Tag: study of sound

Questions Related to study of sound

The walls and ceiling of an auditorium are covered with a fibrous material to

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

  1. Make the auditorium fire proof

  2. Absorb sound and prevent echoes

  3. Give an ethnic look to the auditorium

  4. To prevent leakage of water from the roof

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Correct Option: B
Explanation:

  We know that soft , porous and fibrous materials are good absorbers of sound , so they are used to the walls and ceiling of an auditorium to absorb the sound .

    If this is not done , sound waves after reflections (echoes) from walls and ceiling , will interfere and due to this at some places in auditorium , sound will not be heard (destructive interference) and at some places intensity of sound will be very large (constructive interference)

What will happen if the reverberation time in a big hall is too long?

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

  1. Sound will persist for some time in the hall

  2. The reflection may become an echo.

  3. Multiple reflection may occur

  4. All the above will happen

Show answer
Correct Option: D
Explanation:

The multiple reflection of sound from the walls of a room or hall causing sound to persist for some time is called reverberation. If the hall is too long , the reflection will become an echo.

A man standing between two cliffs hears the first echo of a sound after 2 sec and the second echo 3 sec after the initial sound. If the speed of sound be $330   {m}/{sec}$ the distance between the two cliffs should be

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

  1. 1650 m

  2. 990 m

  3. 825 m

  4. 656 m

Show answer
Correct Option: C
Explanation:

Let $d _{1}$ and $d _{2}$ be the distances of first and second cliff from man .

Echo after 2s will be heard when sound travels a distance of $2d _{1}$ , because an echo comes back to initial point after reflection .
    therefore  by ,   distance=speed$\times$time ,
         or                 $2d _{1}=330\times2=660$ ,
         or                 $d _{1}=660/2=330m$ ,
similarly Echo after 3s will be heard when sound travels a distance of $2d _{2}$ ,  
    therefore   
         or                 $2d _{2}=330\times3=990$ ,
         or                 $d _{2}=990/2=495m$ ,
as the man is in between the cliffs , therefore distance between cliffs willbe ,
             $d=d _{1}+d _{2}=330+495=825m$

(1) The reflected pulse will be in same orientation of incident pulse due to a phase change of $\pi$ radians.
(2) During reflection the wall exerts a force on string in upward direction.
For the above given two statements choose the correct option given below.

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

  1. Only (1) is true

  2. Only (2) is true

  3. Both are true

  4. Both are wrong

Show answer
Correct Option: D
Explanation:

The pulse will be of same orientation as before after the phase change of $2\pi$ radians

During the reflection string tries to move upward but wall applies downward force to stop movement of connected end. So both the statements are wrong.

(A).  The reverberation time is dependent on the shape of the enclosure,  position of the sources and observer.
(B) The unit of absorption coefficient in S.I. system in metric is sabine

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

  1. both (A) and (B) are true

  2. both (A) and (B) are false

  3. (A) is true but (B) is false

  4. (A) is false but (B) is true

Show answer
Correct Option: B
Explanation:
The reverberation time is independent on the shape of the enclosure, position of the sources and observer.
Absorption coefficient is unitless.

A hall has volume 4x6x10 m$^3$. If the total sound absorption of the hall is 27.2metric sabine and 40 visitors are in the hall and each is equivalent to 0.5metric sabine sound absorption then the reverberation time is

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

  1. 0.8644s

  2. 0.05s

  3. 0.72s

  4. 1.8s

Show answer
Correct Option: A
Explanation:

$volume =240m^{3}$
$Area =27.2m^{2}+40\times 0.5m^{2}$
=47.2
$R.T.=0.161\times \frac{V}{A}$
=0.8186s

The reverberation time of a room is t seconds. Another room of double the dimensions with the walls of the same absorption coefficient will have a reverberation time :

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

  1. $t^{2}$

  2. $2 t$

  3. $t/2$

  4. t$^{3}$

Show answer
Correct Option: B
Explanation:

