Tag: conditions for hearing echo

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Multiple choice physics study of sound conditions for hearing echo echo reflection of sound and echo ultrasound and its applications

An echo returned in $3 s$. What is the distance of the reflecting surface from the source, given that the speed of sound is ${ 342ms }^{ -1 }$?

  1. $520 m$
  2. $515 m$
  3. $530 m$
  4. $513 m$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

For the echo, the sound travels distance $2d$ in $3s$. 
Thus, $2d=vt$ or $d=vt/2=\dfrac{342\times 3}{2}=513 m$ 

Multiple choice physics study of sound conditions for hearing echo echo reflection of sound and echo ultrasound and its applications

Sound produced by a thunderstorm is heard 10 s after the lightning is seen. The approximate distance of the thunder cloud :
(Given speed of sound $= 340 m s^{1}$)

  1. 3.4 Km

  2. 340 m

  3. 3.4 m

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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 $3.4 km.$

Multiple choice physics acoustics conditions for hearing echo echo reflection of sound and echo ultrasound and its applications

The ceiling and wall behind the stage of good conference halls or concert halls are made:

  1. Straight

  2. Curved

  3. Circular

  4. Rectangular

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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.

Multiple choice physics study of sound conditions for hearing echo echo reflection of sound and echo ultrasound and its applications

What should be the minimum distance between the listener and the reflector to hear an echo of sound propagating with a speed $v\   m s^{-1}$?

  1. $v/10$ m
  2. $10 v$ m
  3. $v/20$ m
  4. $20v$ m
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
if there is a sound reflector at a distance d from the source, then the time interval between original sound and its echo at the site of the source is given as $T=\dfrac { 2d }{ v } $
 $d=\dfrac { v\times t }{ 2 } = \dfrac { v\times 1 }{ 2\times 10 } $ as Time = 0.1 s.
Hence, $d=\dfrac { v }{ 20 }$.

Multiple choice physics study of sound conditions for hearing echo echo reflection of sound and echo ultrasound and its applications

A man starts from rest and moves with constant acceleration towards a hill which is $648 m$ away. He fires a gun at the instant of starting and hears its echo after $4$ seconds. Velocity of sound in air is $320 m/s$

  1. His acceleration is $1m/s^{2}$
  2. His acceleration is $2m/s^{2}$
  3. His velocity at the instant of hearing echo is $4 m/s$
  4. His velocity at the instant of hearing echo is $8 m/s$
Reveal answer Fill a bubble to check yourself
B,D Correct answer
Explanation

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.

Multiple choice physics acoustics conditions for hearing echo echo reflection of sound and echo ultrasound and its applications

A man fires a gun and hears its echo after 5 s. The man then moves 310 m towards the hill and fires his gun again. This time he hears  the echo after 3 s. Calculate the speed of sound. 

  1. $330\,m\, s^{-1}$
  2. $350\,m\, s^{-1}$
  3. $210\,m\, s^{-1}$
  4. $310\,m\, s^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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)
$v=310 m/s$

Multiple choice physics study of sound conditions for hearing echo echo ultrasound and its applications

An echo will be heard if the minimum distance between the source of sound and the obstacle is :

[assume air to be the medium]

  1. $1\;m$
  2. $10\;m$
  3. $15\;m$
  4. $17\;m$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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. 


Hence, the total distance covered by the sound from the point if generation to
the reflecting surface and back should be at least 

$344 \times 0.1= 34.4 m.$

Thus for hearing distinct echoes, the minimum distance pf the obstacle from 
the source of sound must be half this distance

That is, $\frac{34.4}{2}=17.2\,m \approx 17\,m$

Multiple choice physics study of sound conditions for hearing echo echo ultrasound and its applications

Bats detect the obstacle in their path by receiving the detected

  1. Infrasonic waves

  2. Ultrasonic waves

  3. Both

  4. None
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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.

Multiple choice physics study of sound conditions for hearing echo echo ultrasound and its applications

The technique used by bats to find their way or to locate food is _______.

  1. SONAR

  2. RADAR

  3. Echolocation

  4. Flapping
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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.