This set of Audio Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Audio Devices – Acoustic Reverberation”.
1. What is termed as the persistence of sound even after the source of sound has stopped producing sound?
a) Acoustic reverberation
b) Time reverberation
c) Echo reverberation
d) Audio amplifier
View Answer
Explanation: Reverberation in psychoacoustics and acoustics field is termed as the persistence of sound even after the source of sound has stopped producing sound. A reverberation, is created when a sound or signal is reflected and results in a large number of reflections to build up and then decrease slowly as the sound is absorbed by the surfaces of objects in the space which could include furniture, people and air.
2. Reverberation is limited to indoor spaces.
a) True
b) False
View Answer
Explanation: The above statement is false. Reverberation is directly dependent on the frequency. Reverberation is not only limited to indoor spaces as it exists in forests and other outdoor environments where reflection exists like riverside and mountain caves.
3. What is defined as the time for the acoustic intensity to decrease by a factor of one million?
a) Acoustic reverberation
b) Echo reverberation
c) Reverberation time
d) Audio amplifier
View Answer
Explanation: The reverberation time of a room characterizes how long acoustic energy remains in a room. It is usually defined as the time for the acoustic intensity (or energy density) to decrease by a factor of one million (60 dB).
4. What is the sound pressure level of a loud clap?
a) 100 dB
b) 40 dB
c) 10 dB
d) 500 dB
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Explanation: Since a reasonably loud clap is about 100 dB sound pressure level (SPL) and a whisper is about 40 dB, you can easily estimate the reverberation time for a room by clapping and listening to how long you can still hear some remaining sound from the clap.
5. Which of the following have short reverberation time?
a) Empty room
b) Unfurnished room
c) Cinemas
d) Empty playground
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Explanation: Spaces without sound absorbing materials such as large, unfurnished and empty rooms have long reverberation times and spaces with lots of sound absorbers such as cinemas have short reverberation times.
6. What is the interval when a sound is reflected back from a distant object and hits our ears?
a) 0.1 second
b) 0.8 seconds
c) 0.4 seconds
d) 0.5 seconds
View Answer
Explanation: Echo and reverberation are completely different acoustic phenomena. When a sound is reflected back from a distant object and hits our ears after an interval of 0.1 sec, our brain perceives this as a different sound, which is called an echo.
7. Which phenomenon can happen in an enclosed space?
a) Echo
b) Reverberation
c) Time
d) Acoustic
View Answer
Explanation: It is very important when designing an enclosed space to ensure that no echo can occur, but reverberation happens in any enclosed space, however the reverberation time can be short or long depending upon various factors.
8. What is the recommended reverberation time for speech auditoriums and lecture theatres?
a) 0.4 to 0.5
b) 0.4 to 0.7
c) 0.1 to 0.4
d) 0.7 to 1.0
View Answer
Explanation: A very low reverberation time however can be unbearable, imagine talking in an anechoic chamber. For speech auditoriums, lecture theatres, conference, drama theatres and convention centers the recommended reverberation time is 0.7 to 1.0 seconds.
9. What is the Sabine’s formula for calculation of reverberation time?
a) RT (60) = 0.16 × V / A
b) RT (10) = 0.64 × V / A
c) RT (1000) = 0.32 × V / A
d) RT (360) = 0.16 × V / A
View Answer
Explanation: Over 100 years ago, a Harvard physics professor named Wallace Clement Sabine developed the first equation for reverberation time. Sabine’s simple formula is: RT (60) = 0.16 × V / A
Where,
RT (60) = reverberation time in seconds
V = volume of the room per m3
A = absorption area of the room per m2
10. If the desired reverberation time in a classroom is 0.8 seconds and the dimensions of the classroom are 6 × 10 × 3m and the intention is up to use 45 m2 of absorbing ceiling material, then what is the required absorption coefficient for the product?
a) 0.6
b) 0.45
c) 0.8
d) 0.25
View Answer
Explanation: RT (60) = 0.16 × V / A
Therefore,
A = 0.16 × V / RT
S × α = 0.16 × 180 / 0.8
S × α = 36 m2
α = 36 / S
α = 36 / 45
α = 0.8
Sanfoundry Global Education & Learning Series – Audio Engineering.
To practice all areas of Audio Engineering, here is complete set of 1000+ Multiple Choice Questions and Answers.
