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The laws of Physics Topics are used to explain everything from the smallest subatomic particles to the largest galaxies.

## What is the Persistence of Hearing?

Definition: If the reflected sound is heard separately from an original sound, then the reflected sound is called an echo of the original sound.

We know from practical experiences that the sound produced by drums, crackers, gunshots, etc., reach our ears directly, and also produce echoes after getting reflected from faraway buildings, trees, etc.

Echoes are distinct only when the distance between the listener and the reflecting surface is sufficiently large to allow the reflected sound to reach the listener without overlapping with the original sound.

### Minimum distance of the reflector

Inarticulate sound: Echoes are produced due to reflection of sound. But all reflected sounds cannot produce echo. When we hear an inarticulate sound, its sensation persists in our ear for about \(\frac{1}{10}\)th of a second. This is known as persistence of hearing. So to distinguish echo from the original sound, at least \(\frac{1}{10}\)th of a second should be elapsed between the production of sound and the arrival of the reflected sound in the ear—this is the condition required to hear an echo of inarticulate sound.

We know that the velocity of sound at 0°C in air is about 330 m ᐧ s^{-1}. So, distance travelled by sound in \(\frac{1}{10}\)s is 330 × \(\frac{1}{10}\) = 33 m. Hence, in order to hear an echo, the minimum

distance between the reflecting surface and the listener should be \(\frac{33}{2}\) = 16.5 m = 54 ft. This calculation is valid for inarticulate sounds, like sound produced by a gunshot, clapping, etc.

If the distance of the reflector from the listener is less than 16.5 in or 54 ft. sound takes less than \(\frac{1}{10}\)s to come back to the listener after reflection. So, the listener cannot distinguish between the original sound and the reflected one, i.e., echo is not heard. For this reason, echo is not audible in a room of ordinary size.

Articulate sound: To hear an echo of articulate sound, the minimum distance between the reflector and the listener is greater. This is due to the fact that a man cannot distinctly pronounce more than 5 syllables in one second. Our ears also cannot recognise separately more than 5 syllables in one second. So, for monosyllabic sound, the time interval between the original sound and its echo must not be less than \(\frac{1}{5}\)s.

During this time, sound travels (330 × \(\frac{1}{5}\)) or 66 m. This distance is again twice the distance between the listener and the reflector. So the minimum distance of the listener from the reflector should be \(\frac{66}{2}\) = 33 m, for hearing the echo of a monosyllabic word. For a disyllabic word the distance should be 2 × 33 = 66 m, and so on.

Multiple echoes: When a sound is reflected from a number of suitably placed reflecting surfaces, a number of echoes may be heard. Minimum \(\frac{1}{10}\)s of time interval is required between the original sound and the first echo and also between any two successive echoes which reach our ears. The rumbling and rolling of thunder is really caused by multiple reflections from cloud surfaces.

Successive reflections may also be produced between two reflecting surfaces suitably placed.

Reverberation: It is often noted that when a loud sound is produced near the walls of a big hail, it returns after reflection very quickly (i.e., time interval is less than \(\frac{1}{10}\)s), when the effect of the original sound still persists. On account of successive reflections from the walls, a continuous rolling of sound goes on for some time. This persistence of sound due to multiple reflections from the walls is called reverberation.

This effect is not at all desirable sometimes, because the original sound cannot be recognised distinctly. This effect is minimised by increasing the absorption of sound at the walls by carpeting them with sound absorbing substances.

It is obvious that if there are multiple reflectors, either echo or reverberation of sound will be produced. If the time interval between successive sounds is greater than \(\frac{1}{10}\)s, echo will be heard; if the time interval is less than \(\frac{1}{10}\)s, reverberation will be heard.

### Practical Applications of Echo

1. Measurement of a distance: Distance of a distant hill or of any extended reflector can be determined with the help of echo. In this case, a sharp inarticulate sound, like sound of a gunshot, is produced, and simultaneously a stopwatch is started. The stopwatch is stopped just on hearing the echo from the distant reflector.

Let the distance of the reflector from the place of experiment be D, the time interval between the original sound and the echo heard be t and the velocity of sound in air be V.

∴ Distance travelled by sound = Vt

Since sound travels the distance D and again returns to the place of experiment. the total distance travelled is 2D.

∴ 2D = Vt or, D = \(\frac{V t}{2}\)

Knowing the velocity of sound in air (V) and the time (t) from the stopwatch. the distance (D) of the reflector can be determined.

