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Sound - Speed of Sound

Grade 9CBSE

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

🔑Concepts

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The speed of sound is defined as the distance which a point on a wave, such as a compression or a rarefaction, travels per unit time.

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The speed of sound depends on the properties of the medium through which it travels. In general, the speed of sound decreases as we go from solid to liquid to gaseous state: vsolid>vliquid>vgasv_{solid} > v_{liquid} > v_{gas}.

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In any medium, the speed of sound increases with an increase in temperature. For example, the speed of sound in air is 331 m s−1331 \text{ m s}^{-1} at 0∘C0^\circ\text{C} and 344 m s−1344 \text{ m s}^{-1} at 22∘C22^\circ\text{C}.

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The speed of sound is much slower than the speed of light (c≈3×108 m s−1c \approx 3 \times 10^8 \text{ m s}^{-1}), which is why we see lightning before we hear the thunder.

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Sonic Boom: When an object travels at a speed greater than the speed of sound (supersonic speed), it produces shock waves in air that carry a large amount of energy, leading to a very loud sound.

📐Formulae

v=Distance (d)Time (t)v = \frac{\text{Distance } (d)}{\text{Time } (t)}

v=νλv = \nu \lambda

v=λTv = \frac{\lambda}{T}

Speed (v)=Frequency (ν)×Wavelength (λ)\text{Speed } (v) = \text{Frequency } (\nu) \times \text{Wavelength } (\lambda)

💡Examples

Problem 1:

A sound wave has a frequency of 2 kHz2 \text{ kHz} and a wavelength of 35 cm35 \text{ cm}. How long will it take to travel 1.5 km1.5 \text{ km}?

Solution:

Given: Frequency, ν=2 kHz=2000 Hz\nu = 2 \text{ kHz} = 2000 \text{ Hz}; Wavelength, λ=35 cm=0.35 m\lambda = 35 \text{ cm} = 0.35 \text{ m}; Distance, d=1.5 km=1500 md = 1.5 \text{ km} = 1500 \text{ m}. First, find the speed: v=νλ=2000 Hz×0.35 m=700 m s−1v = \nu \lambda = 2000 \text{ Hz} \times 0.35 \text{ m} = 700 \text{ m s}^{-1}. Now, calculate time: t=dv=1500 m700 m s−1=2.14 st = \frac{d}{v} = \frac{1500 \text{ m}}{700 \text{ m s}^{-1}} = 2.14 \text{ s}.

Explanation:

We first use the wave equation to find the speed of the sound wave in the given medium and then use the basic definition of speed to find the time taken to cover the specified distance.

Problem 2:

A person claps his hands near a cliff and hears the echo after 2 s2 \text{ s}. What is the distance of the cliff from the person if the speed of sound 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 for echo, t=2 st = 2 \text{ s}. In an echo, sound travels to the cliff and back, so the total distance is 2d2d. 2d=v×t=346 m s−1×2 s=692 m2d = v \times t = 346 \text{ m s}^{-1} \times 2 \text{ s} = 692 \text{ m}. Distance of the cliff, d=692 m2=346 md = \frac{692 \text{ m}}{2} = 346 \text{ m}.

Explanation:

For echo problems, the distance traveled by the sound wave is twice the distance between the source and the reflecting surface.

Speed of Sound - Revision Notes & Key Formulas | CBSE Class 9 Science