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Sound - Sound Waves

Grade 9CBSE

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

🔑Concepts

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Sound is a form of energy produced by vibrations and propagates as a longitudinal wave through a material medium (solid, liquid, or gas).

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A sound wave consists of a series of high-pressure regions called compressions (CC) and low-pressure regions called rarefactions (RR).

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Wavelength (λ\lambda) is the distance between two consecutive compressions or two consecutive rarefactions. Its SI unit is the meter (mm).

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Frequency (ν\nu) is the number of complete oscillations or cycles per unit time. Its SI unit is Hertz (Hz\text{Hz}), where 1 Hz=1 s−11 \text{ Hz} = 1 \text{ s}^{-1}.

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The time taken by two consecutive compressions or rarefactions to cross a fixed point is called the Time Period (TT).

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Speed of sound (vv) 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 audible range of sound for human beings extends from about 20 Hz20 \text{ Hz} to 20,000 Hz20,000 \text{ Hz}. Sounds below 20 Hz20 \text{ Hz} are infrasonic and above 20,000 Hz20,000 \text{ Hz} are ultrasonic.

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Echo is the repetition of sound caused by the reflection of sound waves from a surface. To hear a distinct echo, the minimum distance of the obstacle from the source of sound must be half the distance sound travels in 0.1 s0.1 \text{ s} (approx. 17.2 m17.2 \text{ m} at 22∘C22^\circ\text{C}).

📐Formulae

v=νλv = \nu \lambda

T=1νT = \frac{1}{\nu}

v=λTv = \frac{\lambda}{T}

Distance (for Echo)=v×t2\text{Distance (for Echo)} = \frac{v \times t}{2}

💡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: ν=2 kHz=2000 Hz\nu = 2 \text{ kHz} = 2000 \text{ Hz}, 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, calculate the speed (vv): v=λν=0.35×2000=700 m/sv = \lambda \nu = 0.35 \times 2000 = 700 \text{ m/s} Next, calculate the time (tt): t=dv=1500700≈2.14 st = \frac{d}{v} = \frac{1500}{700} \approx 2.14 \text{ s}

Explanation:

To find the time, we first determine the speed of the wave using the relationship between wavelength and frequency. We ensure all units are in SI before calculation.

Problem 2:

A person claps near a cliff and hears an echo after 2 s2 \text{ s}. If the speed of sound is 344 m/s344 \text{ m/s}, calculate the distance of the cliff from the person.

Solution:

Given: Speed v=344 m/sv = 344 \text{ m/s}, Time t=2 st = 2 \text{ s}. The sound travels to the cliff and back, so the total distance covered is 2d2d. Total distance calculation: 344×2688\begin{array}{r} 344 \\ \times 2 \\ \hline 688 \end{array} Total distance 2d=688 m2d = 688 \text{ m}. Distance to the cliff d=6882=344 md = \frac{688}{2} = 344 \text{ m}.

Explanation:

Since the sound must travel to the reflecting surface and back to the observer, the total distance is twice the actual distance between the observer and the cliff. We divide the total distance by 22 to find the one-way distance.