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Wave Behaviour - Travelling Waves

Grade 11IBPhysics

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

🔑Concepts

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A travelling wave is a continuous disturbance that propagates through a medium or vacuum, transferring energy from one point to another without the net transfer of matter.

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Transverse waves are characterized by oscillations that are perpendicular to the direction of energy transfer (e.g., light waves or waves on a string).

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Longitudinal waves are characterized by oscillations that are parallel to the direction of energy transfer, consisting of compressions and rarefactions (e.g., sound waves).

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The displacement xx of a particle is its distance and direction from its equilibrium position. The amplitude AA is the maximum displacement.

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The wavelength λ\lambda is the shortest distance between two points that are in phase (e.g., peak to peak), while the period TT is the time taken for one complete oscillation.

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The intensity II of a wave is the power per unit area (P/AP/A) and is proportional to the square of the amplitude, I∝A2I \propto A^2.

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For a point source, the intensity follows the inverse square law with distance rr, such that I∝1r2I \propto \frac{1}{r^2}.

📐Formulae

v=fλv = f \lambda

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

I∝A2I \propto A^2

I=P4πr2I = \frac{P}{4\pi r^2}

Phase Difference =2πΔxλ\text{Phase Difference } = \frac{2\pi \Delta x}{\lambda}

💡Examples

Problem 1:

A radio station transmits at a frequency of 98.0 MHz98.0 \text{ MHz}. Given that the speed of light is c=3.00×108 m s−1c = 3.00 \times 10^8 \text{ m s}^{-1}, calculate the wavelength λ\lambda of the radio waves.

Solution:

Using the wave equation v=fλv = f \lambda, we rearrange for λ\lambda: λ=vf\lambda = \frac{v}{f} λ=3.00×108 m s−198.0×106 Hz≈3.06 m\lambda = \frac{3.00 \times 10^8 \text{ m s}^{-1}}{98.0 \times 10^6 \text{ Hz}} \approx 3.06 \text{ m}

Explanation:

To find the wavelength, the wave speed (speed of light for EM waves) is divided by the frequency. Ensure the frequency is converted from MHz\text{MHz} to Hz\text{Hz} using 10610^6.

Problem 2:

A wave has an initial intensity I0I_0 and amplitude A0A_0. If the amplitude is increased to 3A03A_0, what is the new intensity in terms of I0I_0?

Solution:

The relationship between intensity and amplitude is I∝A2I \propto A^2. Therefore: InewI0=(3A0A0)2\frac{I_{new}}{I_0} = \left( \frac{3A_0}{A_0} \right)^2 InewI0=32=9\frac{I_{new}}{I_0} = 3^2 = 9 Inew=9I0I_{new} = 9I_0

Explanation:

Since intensity is proportional to the square of the amplitude, tripling the amplitude results in a nine-fold increase in intensity (32=93^2 = 9).

Travelling Waves Grade 11 Notes & Examples | IB Physics