7.5 Loudspeakers and microphones
9 exam questions worth 22 marks. Write your answer, open the mark scheme, then give yourself the marks you earned. Your answers are saved on this device.
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Which effect is used in a loudspeaker?
Choose one answer.
Correct Not quite: the answer is The motor effect
Mark scheme
B: The motor effect
Model answer
The motor effect (B).
What type of potential difference does a moving-coil microphone produce?
Choose one answer.
Correct Not quite: the answer is Alternating (a.c.)
Mark scheme
B: Alternating (a.c.)
Model answer
Alternating (a.c.) (B).
Describe the energy transfer in (a) a loudspeaker (b) a microphone.
Mark scheme
(a) electrical to sound (1); (b) sound to electrical (1)
Model answer
(a) A loudspeaker transfers electrical energy to sound. (b) A microphone transfers sound to electrical energy.
Explain how a loudspeaker makes a sound wave from an alternating current.
Mark scheme
the alternating current flows in a coil in the magnetic field of a magnet (1); this produces a force on the coil (motor effect) (1); the direction of the force alternates, so the coil and cone move back and forth (1); the cone makes the air vibrate, producing a sound wave (1)
Model answer
The a.c. flows in the coil, which is in a magnetic field, so there is a force on it. The force alternates in direction, so the coil and cone vibrate, and this makes the air vibrate to produce a sound wave.
A loudspeaker is supplied with an a.c. of 500 Hz. State the frequency of the sound it produces, and explain how the loudness can be increased.
Mark scheme
500 Hz (1); increase the amplitude of the current (1); the force on the coil is bigger, so the cone moves more (1)
Model answer
The sound has a frequency of 500 Hz. To make it louder the current is increased, so the force is bigger and the cone vibrates with a larger amplitude.
A loudspeaker produces a sound of frequency 660 Hz. The speed of sound in air is 330 m/s. Calculate the wavelength of the sound.
Mark scheme
λ = v ÷ f = 330 ÷ 660 (1); 0.50 m (1)
Model answer
λ = v / f = 330 / 660 = 0.50 m.
Explain how a moving-coil microphone produces a potential difference when a sound wave arrives.
Mark scheme
the sound wave makes the diaphragm vibrate (1); the coil attached to the diaphragm moves in the magnetic field (1); the coil cuts the field lines, so a p.d. is induced (electromagnetic induction) (1); the p.d. alternates with the frequency of the sound (1)
Model answer
The sound makes the diaphragm and the coil vibrate in the field of the magnet. The coil cuts the field lines, which induces a potential difference. It alternates at the same frequency as the sound.
State what happens to the induced potential difference from a microphone when the sound gets louder, and explain why.
Mark scheme
it increases (1); the diaphragm and coil move further and faster, so the coil cuts the field lines at a greater rate (1)
Model answer
It increases, because the diaphragm and coil move further and faster, so the field lines are cut at a higher rate.
Explain why a loudspeaker can also work as a microphone.
Mark scheme
sound waves make the cone and the coil vibrate (1); the coil is moving in a magnetic field, so a p.d. is induced in it (1); the induced a.c. has the same frequency as the sound, so it is an electrical signal of the sound (1)
Model answer
Sound waves make the cone and coil move in the magnetic field. Electromagnetic induction produces an alternating potential difference with the same frequency as the sound, so the loudspeaker acts as a microphone.
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