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A simple harmonic progressive wave is represented by $y=A \sin (100 \pi t+3 x)$. The distance between two points on the wave at a phase difference of $\frac{\pi}{3}$ radian is
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The correct answer is:
$\frac{\pi}{9} \mathrm{~m}$
Equation of the given harmonic progressive wave
$y=A \sin (100 \pi t+3)$
General equation of a harmonic wave
$\mathrm{y}=\mathrm{A} \sin (\mathrm{wt}+\mathrm{kx})$
From equations (i) and (ii),
$\omega=100 \pi, \mathrm{k}=3$
But, $\mathrm{k}=\frac{2 \pi}{\lambda} \Rightarrow 3=\frac{2 \pi}{\lambda}$
$\therefore \quad \lambda=\frac{2 \pi}{3}$
We also know,
Path difference $\Delta \mathrm{x}=\frac{\lambda}{2 \pi} \times$ Phase difference $\Delta \phi$
$\therefore \quad \Delta x=\frac{2 \pi}{3 \times 2 \pi} \times \frac{\pi}{3} \quad \ldots .\left(\right.$ Given $\left.\Delta \phi=\frac{\pi}{3}\right)$
$=\frac{\pi}{9} \mathrm{~m}$
$y=A \sin (100 \pi t+3)$
General equation of a harmonic wave
$\mathrm{y}=\mathrm{A} \sin (\mathrm{wt}+\mathrm{kx})$
From equations (i) and (ii),
$\omega=100 \pi, \mathrm{k}=3$
But, $\mathrm{k}=\frac{2 \pi}{\lambda} \Rightarrow 3=\frac{2 \pi}{\lambda}$
$\therefore \quad \lambda=\frac{2 \pi}{3}$
We also know,
Path difference $\Delta \mathrm{x}=\frac{\lambda}{2 \pi} \times$ Phase difference $\Delta \phi$
$\therefore \quad \Delta x=\frac{2 \pi}{3 \times 2 \pi} \times \frac{\pi}{3} \quad \ldots .\left(\right.$ Given $\left.\Delta \phi=\frac{\pi}{3}\right)$
$=\frac{\pi}{9} \mathrm{~m}$
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