· H2 Physics — Ten-Year Series, worked ← All 8 questions

2023 A-Level · Paper 2 · Question 4

One sound wave meets its reflection, and two loudspeakers meet at a point

Superposition · stationary waves · interference 12 marks Oscilloscope Intensity and amplitude

Twelve marks on superposition. First, a loudspeaker facing a metal plate sets up a stationary wave: how it forms, its frequency read off an oscilloscope, and its node spacing. Then two loudspeakers send equal sound to a point at different distances: is it loud or quiet there, and how big is each wave on arrival? Below is the full solution, and beside it both superpositions running.

The model, live

Wave from the loudspeaker / XReflected wave / YResultantNodes

Part (a)(i)

2 marks

Describe how a stationary wave forms between the loudspeaker and the metal plate.

Approach

A stationary wave needs two progressive waves of the same frequency (and speed) travelling in opposite directions through the same region. Say where the second wave comes from, then say what their superposition produces.

The sound from the loudspeaker is reflected at the metal plate. The incident and reflected waves have the same frequency and speed and travel in opposite directions, so they overlap and superpose. This gives a stationary wave: nodes (zero amplitude) and antinodes (maximum amplitude) at fixed positions.

Watch

"The waves interfere" alone is not enough. The marks are for the reflection (the source of the second wave) and for the opposite directions of travel.

Part (a)(ii)

2 marks

The time base is 0.50 ms cm−10.50\,\text{ms cm}^{-1}. Use the oscilloscope trace to show that the frequency is 400 Hz400\,\text{Hz}.

Approach

Measure one full cycle along the time axis in centimetres, peak to peak, then convert with the time base. Measuring across two cycles and halving is more accurate.

On the trace, adjacent peaks are 5.0 cm5.0\,\text{cm} apart (two cycles span 10.0 cm10.0\,\text{cm}):

T=5.0 cm×0.50 ms cm−1=2.5 msT = 5.0\,\text{cm} \times 0.50\,\text{ms cm}^{-1} = 2.5\,\text{ms}
f=1T=12.5×10−3=400 Hzf = \frac{1}{T} = \frac{1}{2.5\times10^{-3}} = 400\,\text{Hz}

One period is 5.0 cm5.0\,\text{cm} on the screen, so T=5.0×0.50=2.5 msT = 5.0 \times 0.50 = 2.5\,\text{ms} and f=1/T=400 Hzf = 1/T = 400\,\text{Hz}.

Part (a)(iii)

2 marks

The speed of sound is 340 m s−1340\,\text{m s}^{-1}. Find the distance between adjacent nodes.

λ=vf=340400=0.85 m\lambda = \frac{v}{f} = \frac{340}{400} = 0.85\,\text{m}

Adjacent nodes in a stationary wave are half a wavelength apart:

λ2=0.852=0.425 m\frac{\lambda}{2} = \frac{0.85}{2} = 0.425\,\text{m}

distance =0.43 m= 0.43\,\text{m}

Watch

Node to node is λ/2\lambda/2, not λ\lambda. Count the nodes in the model: there are two of them in every 0.85 m0.85\,\text{m}.

Part (b)(i)

2 marks

Two loudspeakers X and Y emit in phase, with wavelength 0.20 m0.20\,\text{m}. P is 1.4 m1.4\,\text{m} from X and 1.8 m1.8\,\text{m} from Y. State and explain whether the intensity at P is a maximum or a minimum.

Approach

For sources in phase, compare the path difference with the wavelength. A whole number of wavelengths means the waves arrive in phase; an odd number of half wavelengths means they arrive in antiphase.

Δ=1.8−1.4=0.40 m=0.400.20λ=2λ\Delta = 1.8 - 1.4 = 0.40\,\text{m} = \frac{0.40}{0.20}\lambda = 2\lambda

Maximum. The path difference is 0.40 m=2λ0.40\,\text{m} = 2\lambda, a whole number of wavelengths, and the sources are in phase, so the waves arrive at P in phase and interfere constructively.

Note

The waves do not cancel even at a minimum here, because they arrive with different amplitudes (part (iii)). A minimum would be quiet, not silent.

Part (b)(ii)

2 marks

The loudspeakers act as point sources. The intensity at P from X is 4.5×10−6 W m−24.5\times10^{-6}\,\text{W m}^{-2}. Find the intensity at P from Y.

Approach

The two loudspeakers emit equal power. For a point source the power spreads over a sphere, so I=P4πr2I = \dfrac{P}{4\pi r^2} and I∝1r2I \propto \dfrac{1}{r^2}.

IYIX=(rXrY)2⟹IY=(4.5×10−6)(1.41.8)2=2.72×10−6 W m−2\frac{I_Y}{I_X} = \left(\frac{r_X}{r_Y}\right)^2 \quad\Longrightarrow\quad I_Y = (4.5\times10^{-6})\left(\frac{1.4}{1.8}\right)^2 = 2.72\times10^{-6}\,\text{W m}^{-2}

intensity =2.7×10−6 W m−2= 2.7\times10^{-6}\,\text{W m}^{-2}

Watch

Y is further away, so its intensity must be smaller. If your ratio gives more than 4.5×10−64.5\times10^{-6}, it is upside down.

Part (b)(iii)

2 marks

Find the ratio of the amplitude of the wave from X to the amplitude of the wave from Y at P.

I∝A2⟹AXAY=IXIY=4.52.72=1.29I \propto A^2 \quad\Longrightarrow\quad \frac{A_X}{A_Y} = \sqrt{\frac{I_X}{I_Y}} = \sqrt{\frac{4.5}{2.72}} = 1.29

To check your answer:

Since I∝1/r2I \propto 1/r^2 and A∝IA \propto \sqrt{I}, the amplitude goes as 1/r1/r. So AXAY=1.81.4=1.29\dfrac{A_X}{A_Y} = \dfrac{1.8}{1.4} = 1.29 directly.

ratio =1.3= 1.3

What this question was really testing

  1. Where the second wave comes from. A stationary wave is a reflection superposing with the incident wave.
  2. Reading an oscilloscope. Divisions × time base = period.
  3. Node spacing is half a wavelength.
  4. Path difference in wavelengths. Whole number for in-phase sources means constructive.
  5. Inverse square, then square root. I∝1/r2I \propto 1/r^2 for a point source and A∝IA \propto \sqrt{I}, so A∝1/rA \propto 1/r.

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