krit.club logo

Sound - Characteristics of Sound Waves

Grade 9CBSE

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

🔑Concepts

Sound is a longitudinal wave that propagates through a medium as a series of compressions (high-pressure regions) and rarefactions (low-pressure regions).

Wavelength (λλ): The distance between two consecutive compressions or two consecutive rarefactions. Its SI unit is meter (mm).

Frequency (νν): The number of complete oscillations per unit time. It represents how frequently the sound wave repeats. Its SI unit is Hertz (HzHz).

Time Period (TT): The time taken by two consecutive compressions or rarefactions to cross a fixed point. It is the reciprocal of frequency: T=1νT = \frac{1}{\nu}.

Amplitude (AA): The magnitude of the maximum disturbance in the medium on either side of the mean value. It determines the loudness of the sound.

Pitch: A characteristic of sound determined by the frequency. A higher frequency corresponds to a higher pitch (shrill sound), while a lower frequency corresponds to a lower pitch (bass sound).

Loudness: A sensation that depends on the intensity of sound and the sensitivity of the ear. It is proportional to the square of the amplitude: LoudnessA2Loudness \propto A^2.

Speed of Sound (vv): The distance which a point on a wave, such as a compression or a rarefaction, travels per unit time.

📐Formulae

ν=1T\nu = \frac{1}{T}

v=λνv = \lambda \nu

v=λTv = \frac{\lambda}{T}

Distance=v×tDistance = v \times t

💡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}.

  1. Calculate Speed (vv): v=λν=0.35 m×2000 Hz=700 m s1v = \lambda \nu = 0.35 \text{ m} \times 2000 \text{ Hz} = 700 \text{ m s}^{-1}.
  2. Calculate Time (tt): t=dv=1500 m700 m s12.14 st = \frac{d}{v} = \frac{1500 \text{ m}}{700 \text{ m s}^{-1}} \approx 2.14 \text{ s}.

Explanation:

First, the speed of the wave is determined using the relationship between wavelength and frequency. Then, the time taken is calculated by dividing the total distance by the speed of the wave.

Problem 2:

Calculate the wavelength of a sound wave whose frequency is 220 Hz220 \text{ Hz} and speed is 440 m/s440 \text{ m/s} in a given medium.

Solution:

Given: ν=220 Hz\nu = 220 \text{ Hz}, v=440 m/sv = 440 \text{ m/s}. Using the formula: v=λνv = \lambda \nu λ=vν=440 m/s220 Hz=2 m\lambda = \frac{v}{\nu} = \frac{440 \text{ m/s}}{220 \text{ Hz}} = 2 \text{ m}.

Explanation:

Wavelength is the ratio of the speed of the sound wave to its frequency.

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