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Waves - Displacement Relation for a Progressive Wave

Grade 11CBSEPhysics

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

🔑Concepts

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A progressive wave is a wave that travels continuously in a medium in the same direction without a change in its amplitude.

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The displacement y(x,t)y(x, t) of a particle in a harmonic progressive wave is a function of both position xx and time tt.

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The general equation is given by 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, ω\omega is the angular frequency, and ϕ\phi is the phase constant.

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The term (kx−ωt+ϕ)(kx - \omega t + \phi) is called the phase of the wave. It describes the state of motion of the particle at position xx and time tt.

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Amplitude (aa) is the maximum displacement of the particles of the medium from their mean position.

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Wavelength (λ\lambda) is the distance between two consecutive points in the same phase. It is related to the wave number by k=2πλk = \frac{2\pi}{\lambda}.

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Period (TT) is the time taken by an element of the medium to complete one full oscillation. It is related to angular frequency by ω=2πT\omega = \frac{2\pi}{T}.

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If the sign between kxkx and ωt\omega t is negative (kx−ωtkx - \omega t), the wave travels in the positive xx-direction. If the sign is positive (kx+ωtkx + \omega t), it travels in the negative xx-direction.

📐Formulae

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

k=2πλk = \frac{2\pi}{\lambda}

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

v=ωk=λT=νλv = \frac{\omega}{k} = \frac{\lambda}{T} = \nu \lambda

Δϕ=2πλΔx\Delta \phi = \frac{2\pi}{\lambda} \Delta x

Δϕ=2πTΔt\Delta \phi = \frac{2\pi}{T} \Delta t

💡Examples

Problem 1:

A wave is represented by the equation y(x,t)=0.03sin⁡(450t−9x)y(x, t) = 0.03 \sin(450t - 9x), where xx and yy are in meters and tt is in seconds. Determine the amplitude, frequency, and speed of the wave.

Solution:

  1. Compare the given equation with the standard form y(x,t)=asin⁡(ωt−kx)y(x, t) = a \sin(\omega t - kx).
  2. Amplitude a=0.03 ma = 0.03 \text{ m}.
  3. Angular frequency ω=450 rad/s\omega = 450 \text{ rad/s}. Frequency ν=ω2π=4502π≈71.62 Hz\nu = \frac{\omega}{2\pi} = \frac{450}{2\pi} \approx 71.62 \text{ Hz}.
  4. Propagation constant k=9 m−1k = 9 \text{ m}^{-1}.
  5. Wave speed v=ωk=4509=50 m/sv = \frac{\omega}{k} = \frac{450}{9} = 50 \text{ m/s}.

Explanation:

By identifying the coefficients of tt and xx in the wave equation, we can extract ω\omega and kk respectively, which then allow for the calculation of frequency and velocity.

Problem 2:

Calculate the phase difference between two points separated by a distance of 0.5 m0.5 \text{ m} in a wave of wavelength λ=2 m\lambda = 2 \text{ m}.

Solution:

The phase difference Δϕ\Delta \phi is related to path difference Δx\Delta x by the formula: Δϕ=2πλΔx\Delta \phi = \frac{2\pi}{\lambda} \Delta x Substituting the values: Δϕ=2π2×0.5=0.5π radians or 90∘\Delta \phi = \frac{2\pi}{2} \times 0.5 = 0.5\pi \text{ radians or } 90^\circ

Explanation:

The phase difference depends on the ratio of the distance between points to the total wavelength, multiplied by the total phase of a circle (2π2\pi).