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Waves - Sound

Grade 12IGCSEPhysics

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

🔑Concepts

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Sound is a longitudinal wave, meaning the particles of the medium vibrate parallel to the direction of wave travel, creating regions of compression and rarefaction.

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Sound requires a physical medium (solid, liquid, or gas) to propagate and cannot travel through a vacuum.

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The speed of sound vv varies depending on the medium: it is generally fastest in solids and slowest in gases because of the proximity and strength of bonds between particles.

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Pitch is directly related to the frequency ff of the wave, whereas loudness is determined by the amplitude AA of the vibration.

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The human range of hearing is approximately 20 Hz20\text{ Hz} to 20,000 Hz20,000\text{ Hz}. Frequencies above this range are called ultrasound.

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Echoes are reflections of sound waves. To calculate distance dd using an echo, the time tt taken for the pulse to return must be halved as it represents a two-way journey.

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The wave equation relates speed, frequency, and wavelength: v=fλv = f\lambda.

📐Formulae

v=fλv = f \lambda

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

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

d=v×t2 (for echo/sonar calculations)d = \frac{v \times t}{2} \text{ (for echo/sonar calculations)}

💡Examples

Problem 1:

A research ship uses sonar to measure the depth of the ocean. It sends an ultrasound pulse that reflects off the seabed and is detected 1.2 s1.2\text{ s} later. If the speed of sound in seawater is 1500 m/s1500\text{ m/s}, calculate the depth of the ocean.

Solution:

d=v×t2d = \frac{v \times t}{2} d=1500 m/s×1.2 s2d = \frac{1500\text{ m/s} \times 1.2\text{ s}}{2} d=1800 m2=900 md = \frac{1800\text{ m}}{2} = 900\text{ m}

Explanation:

Since the sound pulse travels to the seabed and back, the total distance covered is 2d2d. We divide the total calculated distance by 22 to find the one-way depth.

Problem 2:

A tuning fork produces a sound wave with a frequency of 440 Hz440\text{ Hz}. If the speed of sound in air is 340 m/s340\text{ m/s}, determine the wavelength λ\lambda of the sound wave.

Solution:

λ=vf\lambda = \frac{v}{f} λ=340 m/s440 Hz\lambda = \frac{340\text{ m/s}}{440\text{ Hz}} λ≈0.773 m\lambda \approx 0.773\text{ m}

Explanation:

We rearrange the universal wave equation v=fλv = f\lambda to solve for the wavelength λ\lambda.

Sound - Revision Notes & Key Formulas | IGCSE Grade 12 Physics