavelength of the wave in air. [1]
(d) The sound wave in air in (c) enters a pipe that is open at both ends. The diagram shows the displacement, at a particular time \(T\), of the standing wave that is set up in the pipe.
A particular air molecule has its equilibrium position at the point labelled M. On the diagram, at time \(T\),
(i) draw an arrow to indicate the acceleration of this molecule. [1]
(ii) label with the letter C a point in the pipe that is at the centre of a compression. [1]
(e) Sound of frequency \(f=2500\,\text{Hz}\) is emitted from an aircraft that moves with speed \(v=280\,\text{m s}^{-1}\) away from a stationary observer. The speed of sound in still air is \(c=340\,\text{m s}^{-1}\).
(i) Determine the frequency heard by the observer. [2]
(ii) Determine the wavelength measured by the observer. [1]
Most-appropriate topic codes (Physics Guide 2025)
Topic C.1 — Simple harmonic motion (part a)
Topic C.2 — Wave model (parts b, c)
Topic C.3 — Wave phenomena (part b)
Topic C.4 — Standing waves and resonance (part d)
Topic C.5 — Doppler effect (part e)
▶️Answer/Explanation
(a)
For SHM, \(a \propto x\) so \(\dfrac{a_2}{a_1}=\dfrac{x_2}{x_1}\).
\(a_2=7.9\times \dfrac{2.3}{3.2}=5.68\approx 5.7\,\text{m s}^{-2}\).
For SHM, \(a \propto x\) so \(\dfrac{a_2}{a_1}=\dfrac{x_2}{x_1}\).
\(a_2=7.9\times \dfrac{2.3}{3.2}=5.68\approx 5.7\,\text{m s}^{-2}\).
(b)
Use refraction for waves: \(\dfrac{\sin\theta_{\text{metal}}}{\sin\theta_{\text{air}}}=\dfrac{v_{\text{metal}}}{v_{\text{air}}}\).
\(\sin\theta_{\text{air}}=\dfrac{340}{6010}\sin 54^\circ\approx 0.0458\).
\(\theta_{\text{air}}=\sin^{-1}(0.0458)\approx 2.6^\circ\).
Use refraction for waves: \(\dfrac{\sin\theta_{\text{metal}}}{\sin\theta_{\text{air}}}=\dfrac{v_{\text{metal}}}{v_{\text{air}}}\).
\(\sin\theta_{\text{air}}=\dfrac{340}{6010}\sin 54^\circ\approx 0.0458\).
\(\theta_{\text{air}}=\sin^{-1}(0.0458)\approx 2.6^\circ\).
(c)
Frequency does not change at a boundary, so in air \(\lambda=\dfrac{v}{f}=\dfrac{340}{250}=1.36\approx 1.4\,\text{m}\).
Frequency does not change at a boundary, so in air \(\lambda=\dfrac{v}{f}=\dfrac{340}{250}=1.36\approx 1.4\,\text{m}\).
(d)
(i) Acceleration is towards equilibrium (opposite the displacement at \(M\)); draw the arrow at \(M\) pointing left.
(ii) Label \(C\) at the centre of a compression (any correct compression centre on the diagram).
(e)
(i) Source moving away: \(f’ = f\left(\dfrac{c}{c+v}\right)=2500\left(\dfrac{340}{340+280}\right)=1371\approx 1.4\times 10^3\,\text{Hz}\).
(ii) \(\lambda’=\dfrac{c}{f’}=\dfrac{340}{1371}=0.248\approx 0.25\,\text{m}\).
(i) Source moving away: \(f’ = f\left(\dfrac{c}{c+v}\right)=2500\left(\dfrac{340}{340+280}\right)=1371\approx 1.4\times 10^3\,\text{Hz}\).
(ii) \(\lambda’=\dfrac{c}{f’}=\dfrac{340}{1371}=0.248\approx 0.25\,\text{m}\).