Echo and Sound Reflection

An echo is the repetition of sound that results as a reflection from a surface. An echo of the sound produced can be heard only if it reaches our ear after 1 / 10th of a second. This is due to the persistence of sound.

we can hear echoes anywhere, where the distance between the source and the point of reflection is sufficient enough.

As the sensation of sound persists in our brain for about 0.1 s, to hear a distinct echo the time interval between the original sound and the reflected one must be at least 0.1 s. If we take the speed of sound to be 344 m/s at a given temperature, say at 22 $^0C$ in air, sound must go to the obstacle and reach back the ear of the listener on reflection after 0.1s.
 
Hence, the total distance covered by the sound from the point of generation to the reflecting surface and back should be at least (344 m/s) ×0.1 s = 34.4 m. Thus, for hearing distinct echoes, the minimum distance of the obstacle from the source of sound must be half of this distance. I.e., $\frac{34.4}{2} \approx 17 ,m$

As the sensation of sound persists in our brain for about 0.1 s, to hear a distinct echo the time interval between the original sound and the reflected one must be at least 0.1 s. If we take the speed of sound to be 344 m/s at a given temperature, say at 22 $^0C$ in air, sound must go to the obstacle and reach back the ear of the listener on reflection after 0.1s.
 
Hence, the total distance covered by the sound from the point of generation to the reflecting surface and back should be at least (344 m/s) ×0.1 s = 34.4 m. Thus, for hearing distinct echoes, the minimum distance of the obstacle from the source of sound must be half of this distance. I.e., $\frac{34.4}{2} \approx 17 ,m$

The time gap between the two sounds-direct and echo depends upon the medium through which the sound travels. These two sounds can be heard distinctly provided the Distance between the observer and the reflecting surface is large enough to allow the reflected sound to reach him without interfering with the direct sound.

The time gap between the two sounds-direct and echo depends upon the medium through which the sound travels. These two sounds can be heard distinctly provided the Distance between the observer and the reflecting surface is large enough to allow the reflected sound to reach him without interfering with the direct sound.

In rooms having walls less than 17.2 m away from each other, no echo can be heard.

For the production of an echo, the reflecting surface or the obstacle should be rigid such as building, hill, cliff.
Echo can be produced if the size of the obstacle reflecting the sound is quite large.

The minimum distance in air between the observer and the obstacle for an echo to be heard clearly at temperatures higher than $25^o$C is more than 17.2 m.
The speed of sound increases with rise in temperature.

The options are the conditions for an echo to be produced. The obstacle should be rigid at a minimum distance of $17m$ for an echo to be produced.

For the production of an echo, the reflecting surface or the obstacle should be rigid such as building, hill, cliff.

The minimum distance in air between the observer and the obstacle for an echo to be heard clearly is nearly 17.2 m. This condition is followed here but for the production of an echo, the obstacle should be rigid.

$1.$ Echo is simply the reflection of sound that arrives at the listener with a delay after the direct sound.

Rooms having walls less than 17.2 m away from each other, no echo can be heard. here, 159 m is greater than 17.2 m. So an echo will be heard if other conditions are fulfilled.

If the time interval between original sound and reflected sound is more than $\frac{1}{10},s$, then the original sound and the reflected sound can be clearly heard. This reflected sound is known as echo.

Time, after which an echo is heard is given by $t = \dfrac {Distance}{Speed \ \ of \ \ sound \ \ in \ \ air}$
Therefore, since speed of sound in air increases with the increase in temperature, so the time after which echo will be heard decreases. Hence, echo will be heard, sooner than the echo heard when temperature near $44^0C$

In rooms having walls less than 17.2m away from each other, no echo can be heard.

For the production of an echo, the reflecting surface or the obstacle should be rigid such as building, hill, cliff. Echo can be produced if the size of the obstacle reflecting the sound is quite large.

In rooms having walls less than 17.2m away from each other, no echo can be heard. Here,walls of the room are at a distance of 159 m, much greater than 17.2m. So an echo will be heard if other conditions are fulfilled.

