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Waves - General properties of waves

Grade 11IGCSEPhysics

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

🔑Concepts

Waves are oscillations that transfer energy and information from one place to another without the transfer of matter.

Transverse Waves: The vibrations are perpendicular (9090^{\circ}) to the direction of energy transfer. Examples include electromagnetic waves (light, radio) and water ripples.

Longitudinal Waves: The vibrations are parallel to the direction of energy transfer. These consist of compressions (high pressure) and rarefactions (low pressure). Example: Sound waves.

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

Wavelength (λ\lambda): The distance from a point on one wave to the equivalent point on the adjacent wave (e.g., crest to crest or trough to trough).

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

Period (TT): The time taken for one complete wave to pass a fixed point, related to frequency by T=1fT = \frac{1}{f}.

Wavefront: An imaginary line that joins all adjacent points that are in phase (e.g., all the crests in a ripple tank).

Reflection: Occurs when a wave hits a boundary and bounces back. The law of reflection states the angle of incidence is equal to the angle of reflection (i=ri = r).

Refraction: The change in direction of a wave when it transmits from one medium to another due to a change in speed. If a wave slows down, it bends towards the normal.

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

📐Formulae

v=fλv = f \lambda

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

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

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

💡Examples

Problem 1:

A wave on a string has a frequency of 20 Hz20\text{ Hz} and a wavelength of 0.15 m0.15\text{ m}. Calculate the speed of the wave.

Solution:

v=fλv = f \lambda v=20×0.15v = 20 \times 0.15 v=3.0 m/sv = 3.0\text{ m/s}

Explanation:

The speed of a wave is the product of its frequency and its wavelength. Here, we substitute the given values into the wave equation v=fλv = f \lambda.

Problem 2:

Determine the period of a wave if its frequency is 500 Hz500\text{ Hz}.

Solution:

T=1fT = \frac{1}{f} T=1500T = \frac{1}{500} T=0.002 s=2 msT = 0.002\text{ s} = 2\text{ ms}

Explanation:

The period TT is the reciprocal of the frequency ff. Dividing 11 by 500500 gives the time for one oscillation in seconds.

Problem 3:

Water waves with a wavelength of 2.0 cm2.0\text{ cm} approach a gap of width 2.0 cm2.0\text{ cm}. Describe the behavior of the waves as they pass through the gap.

Solution:

The waves will undergo significant diffraction.

Explanation:

Diffraction is the spreading of waves through a gap. It is most prominent when the wavelength (λ=2.0 cm\lambda = 2.0\text{ cm}) is similar in magnitude to the size of the gap (2.0 cm2.0\text{ cm}).

General properties of waves - Revision Notes & Key Formulas | IGCSE Grade 11 Physics