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Physics: Waves, Sound, and Light - Longitudinal and Transverse Waves

Grade 8IB

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

🔑Concepts

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A wave is a periodic disturbance that transfers energy from one place to another without the permanent transfer of matter.

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Transverse Waves: The oscillation of particles is perpendicular (90∘90^{\circ}) to the direction of energy transfer. Examples include light waves, water waves, and S-waves in earthquakes. These waves consist of crests (highest points) and troughs (lowest points).

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Longitudinal Waves: The oscillation of particles is parallel to the direction of energy transfer. Examples include sound waves and P-waves in earthquakes. These waves consist of compressions (regions of high pressure/density) and rarefactions (regions of low pressure/density).

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Amplitude (AA): The maximum displacement of a particle from its equilibrium (rest) position. In sound, higher amplitude means greater loudness.

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Wavelength (λ\lambda): The distance between two consecutive identical points on a wave, such as from crest to crest or compression to compression. It is measured in meters (mm).

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Frequency (ff): The number of complete wave cycles that pass a point per second. It is measured in Hertz (HzHz). In sound, frequency determines pitch.

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Period (TT): The time taken for one complete wave cycle to pass a point, measured in seconds (ss).

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Wave Speed (vv): The speed at which energy is transferred through a medium, measured in meters per second (m/sm/s).

📐Formulae

v=fλv = f \lambda

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

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

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

💡Examples

Problem 1:

A sound wave traveling through air has a frequency of 440 Hz440\ Hz (the note A4). If the speed of sound in air is 343 m/s343\ m/s, calculate the wavelength of the sound wave.

Solution:

Given: f=440 Hzf = 440\ Hz, v=343 m/sv = 343\ m/s. We use the formula v=fλv = f \lambda. To find λ\lambda, we rearrange the formula: λ=vf\lambda = \frac{v}{f} λ=343440≈0.78 m\lambda = \frac{343}{440} \approx 0.78\ m

Explanation:

The wavelength is found by dividing the wave speed by the frequency. This represents the physical distance between two successive compressions in the air.

Problem 2:

A buoy in the ocean moves up and down as a water wave passes. If it takes 2.5 s2.5\ s for the buoy to complete one full oscillation (from crest to trough and back to crest), what is the frequency of the wave?

Solution:

Given: Period T=2.5 sT = 2.5\ s. We use the formula: f=1Tf = \frac{1}{T} f=12.5=0.4 Hzf = \frac{1}{2.5} = 0.4\ Hz

Explanation:

The frequency is the reciprocal of the period. A frequency of 0.4 Hz0.4\ Hz means that 0.40.4 of a wave cycle passes the buoy every second.

Problem 3:

A wave on a string has a wavelength of 0.5 m0.5\ m and a frequency of 10 Hz10\ Hz. Calculate the speed of the wave.

Solution:

Given: λ=0.5 m\lambda = 0.5\ m and f=10 Hzf = 10\ Hz. v=f×λv = f \times \lambda v=10×0.5=5 m/sv = 10 \times 0.5 = 5\ m/s

Explanation:

By multiplying the frequency (cycles per second) by the wavelength (distance per cycle), we calculate the total distance the wave travels per second.