The minimum distance in air between the observer and the obstacle for an echo to be heard clearly is nearly 17.2m. This condition is followed here but for the production of an echo, the obstacle should be rigid.

The minimum distance in air between the observer and the obstacle for an echo to be heard clearly at temperatures higher than ${25}^{0}C$ is more than $17.2m$. The speed of sound increases with rise in temperature.

For the production of an echo, the reflecting surface or the obstacle should be rigid such as building, hill, cliff etc.

The human ear can hear two sounds distinctly only if there is a time interval of $\frac{1}{10}$seconds between them. Now, the speed of sound in air at $20^0$C is $340m/s$, so the distance traveled in $\frac{1}{10}$seconds will be: 
Distance = Speed X time 
$=340 \times \frac{1}{10} \= 34m$

Thus, it is possible to hear the echo at the minimum distance of $\frac{34}{2}= 17m$ between the source of sound and reflecting surface. 

Minimum frequency to the human ear is 20
$\therefore T= \dfrac {1} {F}$
$T=\dfrac{1} {20}$
$d= V \times t$
$d= V\times \dfrac {1} {20}$
$\therefore$ The minimum distance is$ \dfrac {V} {20}$, where V is speed of the sound.

Let the distance between the two cliffs be d. Since, the man is standing
midway between the two cliffs, then the distance of man from either end
is $d/2$.
The distance travelled by sound (in producing an echo)
$2\times \displaystyle\frac{d}{2}=v\times t\Rightarrow d=340\times 1=340$m.

The echo of the first signal will overlap with the second signal. Thus, an echo will not be absent

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)

To hear an echo, the distance between the source and the reflector should be atleast 17 m. Thus the source should be moved 7 m away from the wall

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)

to hear echo of a sound the minimum distance between source and reflecting surface should be 17m.

Let the distance between the source of sound and the obstacle be $x$ metres .

Let the distance between the source of sound and the obstacle be $x$ metres .

According to the Doppler effect, if both the source and observer move at the same velocity in the same direction, the relative velocity between them is zero. Therefore, the frequency heard remains the same as the source frequency.

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

The total distance traveled by the sound is speed times time: 346 m/s * 5 s = 1730 m. Since the sound travels to the cliff and back, the distance to the cliff is half of the total distance: 1730 / 2 = 865 m.

The distance to the first cliff is d1 = (v * t1) / 2 and the second is d2 = (v * t2) / 2. The total distance between the cliffs is d1 + d2 = (v/2) * (t1 + t2) = (336/2) * (3 + 5) = 168 * 8 = 1344 m. This equals 4 * 336 m.

The engine is moving toward the hill at 15 m/s (54 km/h). The frequency of the echo is f' = f * (v + vs) / (v - vs), where v is speed of sound and vs is speed of engine. f' = 160 * (335 + 15) / (335 - 15) = 160 * 350 / 320 = 175 Hz.

Distance = 1020 m. Total sound distance = 2 × 1020 = 2040 m. If echo is heard after 1.5 s (assuming 11/2 means 1.5), time available for driver approach = (2040/340) - 1.5 = 6 - 1.5 = 4.5 s. Speed = 1020 / 4.5 = 31 m/s (approximately).

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.

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 in water is 1400 m/s in the question, the distance traveled by sound in 0.1 second is 35 m, which is calculated from the formula 
$Distance; traveled=velocity; of; sound \times time; taken$. That is, 140 m. This is twice the minimum distance between a source of sound and the reflector. So, if the obstacle is at a distance of 70 m at least, the reflected sound or the echo is heard after 0.1 second, distinctly.

$\displaystyle V _{air}=340m/s, V _{water}=1440m/s$
Echo is the sound that we hear after reflection of sound wave from the reflecting surface.
i.e., total distance travel by sound wave $=2d$
velocity$\displaystyle=\frac{distance}{time}$
$\displaystyle 1440=\frac{2d}{1.5}\Rightarrow d=\frac{1440\times 1.5}{2}=1080.0m=1.08kms$
Hence, option $B$ is the correct answer.

