MP Board · Class 9 · Science · Sound(a) What is reverberation? How can excessive reverberation in a large hall or auditorium be reduced? Explain any three practical methods used for this purpose. (b) Show mathematically how the minimum distance required between the source of sound and an obstacle to hear a distinct echo is determined. (Take the speed of sound in air as $344\text{ m/s}$ at $20^\circ\text{C}$). (c) A sonar device on a submarine sends out an ultrasonic signal towards an underwater obstacle and receives an echo $2.5\text{ seconds}$ later. Calculate the distance of the obstacle from the submarine if the speed of sound in seawater is $1500\text{ m/s}$.

(a) Reverberation and Methods to Reduce It

Reverberation: \nReverberation is the persistence of sound in a big hall or auditorium due to repeated or multiple reflections of sound waves from the walls, ceiling, and floor, even after the source has stopped producing sound. If reverberation is too long, the sound becomes blurred and distorted, making speech unintelligible.

Methods to Reduce Excessive Reverberation:

1. Use of Sound-Absorbing Materials on Walls and Ceilings: The walls and roof of auditoriums are covered with sound-absorbent materials such as compressed fibreboard, rough plaster, or acoustic tiles to absorb excess sound waves.
2. Draperies and Heavy Curtains: Thick, heavy curtains are hung on windows and doors. These soft materials absorb sound reflections rather than reflecting them back.
3. Upholstered Seats and Flooring: Seats in halls are covered with soft, sound-absorbing materials, and floors are carpeted to reduce the reflection of sound caused by empty seats and footsteps.

(b) Derivation of Minimum Distance for an Echo

• The human ear retains the sensation of sound for about $0.1\text{ s}$ (this is known as the persistence of hearing).
• To hear a distinct echo, the reflected sound wave must reach the ear after at least $0.1\text{ s}$ from the original sound. \nLet:
• Speed of sound in air at $20^\circ\text{C}$, $v = 344\text{ m/s}$
• Minimum time interval required, $t = 0.1\text{ s}$
• Distance travelled by sound to the obstacle and back $= 2d$ \nFormula: $$\text{Total distance} = \text{Speed} \times \text{Time}$$ $$2d = v \times t$$ $$2d = 344 \times 0.1 = 34.4\text{ m}$$ $$d = \frac{34.4}{2} = 17.2\text{ m}$$ \nThus, the minimum distance between the source of sound and the reflecting obstacle must be $17.2\text{ metres}$ to hear a clear echo in air at $20^\circ\text{C}$.

(c) Numerical Solution

Given:

• Time interval for echo, $t = 2.5\text{ s}$
• Speed of sound in seawater, $v = 1500\text{ m/s}$

To find:

• Distance of the obstacle, $d$

Formula: $$\text{Distance } (d) = \frac{\text{Speed } (v) \times \text{Time } (t)}{2}$$

Calculation: $$d = \frac{1500\text{ m/s} \times 2.5\text{ s}}{2}$$ $$d = \frac{3750}{2} = 1875\text{ m} = 1.875\text{ km}$$

Answer: The distance of the obstacle from the submarine is $1875\text{ m}$ (or $1.875\text{ km}$).