2. Measurement of the depth of a sea: The depth of a sea can be measured with the help of echo. A source S containing some explosive material is kept at a depth h from one end of a ship and a hydrophone H (an underwater microphone) is kept immersed at the same depth h from the other end [Fig.]. A loud sound is produced due to the explosion at S. Sound can reach the hydrophone along two paths—

- straight Line path SH and
- path SOH after reflection from the bottom O of the sea.

The sound that reaches along SH is called original sound and the sound that reaches along SOH is called reflected sound. The hydrophone electrically records the time intervals in which sound travels along the direct path SH, as well as along the reflected path SOH.

Let the velocity of sound in water be V, the depth of the source and the hydrophone from the surface of the sea be h, the depth of the sea with respect to the source or the hydrophone be NO = d, time taken by the sound to reach the hydrophone directly be t_{1}, and time taken by the echo be t_{2}.

According to the figure, SH = Vt_{1}, i.e., SN = \(\frac{V t_1}{2}\)

Again, SO + OH = Vt_{2}

or, SO = \(\frac{V t_2}{2}\) [∵ SO = OH]

∴ d = ON = \(\sqrt{S O^2-S N^2}\) = \(\sqrt{\left(\frac{V t_2}{2}\right)^2-\left(\frac{V t_1}{2}\right)^2}\)

= \(\frac{V}{2} \sqrt{t_2^2-t_1^2}\)

So, depth of the sea from its surface is,

D = h + d = h + \(\frac{V}{2} \sqrt{t_2^2-t_1^2}\)

So if V is known, the depth D of the sea can be determined. If the depth of the sea is very large, the length of the path SH is very small compared to the path SOH. So, t_{1} may be neglected relative to t_{2} and also h may be neglected relative to d,

i.e., in that case, D = \(\frac{V t_2}{2}\)

This method is called echo depth ranging or SONAR (Sound Navigation And Ranging).

3. Measurement of the altitude of an aeroplane: The altitude of a flying aeroplane can also be measured with the help of echo. Let us consider an aeroplane flying horizontally at a height h along the path AB with a velocity v as shown in Fig. When the plane is at A, a loud sound is produced and the echo is heard after time t when the plane is at B. So, sound reflects from the ground at the point O and reaches the aeroplane along the path AOB, where AO = OB.

Let velocity of the plane be v, velocity of sound in air be V, and time interval between the original sound and the echo be t.

∴ Distance travelled by the plane = AB = vt

i.e., AN = \(\frac{v t}{2}\) [AN = NB]

Again, distance travelled by the sound is AO + OB = Vt

i.e., AO = \(\frac{v t}{2}\) [∵ AO = OB]

So, the altitude of the plane,

h = ON = \(\sqrt{A O^2-A N^2}\) = \(\sqrt{\left(\frac{V t}{2}\right)^2-\left(\frac{v t}{2}\right)^2}\) = \(\frac{t}{2} \sqrt{V^2-v^2}\)

Therefore, knowing v, V and t, the altitude h of the plane can be determined from the above equation.

### Numerical Examples

**Example 1.**

A shell is fired from a motor boat moving along the surface of a sea with a speed of 60 km ᐧ h^{-1} towards

the shore. The echo of the firing is heard in the motor boat after 9 s. What was the distance of the boat from the shore at the time of firing? Velocity of sound 330 m ᐧ s^{-1}.

**Solution:**

Distance travelled by sound in 9 s = 330 × 9

= 2970 m

Now, 60 km ᐧ h^{-1} = \(\frac{60 \times 1000}{60 \times 60}\) = \(\frac{50}{3}\) m ᐧ s^{-1}

So, the distance travelled by the motor boat in 9 s is,

= \(\frac{50}{3}\) × 9 = 150 m

If the distance of the shore at the time of firing is x m, then the sound of firing moves x m and then returns (x – 150) m after reflection from the shore.

So, x + (x – 150) = 2970

or, 2x = 2970 + 150 or, 2x = 3120 or, x = 1560

So, the distance of the motor boat from the shore at the time of firing is 1560m or 1.56 km.

**Example 2.**

What should be the minimum distance of a reflector to hear the echo of a tetrasyllabic word? Velocity of sound = 330 m ᐧ s^{-1}.

**Solution:**

The minimum time interval required between the original sound and the echo of a tetrasyllabic word = 4 × \(\frac{1}{5}\) = \(\frac{4}{5}\)s.

