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Reflection of Sound

Subject: Science

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Overview

Echo is the reflection of sound by a hill or a large building some distance away. When sound falls on a surface, it is partly reflected and partly absorbed. If there are a number of reflecting surfaces near one another, a series of reflection takes place. Such sound is called reverberation. This note provides information about echo, application of echo, absorption of sound, reverberations and some numerals.
Reflection of Sound

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Like light, the sound reflects and obeys the laws of the reflection. Reflection of sound can be heard in long tunnels, deep wells, deep forest, deep ridges of mountains etc.

Echo is, simply, the reflection of sound by a hill or a large building some distance away. You can hear an echo only from a reflecting surface which is far away. This is because when sound reaches in your ears very quickly, your ears will not be able to distinguish it as a separate sound. In fact, your ears can able to distinguish two sounds as separate sounds only if the second sound wave is received 1/15th of a second or more after the first sound.The minimum distance for an echo to be heard is 17 m approximately.

Application of Echo

The principle of reflection of sound is used in an instrument called SONAR which is used to find the depth of the sea at any time. A SONAR produces ultrasonic pulses which are sent down into the sea from a ship. They are received back after being reflected by the bottom of the sea. SONAR stands for Sound Navigation And Ranging.

The depth of the sea is calculated by measuring the time taken by the echo to be received.

Nature also makes use of the principle of reflection of sound. The bat uses the echo to feel its way through and also to detect its prey. It gives out shrieks of ultrasound which gets reflected from any obstacle and avoids it. If echo comes from an insect, it easily recognises and catches it.

Absorption of sound

When sound falls on a surface, it is partly reflected and partly absorbed. Walls of aroom are good reflectors of sound. Wood, carpets, curtains, clothes and even our bodies absorb sound better and reflect less. In fact, soft surfaces are better absorbers of sound whereas hard surfaces are better reflectors of sound. Porous materials, Soft materials, non-metals, etc. are absorbers of sound.

Reverberations

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If there are a number of reflecting surfaces near one another, a series of reflection takes place. Such sound is called reverberation. The reverberation of a thunder is due to thenumber of reflections of sound from clouds, rocks and cliffs, etc. Reflection of sound takes place in a big room, hall or newly constructed room. If the reflecting surface is less than 15m, reverberation distinguishes an echo from the source of a sound. The value of this suitable distance depends on the type of sound produced.

Things to remember
  • Echo is simply the reflection of sound by a hill or a large building some distance away hear an echo only from a reflecting surface which is far away.
  • The principle of reflection of sound is used in an instrument called SONAR which is used to find the depth of the sea at any time.
  • The depth of the sea is calculated by measuring the time taken by the echo to be received.
  • It includes every relationship which established among the people.
  • There can be more than one community in a society. Community smaller than society.
  • It is a network of social relationships which cannot see or touched.
  • common interests and common objectives are not necessary for society.
Videos for Reflection of Sound
Reflection of Sound
Reverberation
Questions and Answers

The phenomenon of returning the incident sound wave from a distance more than 17 m is known as echo.
Its two uses are:
i) It is used to find the depth of the sea.
ii) Bats use echo to find its path and prey.

The phenomenon of multiple reflections of sound when a reflector is at a distance less than 17m is known as reverberation. The conditions for the occurrence of echo or reverberation are : i) The sound should be so loud that after reflection also listener must listen.
ii) For an echo distance of reflecting object must be more than 17m and for reverberation, the distance must be less than 17m.

In an empty room, no absorbers are there,  so that almost all sound is reflected from the wall and heard louder but materials in the room absorb more sound. Hence, sound will be faint.

Echo is not heard in a small room because to occur an echo reflector should be more than 17m distance and in a small room, it becomes less than 17m.

We know that a cinema hall is large. The distance between the wall and the source of sound is more than 17m. So, there may be echo and sound may not be clear. So, to avoid it, sound absorbers are kept on the walls of a cinema hall.

Echo Reverberation
1. The reflected sound is repeated. 1. The reflected sound is prolonged.
2. The distance between the sources and sound and reflecting surface should be more than 17m. 2. The distance between the sources and sound and reflecting surface should be less than 17m.

Given,
Echo time (t) = 3s
Velocity of sound (v) = 332m/s
Distance (d) = ?
For echo,
v = 2d/t
or, 332 = 2d/3
or, 996/2 = d
∴ d = 448m
Hence, the distance between the man and the hill is 448m.

Given,
Distance (d) = 2km = 2000m
Time (t) = 6sec
Speed (v) = ?
We know, Speed (v) = d/t = 2000/6 = 333.3m/s
Hence, speed of sound is 333.3m/s.

Given,
Echo time (t) = 0.4s
Velocity of sound (v) = 332m/s
Distance (d) = ?
For echo,
v =2d/t
or, 332 = 2d/0.4
or, 132.8/2 = d
∴ d = 66.4m
Hence, the distance between Sita and the surface of water in the well is 66.4m.

Echo time (t) = 5s
Velocity of sound (v) = 330m/s
Distance (d) = ?,
v =d/7 or, 332 = d/5 or, 1650 = d ∴ d = 1650m Hence, the distance of flash from the observer is 1650m.

An instrument called SONAR is used to find the depth of the sea.

Solution,

Distance traveled, s = 660 m/s

Speed (v) = distance traveled / time taken

= 660/ 330

= 2 sec#

So the sound is heard after 2 second.

Solution,

Total distance traveled by sound in going and coming back = 2 sec

Therefore, V = distance traveled/ time taken

Or, s = V x t/ 2

Or, s = 350 x ½

Or, s = 175 m#

Thus, the distance of the hill from the boy is 175 m.

Solution,

t = 1.5 s

v = 14000m/s

h= v x t/2

= 14000 x 1.5 /2

= 1050m#

Thus, the depth of the sea is a 1050m for students.

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