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Oscillations and Waves - Displacement Relation for a Progressive Wave

Grade 11ICSEPhysics

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

🔑Concepts

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A progressive wave (or travelling wave) is a disturbance that moves through a medium, transferring energy from one point to another without the bulk transfer of matter.

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The displacement yy of a particle at position xx and time tt for a harmonic progressive wave is given by the relation: y(x,t)=Asin⁡(kx−ωt+ϕ)y(x, t) = A \sin(kx - \omega t + \phi), where AA is the amplitude.

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The term (kx−ωt+ϕ)(kx - \omega t + \phi) is the phase of the wave. If two particles have the same phase, they are in the same state of vibration.

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The propagation constant or angular wave number kk is related to the wavelength λ\lambda by k=2πλk = \frac{2\pi}{\lambda}.

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The angular frequency ω\omega is related to the time period TT and frequency ff by ω=2πT=2πf\omega = \frac{2\pi}{T} = 2\pi f.

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The speed of the wave vv is determined by the properties of the medium and is given by v=fλv = f\lambda or v=ωkv = \frac{\omega}{k}.

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Direction of propagation: If the sign between ωt\omega t and kxkx is negative (e.g., kx−ωtkx - \omega t), the wave travels in the positive xx-direction. If the sign is positive (e.g., 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=2πf\omega = \frac{2\pi}{T} = 2\pi f

v=ωk=fλv = \frac{\omega}{k} = f \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 progressive wave is represented by the equation y=0.5sin⁡(100πt−0.02πx)y = 0.5 \sin(100\pi t - 0.02\pi x), where yy and xx are in meters and tt is in seconds. Find the (i) Amplitude, (ii) Wavelength, (iii) Frequency, and (iv) Velocity of the wave.

Solution:

Comparing the given equation with the standard form y=Asin⁡(ωt−kx)y = A \sin(\omega t - kx):

  1. Amplitude A=0.5A = 0.5 m.
  2. Wave number k=0.02πk = 0.02\pi. Since k=2πλk = \frac{2\pi}{\lambda}, then λ=2π0.02π=100\lambda = \frac{2\pi}{0.02\pi} = 100 m.
  3. Angular frequency ω=100π\omega = 100\pi. Since ω=2πf\omega = 2\pi f, then f=100π2π=50f = \frac{100\pi}{2\pi} = 50 Hz.
  4. Velocity v=fλ=50×100=5000v = f\lambda = 50 \times 100 = 5000 m/s.

Explanation:

By comparing the coefficients of tt and xx in the wave equation to the standard wave parameters ω\omega and kk, we can derive all physical properties of the wave.

Problem 2:

Calculate the phase difference between two points separated by a distance of 2525 cm in a wave of wavelength 11 m.

Solution:

Given path difference Δx=25\Delta x = 25 cm =0.25= 0.25 m and wavelength λ=1\lambda = 1 m. Using the formula Δϕ=2πλΔx\Delta \phi = \frac{2\pi}{\lambda} \Delta x: Δϕ=2π1×0.25=0.5π\Delta \phi = \frac{2\pi}{1} \times 0.25 = 0.5\pi radians.

Explanation:

The phase difference is directly proportional to the ratio of the path difference to the wavelength, scaled by 2π2\pi.