∴ Minimum required distance = \(\frac{330 \times 4 / 5}{2}\) = 132 m.

**Example 3.**

A man is approaching a hill with a velocity of 5 m ᐧ s^{-1}. He fires a gun when he is 3 km away from the hill. When and where will he listen the echo? Velocity of sound = 355 m ᐧ s^{-1}. [HS ’04]

**Solution:**

3 km = 3000 m

Let us consider that the man advances x in after firing the gun and before hearing the echo.

Time required for this purpose, t = \(\frac{x}{5}\)s

Distance travelled by sound in that time

= 3000 + (3000 – x) = (6000 – x) m

∴ t = \(\frac{6000-x}{355}\)

i.e., \(\frac{x}{5}\) = \(\frac{6000-x}{355}\) or 360x = 5 × 6000

or, x = \(\frac{5 \times 6000}{360}\) = \(\frac{250}{3}\) = 83.3 m

∴ t = \(\frac{x}{5}\) = \(\frac{83.3}{5}\) = 16.67 s

So, the man will listen the echo after travelling 83.3 m, 16.67s after firing the gun.

**Example 4.**

A gun is fired from an aeroplane moving horizontally with a velocity of 180 km ᐧ h^{-1} and the echo of the sound after reflection from the ground is heard after 3 s. What is the height of the aeroplane? Velocity of sound = 340 m ᐧ s^{-1}.

**Solution:**

180 km ᐧ h^{-1} = \(\frac{180 \times 1000}{60 \times 60}\) m ᐧ s^{-1} = 50 m ᐧ s^{-1}

In Fig., distance travelled by the plane in 3 s

= AB = 50 × 3 m

So, AN = \(\frac{A B}{2}\) = \(\frac{50 \times 3}{2}\) = 25 × 3 m

Distance travelled by sound in that time

= AO + OB = 340 × 3 m

i.e., AO = \(\frac{340 \times 3}{2}\) = 170 × 3 m

So, height of the plane = ON = \(\sqrt{A O^2-A N^2}\)

= \(\sqrt{(170 \times 3)^2-(25 \times 3)^2}\)

= \(\sqrt{(510)^2-(75)^2}\) ≈ 504.45 m.

**Example 5.**

A man standing 85 m away from a high wall is clapping in a regulated manner. When he claps twice per second, each clap coincides with the echo of its previous clap. Determine the velocity of sound in air.

**Solution:**

Time interval between two claps, t = \(\frac{1}{2}\)s

If V is the velocity of sound in air, 2s = Vt

∴ 2 × 85 = V × \(\frac{1}{2}\) or, V = 85 × 4 = 340 m ᐧ s^{-1}.

**Example 6.**

A ship’s sound measuring equipment sends out pulses looking for echoes from submarines within a 3 km radius. What should be the pulse repetition frequency? Given, velocity of sound in water = 1500 m ᐧ s^{-1}.

**Solution:**

Total path travelled by the pulses

= 3 + 3 = 6 km = 6000 m

Time taken by a pulse for the to and fro journey is,

t = \(\frac{6000}{1500}\) = 4 s

∴ Pulse repetition frequency,

n = \(\frac{1}{4}\) per second = 15 per minute.

**Example 7.**

A man standing at a certain distance from a hill is beating a drum. He observes that if he beats the drum at the rate of 40 the echo is not heard distinctly. He advances 90 m towards the hill and observes that if he beats the drum at the rate of 60 min^{-1}, the echo cannot be heard distinctly. Determine the distance of the first position of the man from the hill and also the velocity of sound.

**Solution:**

Let the distance of the first position of the man from the hill be x and velocity of sound be V.

In the first case, time interval between two consecutive sounds,

t_{1} = \(\frac{60}{40}\) = \(\frac{3}{2}\)s

∴ When echo is just not heard distinctly,

2x = V × \(\frac{3}{2}\)

In the second case, time interval between two consecutive sounds,

t_{2} = \(\frac{60}{60}\) = 1 s

∴ In this case, when the echo is just not heard distinctly,

2(x – 90) = V × 1

Dividing (2) by (1) we have,

\(\frac{2(x-90)}{2 x}\) = \(\frac{V \times 2}{V \times 3}\) or, 3x – 270 = 2x, x = 270 m

From equation (1) we get,

2 × 270 = V × \(\frac{3}{2}\) or, V = 360 m ᐧ s^{-1}.