Equation of a progressive wave is given by
$y = 0.2 \cos \pi \left(0.04 t + .02 x - \frac{\pi}{6}\right)$
The distance is expressed in cm and time in second. What will be the minimum distance between two particles having the phase difference of π π /2
Text Solution
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Comparing with $y = a \cos(\omega t + kx - \phi)$ ,
We get $k = \frac{2\pi}{\lambda} = 0.02 \Rightarrow \lambda = 100 \text{cm}$
Also, it is given that phase difference between particles $\Delta \varphi = \frac{\pi}{2} .$
Hence path difference between them $\Delta = \frac{\lambda}{2 \pi} \times \Delta \varphi = \frac{\lambda}{2 \pi} \times \frac{\pi}{2} = \frac{\lambda}{4} = \frac{100}{4} = 25 \mathrm{cm}$
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