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Physics - Sound Waves and Vibrations

Grade 8IGCSE

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

🔑Concepts

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Sound is a longitudinal wave produced by vibrating objects. It consists of regions of high pressure called compressions and low pressure called rarefactions.

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Sound requires a medium (solid, liquid, or gas) to travel because it relies on particle vibrations. It cannot travel through a vacuum.

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The speed of sound depends on the medium: it travels fastest in solids, slower in liquids, and slowest in gases (vsolid>vliquid>vgasv_{solid} > v_{liquid} > v_{gas}). In air, the speed is approximately 330 m/s330\text{ m/s} to 340 m/s340\text{ m/s}.

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Pitch is determined by the frequency (ff) of the wave. Higher frequency results in a higher pitch. Frequency is measured in Hertz (Hz\text{Hz}), where 1 Hz=1 vibration per second1\text{ Hz} = 1\text{ vibration per second}.

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Loudness is determined by the amplitude (AA) of the wave. A larger amplitude carries more energy and sounds louder.

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The human range of hearing is typically between 20 Hz20\text{ Hz} and 20,000 Hz20,000\text{ Hz} (20 kHz20\text{ kHz}). Frequencies above 20 kHz20\text{ kHz} are called ultrasound.

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An echo is the reflection of sound waves off a surface. The time taken for an echo to return can be used to measure distances (SONAR).

📐Formulae

v=fλv = f \lambda

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

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

d=v×t2d = \frac{v \times t}{2}

💡Examples

Problem 1:

A tuning fork vibrates at a frequency of 440 Hz440\text{ Hz}. If the speed of sound in air is 340 m/s340\text{ m/s}, calculate the wavelength (λ\lambda) of the sound produced.

Solution:

λ=vf=340440≈0.77 m\lambda = \frac{v}{f} = \frac{340}{440} \approx 0.77\text{ m}

Explanation:

Using the wave equation v=fλv = f \lambda, we rearrange to solve for wavelength. Dividing the speed of sound by the frequency gives the distance between consecutive compressions.

Problem 2:

A girl stands 165 m165\text{ m} away from a high wall and claps her hands. How long will it take for her to hear the echo? (Assume speed of sound v=330 m/sv = 330\text{ m/s})

Solution:

t=2dv=2×165330=330330=1.0 st = \frac{2d}{v} = \frac{2 \times 165}{330} = \frac{330}{330} = 1.0\text{ s}

Explanation:

For an echo, the sound must travel to the wall and back, covering a total distance of 2d2d. We use the formula t=total distancespeedt = \frac{\text{total distance}}{\text{speed}}.

Problem 3:

A wave pulse on an oscilloscope shows 22 complete cycles over a horizontal distance of 4 ms4\text{ ms} (milliseconds). Calculate the frequency of the sound.

Solution:

T=4 ms2=2 ms=0.002 sT = \frac{4\text{ ms}}{2} = 2\text{ ms} = 0.002\text{ s} f=1T=10.002=500 Hzf = \frac{1}{T} = \frac{1}{0.002} = 500\text{ Hz}

Explanation:

First, find the time period (TT) for one full wave. Since 22 waves take 4 ms4\text{ ms}, one wave takes 2 ms2\text{ ms}. Convert milliseconds to seconds (1 ms=10−3 s1\text{ ms} = 10^{-3}\text{ s}) and use f=1/Tf = 1/T.