MP Board · Class 9 · Science · Sound(a) State the Laws of Reflection of Sound. (b) Define Reverberation. Why is excessive reverberation undesirable in big concert halls or auditoriums? Describe three methods used to reduce reverberation in such halls. (c) Numerical: A person standing between two tall parallel cliffs fires a pistol. He hears the first echo after $1.5\text{ seconds}$ and the second echo after $2.5\text{ seconds}$. Calculate the distance between the two cliffs. (Take speed of sound in air = $340\text{ m/s}$).

Part (a): Laws of Reflection of Sound

\nSound waves follow the same laws of reflection as light waves do:

1. First Law: The incident sound wave, the reflected sound wave, and the normal at the point of incidence on the reflecting surface all lie in the same plane.
2. Second Law: The angle of incidence ($\angle i$) is always equal to the angle of reflection ($\angle r$).

$$\angle i = \angle r$$

Part (b): Reverberation and Methods to Control It

• Definition of Reverberation: The repeated reflection of sound from walls, ceiling, and floor of a large enclosure, which results in the persistence of sound even after the source has stopped producing sound, is called reverberation.

• Why Excessive Reverberation is Undesirable: If the reverberation time is too long, successive sounds overlap with each other, making the original speech or music blurred, distorted, and difficult to understand clearly for the audience.

• Methods to Reduce Reverberation:

  1. Sound-Absorbent Wall Coverings: The walls and ceiling of halls are covered with sound-absorbing materials like compressed fiberboard, acoustic tiles, rough plaster, or padded fabrics.
  2. Heavy Draperies and Curtains: Hanging thick, heavy curtains on doors, windows, and walls absorbs sound waves instead of reflecting them.
  3. Padded Seating Material: Seats in the hall are made using sound-absorbing materials (like cushioned fabrics) so that sound is absorbed whether seats are occupied or vacant.

Part (c): Step-by-Step Numerical Solution

Given:

• Speed of sound in air ($v$) = $340\text{ m/s}$
• Time for 1st echo from nearer cliff ($t_1$) = $1.5\text{ s}$
• Time for 2nd echo from farther cliff ($t_2$) = $2.5\text{ s}$

Formula:

• Distance to 1st cliff ($d_1$): $d_1 = \frac{v \times t_1}{2}$
• Distance to 2nd cliff ($d_2$): $d_2 = \frac{v \times t_2}{2}$
• Total distance between cliffs ($D$) = $d_1 + d_2$

Calculation: $$d_1 = \frac{340 \times 1.5}{2} = 170 \times 1.5 = 255\text{ m}$$ $$d_2 = \frac{340 \times 2.5}{2} = 170 \times 2.5 = 425\text{ m}$$

$$\text{Total Distance } (D) = d_1 + d_2 = 255\text{ m} + 425\text{ m} = 680\text{ m}$$

Answer:\nThe distance between the two cliffs is $680\text{ meters}$.