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Physics: Forces and Energy - Wave Properties (Light and Sound)

Grade 8IB

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

🔑Concepts

Waves transfer energy from one place to another without the transfer of matter.

Transverse waves (e.g., light) oscillate at right angles (9090^\circ) to the direction of energy transfer, consisting of crests and troughs.

Longitudinal waves (e.g., sound) oscillate parallel to the direction of energy transfer, consisting of compressions and rarefactions.

The Amplitude (AA) is the maximum displacement from the equilibrium position; it determines the loudness of sound and brightness of light.

Wavelength (λ\lambda) is the distance between two successive identical points on a wave (e.g., crest to crest).

Frequency (ff) is the number of waves passing a point per second, measured in Hertz (HzHz). For sound, frequency determines pitch.

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

Light is an electromagnetic wave that travels at approximately 3.0×108 m/s3.0 \times 10^8 \text{ m/s} in a vacuum and can travel through empty space.

Law of Reflection: The angle of incidence (θi\theta_i) is equal to the angle of reflection (θr\theta_r).

Refraction occurs when a wave changes speed as it enters a medium of different optical density, causing it to bend.

📐Formulae

v=fλv = f \lambda

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

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

Speed of Light (c)3.0×108 m/s\text{Speed of Light (c)} \approx 3.0 \times 10^8 \text{ m/s}

💡Examples

Problem 1:

Calculate the speed of a sound wave that has a frequency of 256 Hz256 \text{ Hz} and a wavelength of 1.32 m1.32 \text{ m}.

Solution:

v=256 Hz×1.32 m=337.92 m/sv = 256 \text{ Hz} \times 1.32 \text{ m} = 337.92 \text{ m/s}

Explanation:

By applying the wave equation v=fλv = f \lambda, we multiply the frequency by the wavelength to determine the wave speed.

Problem 2:

A wave takes 0.005 s0.005 \text{ s} to complete one full cycle. What is its frequency?

Solution:

f=10.005 s=200 Hzf = \frac{1}{0.005 \text{ s}} = 200 \text{ Hz}

Explanation:

Frequency is the reciprocal of the period (TT). Using f=1Tf = \frac{1}{T} gives the number of cycles per second.

Problem 3:

If a lightning strike is seen and the thunder is heard 3 seconds3 \text{ seconds} later, how far away is the storm? (Assume speed of sound v=340 m/sv = 340 \text{ m/s})

Solution:

d=v×t=340 m/s×3 s=1020 md = v \times t = 340 \text{ m/s} \times 3 \text{ s} = 1020 \text{ m}

Explanation:

Since light travels much faster than sound, we assume the light reaches the observer almost instantly. The distance is found by multiplying the speed of sound by the time delay.

Wave Properties (Light and Sound) - Revision Notes & Key Formulas | IB Grade 8 Science