MP Board · Class 9 · Science · Sound(a) State the laws of reflection of sound. Describe two practical applications based on the multiple reflection of sound. (b) A person stands between two parallel vertical cliffs and is at a distance of $680\text{ m}$ from the nearer cliff. Upon firing a shot, he hears the first echo after $4\text{ seconds}$ and the second echo after $6\text{ seconds}$. Calculate: The speed of sound in air. The distance between the two cliffs. The time interval after firing when he will hear the third echo.

(a) Laws of Reflection of Sound & Practical Applications

Laws of Reflection of Sound:

1. 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. The angle of incidence ($ \angle i$) is equal to the angle of reflection ($ \angle r$).

Practical Applications of Multiple Reflection of Sound:

• Stethoscope: A medical instrument used by doctors to listen to sounds produced within the body (like heartbeats). The sound of the heartbeat reaches the doctor's ears through multiple reflections along the inner tube of the stethoscope.
• Megaphone / Loudspeaker: Horn-shaped tubes designed to direct sound in a particular direction without spreading it all around. The sound waves undergo successive multiple reflections along the conical boundary and travel forward effectively.

(b) Numerical Calculation

\nLet:

• Distance to nearer cliff $d_1 = 680\text{ m}$
• Time for 1st echo $t_1 = 4\text{ s}$
• Time for 2nd echo $t_2 = 6\text{ s}$

1. Speed of Sound ($v$):\nFor the first echo from the nearer cliff, sound travels to the cliff and returns (total distance $= 2 d_1$).

$$ \text{Distance} = v \times t_1 $$ $$ 2 d_1 = v \times t_1 $$ $$ 2 \times 680 = v \times 4 $$ $$ 1360 = 4v \implies v = \frac{1360}{4} = 340\text{ m/s} $$ The speed of sound in air is $340\text{ m/s}$.

2. Distance Between the Two Cliffs ($D$):\nLet $d_2$ be the distance to the farther cliff.\nFor the second echo coming from the farther cliff:

$$ 2 d_2 = v \times t_2 $$ $$ 2 d_2 = 340 \times 6 $$ $$ 2 d_2 = 2040 \implies d_2 = \frac{2040}{2} = 1020\text{ m} $$ \nTotal distance between the two cliffs ($D$): $$\nD = d_1 + d_2 = 680\text{ m} + 1020\text{ m} = 1700\text{ m} $$ The distance between the two cliffs is $1700\text{ m}$.

3. Time for Third Echo ($t_3$):\nThe third echo is produced when the sound reflects off both cliffs consecutively (or travels across the total distance $D$ twice, i.e., $2D$).

$$ 2 D = v \times t_3 $$ $$ 2 \times 1700 = 340 \times t_3 $$ $$ 3400 = 340 \times t_3 \implies t_3 = \frac{3400}{340} = 10\text{ s} $$ (Alternatively, $t_3 = t_1 + t_2 = 4\text{ s} + 6\text{ s} = 10\text{ s}$).

The person will hear the third echo after $10\text{ seconds}$.