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Waves - Sound

Grade 11A LevelPhysics

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

🔑Concepts

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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.

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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).

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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).

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

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

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

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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}.

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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=6.0×10−4 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.