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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

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

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Transverse waves (e.g., light) oscillate at right angles (90∘90^\circ) to the direction of energy transfer, consisting of crests and troughs.

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

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

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

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

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

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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.

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

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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.