Tag: study of sound

Quizzes (14) Questions (314)

Questions Related to study of sound

Multiple choice physics study of sound reverberation applications of reflection of sound multiple reflection of sound and reverberation applications of ultrasound

For good absorption of sound in auditorium it requires

  1. a few windows to be opened

  2. all closed windows

  3. maps hanging from walls

  4. hanging light curtains.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A few window should be kept opened for the good absorption of sound in auditorium. Because if we open few windows, more sound will absorbed and more will be the sound clarity and better sound effect can be experienced. 

Multiple choice physics study of sound reverberation applications of reflection of sound multiple reflection of sound and reverberation applications of ultrasound

In an auditorium the big curtain when falls suddenly then the reverberation time

  1. increases

  2. decreases

  3. Remains same

  4. becomes zero
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The reverberation time will get decreased because curtain will abrosh the sound and more sound will get abrashed then the earlier.  

Multiple choice physics study of sound reverberation applications of reflection of sound multiple reflection of sound and reverberation applications of ultrasound

Which of the following has largest absorption coefficient?

  1. heavy curtain

  2. cork

  3. marble

  4. open window
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The absorption coefficient of a body is the ratio of the heat radiation absorbed by the body to the total heat radiation incident on it. Since an open window can allow all the radiation into it, it is a good absorber. Hence its coefficient is highest.

Multiple choice physics study of sound reverberation applications of reflection of sound multiple reflection of sound and reverberation applications of ultrasound

The total absorption of a volume $12000 m^3$ is equal to $1000 m^2$ in the units of an open window. Total absorption of the hall is increased to $2000m^2$ due to entry of audience. The change in its reverberating time is

  1. $1s$
  2. $2s$
  3. $3s$
  4. $4s$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Reverberation time (T) is given by T = 0.161 * V / A. Initially, T1 = 0.161 * 12000 / 1000 = 1.932s. After adding audience, A2 = 2000, so T2 = 0.161 * 12000 / 2000 = 0.966s. The change is approximately 1s.

Multiple choice physics study of sound reverberation applications of reflection of sound multiple reflection of sound and reverberation applications of ultrasound

In theatres, big halls., the reverberation of sound is a common problem.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

since the halls and theatres are big enough so that we can distinguish between the generated sound and the reflected sound so the reverbereation  easily happens .

so the answer is A.

Multiple choice physics study of sound reverberation applications of reflection of sound multiple reflection of sound and reverberation applications of ultrasound

A meeting hall of volume $100\times30\times10 m^3$ has a reverberation time of 3 seconds. If 1000 visitors are in the hall. The absorption of total visitors if the sound absorption of each visitor is 0.5 is:

  1. 500metric sabin

  2. 600metric sabin

  3. 700metric sabin

  4. 800metric sabin
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

absorption of each visitor is 0.5
so, absorption for 1000 visitor is $1000\times 0.5$
$=500 metric sabine$

Multiple choice physics study of sound reverberation applications of reflection of sound multiple reflection of sound and reverberation applications of ultrasound

A source of sound A emitting waves of frequency 1800 Hz is falling towards ground with a terminal speed v. The observer B on the ground directly beneath the source receives wave of frequency 2150hz. The source A receives waves, reflected from frequency nearly: (Speed of sound = 343 m/s)

  1. 2150 Hz

  2. 2500Hz

  3. 1800Hz

  4. 2400Hz
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Frequency received by source A is
$f=1800\left( \dfrac { 343+V }{ 343-V }  \right) $
$for\quad V;$
$2150=1800\left( \dfrac { 343 }{ 343-V }  \right) $
$343-V=\dfrac { 1800\times 343 }{ 2150 } $
$V=56\quad m/s$
$\therefore \quad \quad f=1800\left( \dfrac { 399 }{ 287 }  \right) \simeq 2500Hz$