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Oscillations and Waves - Transverse and Longitudinal Waves

Grade 11ICSEPhysics

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

🔑Concepts

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A wave is a periodic disturbance in a medium that transfers energy and momentum from one point to another without the actual transfer of matter.

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In Transverse Waves, the particles of the medium vibrate perpendicular to the direction of wave propagation. They consist of crests (points of maximum positive displacement) and troughs (points of maximum negative displacement). Examples include light waves and waves on a plucked string.

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In Longitudinal Waves, the particles of the medium vibrate parallel to the direction of wave propagation. They consist of compressions (high-pressure regions) and rarefactions (low-pressure regions). Examples include sound waves in air.

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Amplitude (AA): The maximum displacement of a particle from its mean position.

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Wavelength (λ\lambda): The distance between two consecutive points in the same phase (e.g., two successive crests or compressions).

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Frequency (ff or ν\nu): The number of oscillations completed per second, measured in HertzHertz (HzHz).

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Time Period (TT): The time taken to complete one full oscillation, related to frequency by T=1fT = \frac{1}{f}.

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Phase (ϕ\phi): It describes the state of motion of a particle at any instant. Two particles are in phase if their displacement and direction of motion are identical.

📐Formulae

v=fλv = f\lambda

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

ω=2πf=2πT\omega = 2\pi f = \frac{2\pi}{T}

k=2πλk = \frac{2\pi}{\lambda}

y(x,t)=Asin⁡(kx−ωt+ϕ)y(x,t) = A \sin(kx - \omega t + \phi)

v=Tμ (Speed of transverse wave on a string)v = \sqrt{\frac{T}{\mu}} \text{ (Speed of transverse wave on a string)}

v=γPρ (Laplace’s correction for speed of sound in gas)v = \sqrt{\frac{\gamma P}{\rho}} \text{ (Laplace's correction for speed of sound in gas)}

💡Examples

Problem 1:

A radio station broadcasts at a frequency of 98.3 MHz98.3 \text{ MHz}. If the speed of the electromagnetic waves is 3×108 m/s3 \times 10^8 \text{ m/s}, calculate the wavelength of the waves.

Solution:

Given: f=98.3 MHz=98.3×106 Hzf = 98.3 \text{ MHz} = 98.3 \times 10^6 \text{ Hz}, v=3×108 m/sv = 3 \times 10^8 \text{ m/s}. Using the formula v=fλv = f\lambda: λ=vf\lambda = \frac{v}{f} λ=3×10898.3×106≈3.05 m\lambda = \frac{3 \times 10^8}{98.3 \times 10^6} \approx 3.05 \text{ m}

Explanation:

The wavelength is found by dividing the wave speed by the frequency. Ensure frequency is converted to the SI unit (HzHz) before calculation.

Problem 2:

A steel wire 0.72 m0.72 \text{ m} long has a mass of 5×10−3 kg5 \times 10^{-3} \text{ kg}. If the wire is under a tension of 60 N60 \text{ N}, what is the speed of transverse waves on the wire?

Solution:

Linear mass density μ=masslength=5×10−3 kg0.72 m≈6.94×10−3 kg/m\mu = \frac{\text{mass}}{\text{length}} = \frac{5 \times 10^{-3} \text{ kg}}{0.72 \text{ m}} \approx 6.94 \times 10^{-3} \text{ kg/m}. Tension T=60 NT = 60 \text{ N}. Using v=Tμ=606.94×10−3≈8645.5≈93 m/sv = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{60}{6.94 \times 10^{-3}}} \approx \sqrt{8645.5} \approx 93 \text{ m/s}

Explanation:

The speed of a transverse wave on a stretched string depends on the tension and the linear mass density (mass per unit length) of the string.

Transverse and Longitudinal Waves - Revision Notes & Key Formulas | ICSE Class 11 Physics