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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 } $
$V=56\quad m/s$
$\therefore \quad \quad f=1800\left( \dfrac { 399 }{ 287 }  \right) \simeq 2500Hz$
Multiple choice physics study of sound reverberation applications of reflection of sound multiple reflection of sound and reverberation applications of ultrasound

The time of reverberation of a room A is one second. What will be the time (in seconds) of reverberation of a room, having all the dimensions double of those of room A:

  1. 2

  2. 4
  3. $\frac{1} {2}$

  4. 1
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Time of reverberation $\propto$ $\frac{V} {A}$ (sabine's formula)

Where V = volume of room and A = area of room
Area of new room becomes 4 times of A and Volume becomes 8 times of V
Time of reverberation will be 2 seconds

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

An auditorium of dimensions $(100\times40\times10)\space m^3$ contains $1000\space m^2$ curtains of absorption coefficient $0.2\space m^{-2},\space 2000 m^2$ of carpets of absorption coefficient $0.7\space m^{-2}$. If $1000$ men of absorption coefficients $0.9$ per person are sitting in the hall, then reverberation time is

  1. $2.7\space s$
  2. $7.2\space s$
  3. $3.5\space s$
  4. $3.7\space s$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Volume of the room          $V = 100\times 40 \times 10 = 4\times 10^4      m^3$

Effective surface area of the room           $A = (0.2) 1000  +  (0.7) 2000  +  (0.9) 1000  =  2500$
Now reverberation time       $T _r = 0.161  \dfrac{V}{A}$
$T _r = 0.161  \dfrac{4\times 10^4}{2500} = 2.576    s$