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Sound - Reflection of Sound and Echo

Grade 7ICSE

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

🔑Concepts

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Reflection of Sound: The phenomenon in which a sound wave, on striking a hard surface, bounces back into the same medium is called the reflection of sound.

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Laws of Reflection: 1. The angle of incidence ∠i\angle i is always equal to the angle of reflection ∠r\angle r. 2. The incident sound wave, the reflected sound wave, and the normal at the point of incidence all lie in the same plane.

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Echo: An echo is the sound heard after reflection from a distant, hard, and rigid obstacle (such as a cliff, a hillside, or a wall), after the original sound has ceased.

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Persistence of Hearing: The sensation of sound persists in our brain for about 0.1 s0.1 \text{ s}. To hear a distinct echo, the reflected sound must reach the ear after at least 0.1 s0.1 \text{ s} of the original sound.

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Conditions for an Echo: To hear a clear echo in air (where speed of sound v≈340 m/sv \approx 340 \text{ m/s}), the minimum distance between the source of sound and the obstacle must be d=v×t2=340×0.12=17 md = \frac{v \times t}{2} = \frac{340 \times 0.1}{2} = 17 \text{ m}.

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Reverberation: If the reflecting surface is at a distance less than 17 m17 \text{ m}, the reflected sound merges with the original sound, causing a prolonged sound sensation called reverberation.

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SONAR: Stands for Sound Navigation and Ranging. It uses the principle of reflection of ultrasonic sound waves (>20,000 Hz> 20,000 \text{ Hz}) to measure the depth of the sea or locate underwater objects.

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Medical Use: Ultrasonography uses the reflection of sound to produce images of internal organs.

📐Formulae

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

d=v×t2d = \frac{v \times t}{2}

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

💡Examples

Problem 1:

A person stands at a distance of 680 m680 \text{ m} from a cliff and fires a gun. After what time interval will he hear the echo? (Take speed of sound v=340 m/sv = 340 \text{ m/s})

Solution:

Given: Distance d=680 md = 680 \text{ m}, Speed v=340 m/sv = 340 \text{ m/s}. Formula for time: t=2dvt = \frac{2d}{v}. Calculation: t=2×680340=1360340=4 st = \frac{2 \times 680}{340} = \frac{1360}{340} = 4 \text{ s}.

Explanation:

The sound travels to the cliff and back to the person, covering a total distance of 2d2d. Thus, the time taken is twice the distance divided by the speed.

Problem 2:

A RADAR signal is reflected from an airplane and received back in 0.002 s0.002 \text{ s}. If the speed of the signal is 3×108 m/s3 \times 10^8 \text{ m/s}, calculate the distance of the airplane.

Solution:

Given: Time t=0.002 st = 0.002 \text{ s}, Speed v=3×108 m/sv = 3 \times 10^8 \text{ m/s}. Formula: d=v×t2d = \frac{v \times t}{2}. Calculation: d=3×108×0.0022=3×108×2×10−32=3×105 m=300 kmd = \frac{3 \times 10^8 \times 0.002}{2} = \frac{3 \times 10^8 \times 2 \times 10^{-3}}{2} = 3 \times 10^5 \text{ m} = 300 \text{ km}.

Explanation:

Using the echo principle, the distance is half of the total path traveled by the signal (v×tv \times t).

Problem 3:

Calculate the minimum distance required to hear an echo in water, where the speed of sound is 1500 m/s1500 \text{ m/s}.

Solution:

Given: Speed v=1500 m/sv = 1500 \text{ m/s}, Persistence of hearing t=0.1 st = 0.1 \text{ s}. Formula: d=v×t2d = \frac{v \times t}{2}. Calculation: d=1500×0.12=1502=75 md = \frac{1500 \times 0.1}{2} = \frac{150}{2} = 75 \text{ m}.

Explanation:

In water, sound travels much faster than in air, so the obstacle must be further away (75 m75 \text{ m} instead of 17 m17 \text{ m}) to distinguish the echo from the original sound.