Tag: applications of ultrasound

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

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

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

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 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 $T _2/T _1=2/3$.

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$
 

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.

When a reflecting surface is less than 17 m away from the source, the echo that returns appears to be the original sound, just prolonged. This effect is called reverberation. The time interval between the original sound and the returning echo must be less than 0.1 seconds. It depends on the adsorption coefficient of the material which usually defines at 2048 Hz frequency.

Answer is A.
Ultrasound is sound with a frequency higher than 20 kHz. This is above the human range of hearing. The most common use of ultrasound, creating images, has industrial and medical applications.
Sonar illustrates how a ship on the ocean utilizes the reflecting properties of ultrasound waves to determine the depth of the ocean. A ultrasound wave is transmitted and bounces off the seabed. Because the speed of sound is known and the time lapse between sending and receiving the sound can be measured, the distance from the ship to the bottom of the ocean can be determined. This technique is called sonar (originally an acronym for SOund Navigation And Ranging).