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Waves - General properties of waves

Grade 12IGCSEPhysics

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

🔑Concepts

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Waves are oscillations that transfer energy from one place to another without transferring matter.

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Transverse waves: The oscillations are at 90∘90^{\circ} (perpendicular) to the direction of energy transfer. Examples include light, electromagnetic waves, and water waves.

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Longitudinal waves: The oscillations are parallel to the direction of energy transfer, consisting of compressions and rarefactions. Example: sound waves.

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Amplitude (AA): The maximum displacement of a point on the wave from its undisturbed (equilibrium) position.

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Wavelength (λ\lambda): The distance between two consecutive identical points on a wave, such as crest to crest or compression to compression.

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Frequency (ff): The number of complete wave cycles passing a fixed point per second, measured in Hertz (HzHz).

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Period (TT): The time taken for one complete wave cycle to pass a point, expressed as T=1fT = \frac{1}{f}.

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Wavefront: A line representing the points on a wave that are in phase (e.g., all crests). The direction of wave travel is always perpendicular to the wavefront.

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Reflection: When a wave hits a boundary, it bounces back. The law of reflection states that the angle of incidence θi\theta_i equals the angle of reflection θr\theta_r.

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Refraction: The change in direction of a wave as it crosses a boundary between two different media, caused by a change in wave speed. Frequency remains constant, but λ\lambda changes.

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Diffraction: The spreading of waves as they pass through a gap or around an obstacle. Diffraction is most significant when the size of the gap is approximately equal to the wavelength λ\lambda.

📐Formulae

v=fλv = f \lambda

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

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

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

💡Examples

Problem 1:

A radio station transmits waves with a frequency of 98×106 Hz98 \times 10^6\text{ Hz}. Given that the speed of electromagnetic waves is 3.0×108 m/s3.0 \times 10^8\text{ m/s}, calculate the wavelength λ\lambda.

Solution:

λ=vf=3.0×10898×106≈3.06 m\lambda = \frac{v}{f} = \frac{3.0 \times 10^8}{98 \times 10^6} \approx 3.06\text{ m}

Explanation:

We use the wave equation v=fλv = f \lambda and rearrange it to solve for λ\lambda by dividing the speed of light by the frequency.

Problem 2:

A ripple tank produces water waves where the distance between 55 consecutive crests is 20 cm20\text{ cm}. If the waves take 2 seconds2\text{ seconds} to travel this distance, find the frequency ff.

Solution:

11 wavelength λ=20 cm4=5 cm=0.05 m\lambda = \frac{20\text{ cm}}{4} = 5\text{ cm} = 0.05\text{ m}. Speed v=0.20 m2 s=0.1 m/sv = \frac{0.20\text{ m}}{2\text{ s}} = 0.1\text{ m/s}. Frequency f=vλ=0.10.05=2 Hzf = \frac{v}{\lambda} = \frac{0.1}{0.05} = 2\text{ Hz}.

Explanation:

First, identify that 55 crests encompass 44 wavelengths. Calculate the speed using v=dtv = \frac{d}{t} and then apply f=vλf = \frac{v}{\lambda}.

Problem 3:

A sound wave has a period of 0.005 s0.005\text{ s}. Calculate its frequency and state if it is audible to a human with a range of 20 Hz20\text{ Hz} to 20,000 Hz20,000\text{ Hz}.

Solution:

f=1T=10.005=200 Hzf = \frac{1}{T} = \frac{1}{0.005} = 200\text{ Hz}. This is audible.

Explanation:

The frequency is calculated using the reciprocal of the period. Since 200 Hz200\text{ Hz} falls within the standard human hearing range, it is audible.

General properties of waves - Revision Notes & Key Formulas | IGCSE Grade 12 Physics