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Sound - Reflection of Sound (Echo and Reverberation)

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Reflection of Sound: Like light, sound reflects off the surface of a solid or a liquid and follows the same laws of reflection where the angle of incidence ∠i\angle i is equal to the angle of reflection ∠r\angle r.

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Echo: A distinct reflected sound heard separately from the original sound. To hear a distinct echo, the time interval between the original sound and the reflected one must be at least 0.1 s0.1\text{ s}.

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Persistence of Hearing: The sensation of sound persists in our brain for about 0.1 s0.1\text{ s}. If a reflected sound arrives within this timeframe, it blends with the original sound.

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Minimum Distance for Echo: At 22∘C22^\circ\text{C} in air (speed of sound v≈344 m s−1v \approx 344\text{ m s}^{-1}), the minimum distance of the reflecting surface must be 17.2 m17.2\text{ m} (d=344×0.12d = \frac{344 \times 0.1}{2}) to hear an echo.

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Reverberation: The persistence of sound in an enclosed space due to multiple reflections. Excessive reverberation is undesirable and is reduced by using sound-absorbent materials like compressed fibreboard, heavy curtains, or carpets.

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Applications of Reflection: Megaphones, horns, stethoscopes, and the curved ceilings of concert halls (soundboards) are all designed to utilize multiple reflections of sound.

📐Formulae

v=2dtv = \frac{2d}{t}

d=v×t2d = \frac{v \times t}{2}

Condition for Echo: t≥0.1 s\text{Condition for Echo: } t \ge 0.1 \text{ s}

Minimum Distance (dmin)=v×0.12\text{Minimum Distance } (d_{min}) = \frac{v \times 0.1}{2}

💡Examples

Problem 1:

A person claps his hands near a cliff and hears the echo after 5 s5\text{ s}. What is the distance of the cliff from the person if the speed of the sound, vv, is taken as 346 m s−1346\text{ m s}^{-1}?

Solution:

v=346 m s−1v = 346\text{ m s}^{-1} t=5 st = 5\text{ s} d=v×t2d = \frac{v \times t}{2} d=346×52=1730/2=865 md = \frac{346 \times 5}{2} = 1730 / 2 = 865\text{ m}

Explanation:

Since the sound has to travel to the cliff and back to the observer, the total distance covered is 2d2d. Therefore, the distance of the cliff is half the product of speed and total time.

Problem 2:

Calculate the minimum distance required from a reflecting surface to hear an echo when the temperature of air is such that the speed of sound is 340 m s−1340\text{ m s}^{-1}.

Solution:

v=340 m s−1v = 340\text{ m s}^{-1} Persistence of hearing (t)=0.1 s\text{Persistence of hearing } (t) = 0.1\text{ s} d=v×t2=340×0.12=342=17 md = \frac{v \times t}{2} = \frac{340 \times 0.1}{2} = \frac{34}{2} = 17\text{ m}

Explanation:

To hear an echo, the reflected sound must reach the ear after 0.1 s0.1\text{ s}. Using the speed of sound in the given conditions, the distance is calculated as 17 m17\text{ m}.