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Sound - Reflection of Sound

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 surfaces of solids and liquids. The incident sound wave, the reflected sound wave, and the normal at the point of incidence lie in the same plane.

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Laws of Reflection: The angle of incidence of sound is always equal to the angle of reflection (i=ri = r).

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Echo: A distinct reflected sound heard after the original sound has ceased. 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} due to the persistence of hearing.

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Minimum Distance for Echo: At 22∘C22^\circ\text{C} where speed of sound is 344 m s−1344 \text{ m s}^{-1}, the total distance traveled by sound to be heard as an echo must be at least d=344 m s−1×0.1 s=34.4 md = 344 \text{ m s}^{-1} \times 0.1 \text{ s} = 34.4 \text{ m}. Thus, the obstacle must be at a distance of half of this, which is 17.2 m17.2 \text{ m}.

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Reverberation: The persistence of sound in a big hall due to repeated reflections. It is reduced by using sound-absorbent materials like compressed fiberboards, rough plaster, or draperies.

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SONAR (Sound Navigation and Ranging): A device that uses ultrasonic waves to measure the distance, direction, and speed of underwater objects. It works on the principle of echo-ranging.

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Ultrasound Applications: Used in medical imaging (Echocardiography, Ultrasonography), cleaning hard-to-reach parts, and detecting cracks in metal blocks.

📐Formulae

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

d=v×t2d = \frac{v \times t}{2}

Total distance traveled=2×Depth/Distance\text{Total distance traveled} = 2 \times \text{Depth/Distance}

💡Examples

Problem 1:

A person claps their 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:

Given: Speed of sound v=346 m s−1v = 346 \text{ m s}^{-1}, Time taken t=5 st = 5 \text{ s}. We need to find distance dd. Using the formula d=v×t2d = \frac{v \times t}{2}, we get d=346×52d = \frac{346 \times 5}{2}. First, calculate the total distance (2d2d): 346×51730\begin{array}{r} 346 \\ \times 5 \\ \hline 1730 \end{array} Now, d=17302=865 md = \frac{1730}{2} = 865 \text{ m}.

Explanation:

Since the sound travels to the cliff and back to the observer, the total distance covered is 2d2d. To find the one-way distance, we divide the total distance by 22.

Problem 2:

A SONAR device on a submarine sends out a signal and receives an echo 5 s5 \text{ s} later. Calculate the speed of sound in water if the distance of the object from the submarine is 3625 m3625 \text{ m}.

Solution:

Given: Distance d=3625 md = 3625 \text{ m}, Time t=5 st = 5 \text{ s}. Formula for speed is v=2dtv = \frac{2d}{t}. Substituting the values: v=2×36255v = \frac{2 \times 3625}{5}. First, 2×3625=72502 \times 3625 = 7250. Then, v=72505=1450 m s−1v = \frac{7250}{5} = 1450 \text{ m s}^{-1}.

Explanation:

In SONAR, the ultrasonic waves travel to the seabed and reflect back. The formula v=2dtv = \frac{2d}{t} accounts for the round-trip distance.