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Waves - Sound

Grade 11IGCSEPhysics

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

🔑Concepts

Sound is a longitudinal wave produced by vibrating sources that requires a medium (solid, liquid, or gas) to travel; it cannot travel through a vacuum.

Sound waves consist of compressions (regions of high pressure where particles are close together) and rarefactions (regions of low pressure where particles are spread apart).

The speed of sound depends on the medium: it is fastest in solids, slower in liquids, and slowest in gases (typically around 330 m/s330 \text{ m/s} to 350 m/s350 \text{ m/s} in air).

The pitch of a sound is determined by its frequency (ff); a higher frequency results in a higher pitch.

The loudness of a sound is determined by its amplitude (AA); a larger amplitude results in a louder sound.

The audible frequency range for a healthy human ear is approximately 20 Hz20 \text{ Hz} to 20,000 Hz20,000 \text{ Hz}.

Ultrasound is defined as sound waves with a frequency higher than the upper limit of human hearing, i.e., greater than 20,000 Hz20,000 \text{ Hz}.

An echo is the reflection of sound waves from a surface. To calculate the distance to a surface using an echo, the sound travels a total distance of 2d2d.

📐Formulae

v=fλv = f \lambda

f=1Tf = \frac{1}{T}

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

v=2dt (for echo calculations)v = \frac{2d}{t} \text{ (for echo calculations)}

💡Examples

Problem 1:

A student stands 170 m170 \text{ m} away from a large wall and claps their hands. If the speed of sound in air is 340 m/s340 \text{ m/s}, calculate the time interval before the student hears the echo.

Solution:

t=2dv=2×170340=1.0 st = \frac{2d}{v} = \frac{2 \times 170}{340} = 1.0 \text{ s}

Explanation:

Since the sound must travel to the wall and back, the total distance is 2d2d. Rearranging the speed formula v=2dtv = \frac{2d}{t} allows us to solve for time tt.

Problem 2:

An ultrasound scanner emits a wave with a frequency of 2.5 MHz2.5 \text{ MHz}. If the speed of sound in human tissue is 1500 m/s1500 \text{ m/s}, find the wavelength of the ultrasound wave.

Solution:

λ=vf=15002.5×106=6.0×104 m=0.6 mm\lambda = \frac{v}{f} = \frac{1500}{2.5 \times 10^6} = 6.0 \times 10^{-4} \text{ m} = 0.6 \text{ mm}

Explanation:

First, convert the frequency from 2.5 MHz2.5 \text{ MHz} to 2.5×106 Hz2.5 \times 10^6 \text{ Hz}. Then, use the wave equation v=fλv = f \lambda rearranged to solve for λ\lambda.

Sound - Revision Notes & Key Formulas | IGCSE Grade 11 Physics