Loudspeakers (physics only) (HT only) - AQA GCSE Physics Exam Questions

Practise 19 exam-style questions on Loudspeakers (physics only) (HT only) for AQA GCSE Physics. Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

19 questions
1–6 marks each
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19 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

A loudspeaker is a device that converts one type of variation into another.

State what a loudspeaker converts, and what it converts it into.

Model answer

A loudspeaker converts variations in the current in an electrical circuit into the pressure variations of sound waves in the air.

Explanation

The electrical signal is a current that varies in step with the sound to be produced. The loudspeaker turns these current variations into movements of a cone, which create compressions and rarefactions in the air.

Mark scheme

1 mark
it converts variations in current (in an electrical circuit)
allow an alternating current / electrical signal
1 mark
into pressure variations in (the air), which are sound waves
allow into sound waves / vibrations of the air

Question 2

3 marks

A loudspeaker is connected to an alternating current with a frequency of 550 Hz.

The speed of sound in air is 330 m/s.

Calculate the wavelength of the sound wave produced.

Model answer

The cone vibrates at the same frequency as the current, so f = 550 Hz.
v = f λ
330 = 550 × λ
λ = 330 ÷ 550 = 0.60 m

Explanation

The loudspeaker cone vibrates at the frequency of the current, so the sound has that frequency too. Then use wave speed = frequency × wavelength, rearranged to λ = v ÷ f.

Mark scheme

1 mark
frequency of the sound = 550 Hz (the same as the current)
an answer of 0.60 (m) scores 3 marks
1 mark
330 = 550 × λ
1 mark
λ = 0.60 m
allow 0.6 m

Question 3

3 marks

The alternating current in a loudspeaker coil has a frequency of 250 Hz.

Use the equation:
period = 1 ÷ frequency

Calculate the time taken for one complete vibration of the loudspeaker cone. Give your answer in milliseconds.

Model answer

The cone vibrates at the same frequency as the current, 250 Hz.
T = 1 ÷ f = 1 ÷ 250
T = 0.0040 s
T = 4.0 ms

Explanation

The period is the time for one complete vibration. Multiply seconds by 1000 to convert to milliseconds.

Mark scheme

1 mark
T = 1 ÷ 250
an answer of 4.0 (ms) scores 3 marks
1 mark
T = 0.0040 s
1 mark
T = 4.0 ms
allow ecf from an incorrect time in seconds

Question 4

4 marks

The coil of a loudspeaker has 50 turns. Each turn has a circumference of 6.0 cm. The magnetic field is at right angles to the wire all the way round the coil, and has a flux density of 0.90 T. The maximum current in the coil is 0.40 A.

Use the equation:
force = magnetic flux density × current × length

Calculate the maximum force on the coil.

Model answer

Length of one turn = 6.0 cm = 0.060 m
Total length of wire in the field = 50 × 0.060 = 3.0 m
F = B I l = 0.90 × 0.40 × 3.0
F = 1.08 N

Explanation

In a loudspeaker the field is arranged so that it is at right angles to the wire everywhere round the coil, so the whole length of wire counts. Find the total length first: number of turns × length of one turn.

Mark scheme

1 mark
6.0 cm = 0.060 m
an answer of 1.08 (N) scores 4 marks; allow 1.1 N
1 mark
total length of wire in the field = 50 × 0.060 = 3.0 m
allow ecf from an incorrect conversion
1 mark
F = 0.90 × 0.40 × 3.0
allow ecf from an incorrect length
1 mark
F = 1.08 N
allow 1.1 N; an answer of 0.0216 N (one turn only) scores 2 marks

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