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 this case, the girl is seated in the middle of the park. So, the sound travels through a distance of 6 m up to the girl and then another 6 m up to the building. The total distance is just 12 m which is less than the minimum distance required for the sound to echo. Hence, the echo cannot be heard in this case. 

The sensation of any sound persists in our ear for about 0.1 seconds.  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 travelled 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.

Echo is not heard distinctly, when next beat overlaps with echo simultaneously.
Time per beat $=$ time taken by reflected beat to reach listener.
d being the distance and drum beating at rate of $40$ per minute.
$\dfrac{2d}{v}=\dfrac{60}{40}=\dfrac{3}{2}$
and $\dfrac{2d-80}{v}=1$
Solving we get $d=240m$

An echo is heard when a sound takes minimum of $0.1 \ s$ to reach the observer after getting reflected from the obstacle.
Let the distance between the distance between the source (or observer) and the obstacle be $x$.
Thus distance traveled by sound to hear echo is $2x$.
Speed of sound  $v = 340 \ m/s$
$\therefore$  $2x = vt$
Or  $2x = 340\times 0.1$
$\implies  \ x = 17 \ m$

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 .

Time taken by the sound to go from the boy to the cliff, $t=\dfrac{4}{2}s=2s$

Speed of sound in air, $v=332m{s}^{-1}$ 

So, distance of the cliff from the boy, $s=v\times{t}=332\times2m=664m$.

Let distance of hill from the position of the body be $d$.

Echo is heard after $6s$

Speed of sound, $v=340m/s$

Time taken by sound to travel from the source to the deflecting surface $t=\dfrac{3}{2}=1.5s$

Speed of sound $v=340m/s$

So, distance between the surface and the source $=v \times t=340 \times 1.5=510m$

If $d$ be the distance between reflecting surface and the source. 

Lightning and thunder are produced simultaneously, but the thunder is heard a few seconds after the lightning is seen. This is because the speed of light in air is more than the speed of sound in air.
The speed of light is given as $3\times { 10 }^{ 8 }$ m/s. The speed of sound is 340 m/s. Thus, the entire time of 10 seconds (as mentioned in the question) between seeing the light and hearing the thunderstorm is taken by the sound to travel to the observer. Hence, the distance traveled by the thunder is given as follows.
$s=vt$; where,
s is the distance traveled, v is the velocity of sound and t is the time taken.
That is,  $330m/s\times 10s = 3400 m = 3.4 km.$
Thus, the distance traveled by the thunder is $3.4 km.$

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.

Using s = 0.5 * a * t^2, the distance the man travels is 0.5 * a * 16 = 8a. The sound travels to the hill (648 m) and back to the man (648 - 8a). Total distance = 648 + (648 - 8a) = 1296 - 8a. Also, distance = v_sound * t = 320 * 4 = 1280. 1296 - 8a = 1280 implies 8a = 16, so a = 2 m/s^2. Velocity v = a * t = 2 * 4 = 8 m/s.

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)
$v=310 m/s$

As the sensation of sound persists in our brain for about 0.1 s, to hear a distinct echo the time interval between the original sound and the reflected one must be at least 0.1 s. If we take the speed of the sound to be 344 m/s at a given temperature, say at 22 $^0 C $ in air, sound must go to the obstacle and reach back the ear of the listener on reflection after 0.1 s. 

The persistence of audibility is 0.1 s

Bats can produce ultrasonic waves by flapping their wings. They can also detect these waves.these waves produced by the bats after reflection from the obstacles like building guide them to  remain away from the obstacles during night with the help of ultrasonic waves.

Bats can produce ultrasonic waves by flapping their wings. They can also detect these waves.these waves produced by the bats after reflection from the obstacles like building guide them to  remain away from the obstacles during night with the help of ultrasonic waves.

Among the four creatures birds voice has higher pitch than other three. Higher pitch means higher frequency. Pitch determines the shrillness of sound.

Has been seen that the behaviour of animals becomes unusual just before the tremor is felt.This is because animals are able to detect infrasonic waves produced at the time of tremor. Infrasonic waves are waves of frequency less than 20Hz.