Tag: acoustics

Distance required for an echo to be heard is 17 m. Thus, $17 =  0 + \dfrac{0.5 t^2}{2} \implies t= 2\sqrt{17}=8.2 s$. 

The correct option is (b)

distance moved by the source = 17 m for the echo to be heard

Thus, $17 = v(4) \implies v = 17/4 = 4.25 $ m/s

The correct option is (c)

$\begin{array}{l} For\, \, source\, \, vS=r\omega =0.70\times 2\pi \times 5=22\, m/s \ the\, \, source\, \, is\, \, receding\, \, the\, \, man.\, it\, \, is\, \, given\, \, by\, \, { \eta _{ \min   } }=n\frac { v }{ { v+{ v _{ s } } } } Hz \ Minimum\, \, frequency\, \, is\, \, heard\, \, when=1000\times \frac { { 352 } }{ { 352+22 } } =941 \end{array}$

Like all waves, sound waves can be reflected. Sound waves suffer reflection from the large obstacles. As a result of reflection of sound wave from a large obstacle, the sound is heard which is named as an echo. Ordinarily echo is not heard as the reflected sound gets merged with the original sound. Certain conditions have to be satisfied to hear an echo distinctly (as a separate sound).
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.
As the speed of sound is 340 m/s, the distance traveled by sound in 0.1 second is 34 m. This is twice the minimum distance between a source of sound and the reflector. So, if the obstacle is at a distance of 17 m at least, the reflected sound or the echo is heard after 0.1 second, distinctly. 
In the question, the wall is 25 m away from the man and hence, the man will be able to hear the echo distinctly.

When sound waves travel in a medium, it gets reflected from the obstacle and we hear the same sound produced even after some time. This reflected sound is known as echo.
Human ear can hear an echo if the reflected sound reaches the ear after at least $0.1 \ s$. If the reflected sound reaches the human ear in less than $0.1\ s$, then we cannot hear the echo.

The minimum distance between source of sound and obstacle , to hear an echo is 17m , as the dimensions of the park are $l=6m  , b=6m$ , therefore  distance between road and building is 6m , therefore echo of the bomb sound will not be produced as distance between source of sound (bomb) and building is less than 17m i.e. echo will not be heard .

In concert or conference halls, music or other sounds that are produced on the stage must be carried through the air to people in the crowd. Some of these sound waves go directly to the people in the crowd, without bouncing off on anything first. However, other sound waves from the stage first go to areas like walls and ceilings, and the sound either bounces off or is absorbed. When sound waves bounce off a surface, the new direction they travel is related to the angle they strike the surface. 
Thus, acoustical engineers and architects sometimes need to place panels on the ceilings and walls to reflect sound in a specific direction, that is, back to the audience. Using panels that are curved have a similar effect. Hence, ceiling and wall behind the stage of good conference halls or concert halls made curved so that sound after reflection reaches all the corners of the hall and the whole audience.

The speed of the sound is calculated as follows.
Let d be the distance between the man and the hill in the beginning.
$v=\dfrac { 2\times d }{ t } $ = $\dfrac { 2\times d }{ 5 } \rightarrow eqn\quad 1$
He moves 310 m towards the hill. Therefore the distance will be (d - 310) m. Therefore,
$v=\dfrac { 2\left( d-310 \right)  }{ 3 } \rightarrow eqn\quad 2$
Since velocity of sound is same, equating (1) and (2), we get
$\dfrac { 2d }{ 5 } =\dfrac { 2\left( d-310 \right)  }{ 3 } $
3d = 5d - 1550
2d = 1550
d = 775 m
Hence, the velocity of sound v=$\dfrac { 2\times 775 }{ 5 } $ (substituting in equation 1)
$v=310 m/s$