(a) Define an echo. What is the minimum distance required between the source of sound and the reflecting body to hear a clear echo in air at $22^\circ ext{C}$? Show the calculation (speed of sound in air $= 344 ext{ m/s}$). (b) What is reverberation? List two methods used to reduce reverberation in large halls. (c) A person stands in front of a high mountain wall, claps, and hears the echo after $4 ext{ seconds}$. Calculate the distance of the wall from the person if the speed of sound in air is $344 ext{ m/s}$.

Q: (a) Define an echo. What is the minimum distance required between the source of sound and the reflecting body to hear a clear echo in air at $22^\circ\text{C}$? Show the calculation (speed of sound in air $= 344\text{ m/s}$). (b) What is reverberation? List two methods used to reduce reverberation in large halls. (c) A person stands in front of a high mountain wall, claps, and hears the echo after $4\text{ seconds}$. Calculate the distance of the wall from the person if the speed of sound in air is $344\text{ m/s}$.

(a) Echo and Calculation of Minimum Distance Echo: The repetition of sound caused by the reflection of sound waves from a distant, hard surface (like a mountain or wall) back to the listener is called an echo. Persistence of Hearing: The sensation of sound persists in our brain for about $0.1\text{ seconds}$. To hear a distinct echo, the reflected sound must reach the ear after at least $0.1\text{ s}$. Calculation: Speed of sound in air ($v$) $= 344\text{ m/s}$ Minimum time interval ($t$) $= 0.1\text{ s}$ Total distance covered by sound to go and return $= 2d$ Using formula: $2d = v \times t$ $2d = 344\text{ m/s} \times 0.1\text{ s} = 34.4\text{ m}$ $d = \frac{34.4}{2} = 17.2\text{ m}$ Result: The minimum distance of the reflecting obstacle must be $17.2\text{ meters}$. --- (b) Reverberation and Its Reduction Reverberation: The persistence of sound in a big hall or auditorium due to repeated, multiple reflections from walls, ceiling, and floor is known as reverberation. Excessive reverberation causes sound to overlap and become unclear. Methods to Reduce Reverberation: Covering the walls and ceiling of the auditorium with sound-absorbing materials such as compressed fibreboard or acoustic plaster. Providing heavy draperies/curtains on doors and windows and using upholstered seats to absorb excess sound energy. --- (c) Numerical Problem Solution Given: Time taken to hear the echo ($t$) $= 4\text{ s}$ Speed of sound ($v$) $= 344\text{ m/s}$ To Find: Distance of the wall ($d$) Formula: $$\text{Total distance travelled by sound } (2d) = v \times t$$ Step-by-Step Calculation: $$2d = 344\text{ m/s} \times 4\text{ s}$$ $$2d = 1376\text{ m}$$ $$d = \frac{1376}{2} = 688\text{ m}$$ Answer: The distance of the mountain wall from the person is $688\text{ meters}$.

Question

(a) Define an echo. What is the minimum distance required between the source of sound and the reflecting body to hear a clear echo in air at $22^\circ	ext{C}$? Show the calculation (speed of sound in air $= 344	ext{ m/s}$). (b) What is reverberation? List two methods used to reduce reverberation in large halls. (c) A person stands in front of a high mountain wall, claps, and hears the echo after $4	ext{ seconds}$. Calculate the distance of the wall from the person if the speed of sound in air is $344	ext{ m/s}$.

Accepted Answer

(a) Echo and Calculation of Minimum Distance

Echo: The repetition of sound caused by the reflection of sound waves from a distant, hard surface (like a mountain or wall) back to the listener is called an echo.
Persistence of Hearing: The sensation of sound persists in our brain for about $0.1\text{ seconds}$. To hear a distinct echo, the reflected sound must reach the ear after at least $0.1\text{ s}$.
Calculation:
- Speed of sound in air ($v$) $= 344\text{ m/s}$
- Minimum time interval ($t$) $= 0.1\text{ s}$
- Total distance covered by sound to go and return $= 2d$
- Using formula: $2d = v \times t$
- $2d = 344\text{ m/s} \times 0.1\text{ s} = 34.4\text{ m}$
- $d = \frac{34.4}{2} = 17.2\text{ m}$
- Result: The minimum distance of the reflecting obstacle must be $17.2\text{ meters}$.

(b) Reverberation and Its Reduction

Reverberation: The persistence of sound in a big hall or auditorium due to repeated, multiple reflections from walls, ceiling, and floor is known as reverberation. Excessive reverberation causes sound to overlap and become unclear.
Methods to Reduce Reverberation:
1. Covering the walls and ceiling of the auditorium with sound-absorbing materials such as compressed fibreboard or acoustic plaster.
2. Providing heavy draperies/curtains on doors and windows and using upholstered seats to absorb excess sound energy.

(c) Numerical Problem Solution

Given:

Time taken to hear the echo ($t$) $= 4\text{ s}$
Speed of sound ($v$) $= 344\text{ m/s}$

To Find:

Distance of the wall ($d$)

Formula: $$\text{Total distance travelled by sound } (2d) = v \times t$$

Step-by-Step Calculation: $$2d = 344\text{ m/s} \times 4\text{ s}$$ $$2d = 1376\text{ m}$$ $$d = \frac{1376}{2} = 688\text{ m}$$

Answer: The distance of the mountain wall from the person is $688\text{ meters}$.