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Waves — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsWaves3 Marks
Question
Define kinetic energy density of a wave. Derive an expression for its maximum value. Using it, prove that the intensity of the wave is, $\text{I}=\frac{1}{2}\rho \omega^2\text{A}^2\nu.$
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Answer

For a travalling wave, $\text{y}=\text{A}\sin\sin(\omega\text{t}-\text{Kx}),$ be the displacement. The velocity of particles is given by $\nu_\text{p}=\frac{\text{dy}}{\text{dt}}=\omega\text{A}(\omega\text{t}-\text{Kx})$
K.E. per unit valume
$=\frac{1}{2}\frac{\text{m}}{\text{V}}\nu^2=\frac{1}{2}\rho\nu^2$
$=\frac{1}{2}\rho\omega^2\text{A}^2\cos^2(\omega\text{t}-\text{Kx})$
$\therefore$ K.E. density defined as K.E. per unit volume is given by $\frac{1}{2}\rho\omega^2\text{A}^2\cos^2(\omega\text{t}-\text{Kx}).$
Maximum value of energy density $=\frac{1}{2}\rho\omega^2\text{A}^2.$
Intensity is the energy falling per unit area per unit time.
So, $\text{T}=\frac{\Delta \text{E}}{\Delta \text{ts}}=\frac{\text{P}}{\text{S}}$
$\therefore \text{I}=\frac{\frac{1}{2}\rho\omega^2\text{A}^2.\text{S}\Delta\text{x}}{\text{S}\Delta\text{t}}$
$=\frac{1}{2}\rho \omega^2\text{A}^2\nu$

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