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Waves - Transverse and Longitudinal Waves

Grade 11CBSEPhysics

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

🔑Concepts

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A wave is a disturbance that travels through a medium, transporting energy and momentum without the actual physical transfer of matter.

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In a Transverse Wave, the particles of the medium oscillate perpendicular to the direction of wave propagation. These waves consist of 'crests' (highest points) and 'troughs' (lowest points). Examples include waves on a plucked string or electromagnetic waves.

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In a Longitudinal Wave, the particles of the medium oscillate parallel to the direction of wave propagation. These waves consist of 'compressions' (regions of high pressure/density) and 'rarefactions' (regions of low pressure/density). Examples include sound waves in air.

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Wavelength (λ\lambda) is the distance between two consecutive points in the same phase, such as two consecutive crests or compressions.

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Frequency (ν\nu or ff) is the number of oscillations per unit time, while the Time Period (TT) is the time taken for one complete oscillation, related by T=1/νT = 1/\nu.

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The displacement of a sinusoidal wave traveling along the positive xx-axis is given by the function y(x,t)=Asin⁡(kx−ωt+ϕ)y(x, t) = A \sin(kx - \omega t + \phi), where AA is the amplitude, kk is the angular wave number, and ω\omega is the angular frequency.

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Transverse waves can only propagate through solids and on the surface of liquids because they require shear strength. Longitudinal waves can propagate through solids, liquids, and gases.

📐Formulae

v=νλv = \nu \lambda

T=1νT = \frac{1}{\nu}

ω=2πν=2πT\omega = 2\pi\nu = \frac{2\pi}{T}

k=2πλk = \frac{2\pi}{\lambda}

v=ωkv = \frac{\omega}{k}

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

💡Examples

Problem 1:

A harmonic wave is described by the equation y(x,t)=0.03sin⁡(40x−2t)y(x, t) = 0.03 \sin(40x - 2t), where xx and yy are in meters and tt is in seconds. Determine the amplitude, wavelength, and speed of the wave.

Solution:

Comparing with the standard equation y(x,t)=Asin⁡(kx−ωt)y(x, t) = A \sin(kx - \omega t), we get:

  1. Amplitude A=0.03 mA = 0.03\text{ m}.
  2. Wave number k=40 rad/mk = 40\text{ rad/m}. Since k=2πλk = \frac{2\pi}{\lambda}, we have λ=2π40=π20≈0.157 m\lambda = \frac{2\pi}{40} = \frac{\pi}{20} \approx 0.157\text{ m}.
  3. Angular frequency ω=2 rad/s\omega = 2\text{ rad/s}. Wave speed v=ωk=240=0.05 m/sv = \frac{\omega}{k} = \frac{2}{40} = 0.05\text{ m/s}.

Explanation:

The parameters are extracted directly by comparing the given wave function with the general wave displacement formula.

Problem 2:

Calculate the frequency of a sound wave (longitudinal) traveling in air if its wavelength is 0.5 m0.5\text{ m} and the speed of sound is 340 m/s340\text{ m/s}.

Solution:

Given v=340 m/sv = 340\text{ m/s} and λ=0.5 m\lambda = 0.5\text{ m}. Using the relation v=νλv = \nu \lambda, we find: ν=vλ\nu = \frac{v}{\lambda} ν=3400.5=680 Hz\nu = \frac{340}{0.5} = 680\text{ Hz}

Explanation:

The frequency is the ratio of wave speed to wavelength, representing how many compressions pass a fixed point per second.