CAIE A-Level Physics AS 8.1 Stationary Waves Questions

Practise explaining and investigating stationary waves with superposition, nodes, antinodes and resonance, then finding wavelength or phase from data.

Exam points

apply superposition to calculate or sketch resultant displacements and explain constructive/destructive combinations
explain and investigate stationary waves using opposite travelling waves, nodes, antinodes and experimental observations
determine wavelength and phase information from node/antinode positions or stationary-wave graphs

Covered syllabus points

8.1.1—And use the principle of superposition 8.1.2—Experiments that demonstrate stationary waves using microwaves, stretched 8.1.3—The formation of a stationary wave using a graphical method, and identify 8.1.4—Wavelength may be determined from the positions of nodes or antinodes of a

[Maximum number: 5]

(a)

A microwave of intensity $I_{0}$ and amplitude $A_{0}$ meets another microwave of the same frequency and of intensity $\frac{1}{4} I_{0}$ travelling in the opposite direction. Both microwaves are vertically plane polarised and superpose where they meet.

[ 5 ]

(i)

Explain, without calculation, why these two waves cannot form a stationary wave with zero amplitude at its nodes.

[ 2 ]

(ii)

Determine, in terms of $A_{0}$ , the maximum amplitude of the wave formed.
maximum amplitude = $A_{0}[3]$

[ 3 ]

$I \propto A^{2}$ $A^{2} / A_{0}{ }^{2}=\left(I_{0} / 4\right) / I_{0}$ $A=0.5 A_{0}$
Marking guidance:

maximum amplitude $=A_{0}+0.5 A_{0}$ $=1.5 A_{0}$

[Maximum number: 7]

(a)

A tube is open at both ends. A loudspeaker, emitting sound of a single frequency, is placed near one end of the tube, as shown in Fig. 3.1.

Fig. 3.1

The speed of the sound in the tube is $340 \mathrm{~ms}^{-1}$ . The length of the tube is 0.60 m .
A stationary wave is formed with an antinode A at each end of the tube and two antinodes inside the tube.

[ 7 ]

(i)

State what is meant by an antinode of the stationary wave.

[ 1 ]

(ii)

State the distance between a node and an adjacent antinode.

distance =m [1]

[ 1 ]

distance $=0.10 \mathrm{~m}$

(iii)

Determine, for the sound in the tube,
1. the wavelength,
wavelength = m
2. the frequency.
frequency = Hz

[ 3 ]

1. $\lambda=0.60 / 1.5$ $=0.40 \mathrm{~m}$
2. $v=f \lambda$ $\begin{aligned} f =340 / 0.40 =850 \mathrm{~Hz} \end{aligned}$

(iv)

Determine the minimum frequency of the sound from the loudspeaker that produces a stationary wave in the tube.
minimum frequency = Hz

[ 2 ]

$\lambda=2 \times 0.60$ or $\lambda=3 \times 0.40$ or f=850 / 3
f=280(283) Hz

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Covered syllabus points

Question 1

Question (a)

Question (i)

Question (ii)

Question 2

Question (a)

Question (i)

Question (ii)

Question (iii)

Question (iv)