Given: $\dfrac{V _{1}}{V _{2}}=\dfrac{1}{8} \
\ \dfrac{A _{1}}{A _{2}}=\dfrac{1}{4}$
$R.T. _{1}=t $
$R.T=0.161\dfrac{v}{A}$
$\dfrac{R.T. _{1}}{R.T. _{2}}=\dfrac{0.161 \times \dfrac{v _1}{A _1}}{0.161 \times \dfrac{v _2}{A _2}}=\dfrac{v _{1}A _{2}}{v _{2}A _{1}}=\dfrac{1}{8}\times \dfrac{4}{1} =\dfrac{1}{2}$
$R.T _{2}=2R.T _{1}$
          $=2\times t$
          $=2t$ 

The ratio of the absorption of an auditorium to that of a person is 1000. If 500 persons are accommodated in that auditorium then the ratio of the reverberation times with and without
audience is:

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

  1. 1/3

  2. 2/3

  3. 3/2

  4. 3/4

Show answer
Correct Option: B
Explanation:

The reverberation time of the auditorium with volume V and total absorption coefficient A is given by:
$T _1=

\dfrac{0.17V}{A}$. Now when we will include n number of people each

having an absorption coefficient $A _p$ in the auditorium than the

reverberation time of the auditorium will be:
$T _2= \dfrac{0.17V}{A+nA _p}$. Now, here $A _p/A=1/1000$ and $n=500$ Using this two values the $T _2/T _1=2/3$.

The reverberation times in a cinema theatre are 3s, 2s when it is empty, filled with audience respectively.  The reverberation time when the theatre is half filled  with audience is

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

  1. 2.3 s

  2. 2.4 s

  3. 2.5 s

  4. 2.6 s

Show answer
Correct Option: B
Explanation:

let Area be A, when empty
let Area of audience be $A _{2}$ when full
so Area of cinema theatre when full $=A _{1}+A _{2}$
Area when half full $=A _{1}+\frac{A _{2}}{2}$
$R.T. \alpha \frac{1}{A}  ; \frac{R.T _{1}}{R.T _{2}}=\frac{A _{2}+A _{1}}{A _{1}}$
$3A _{1}=2A _{1}+2A _{2}  ; A _{1}=2A _{2} ; A _{2}=^{A _{1}}/ _{2}$
$R.T _{3}\alpha \frac{1}{A _{1}\frac{A _{2}}{2}} =\frac{4}{5 A _{1}}$
$ this \frac{4}{5} times     3 sec =2.4 sec$
 

A pendulum has a frequency of 5 vibrations per second. An observer starts the pendulum and fire a gun simultaneously. He hears echo from the cliff after 8 vibrations of the pendulum. If the velocity of sound in air is 340  $ms^{-1}$, find the distance between the cliff and the observer :

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

  1. 252 m

  2. 240 m

  3. 272 m

  4. 182 m

Show answer
Correct Option: C
Explanation:

The sensation of any sound persists in our ear for about 0.1 seconds. This is known as the persistence of hearing. If the echo is heard within this time interval, the original sound and its echo cannot be distinguished. So the most important condition for hearing an echo is that the reflected sound should reach the ear only after a lapse of at least 0.1 second after the original sound dies off. 
It is given that a pendulum has a frequency of $5$ vibrations per second. An observer starts the pendulum and fire a gun simultaneously. He hears echo from the cliff after $8$ vibrations of the pendulum which means that he hears the echo after 1.6 seconds. The speed of sound is given as $340 m/s$.
The distance traveled by sound of the gun shot in $1.6 $ seconds is calculated from the formula 
Distance traveled $=velocity\quad of\quad sound\times time\quad taken$. That is, $340 \times 1.6 $ = 544 m. This is twice the minimum distance between a source of sound (man) and the reflector (cliff) as it is reflected sound. 
So, the cliff is at a distance of 272 m least from the man, for the reflected sound or the echo to be heard distinctly in 1.6 seconds.