Solution:
The wave number is \(k = 0.02\pi\text{ cm}^{-1}\). Since phase difference \(\Delta \phi = k \Delta x\), we have \(\Delta x = \frac{\Delta \phi}{k} = \frac{\pi/2}{0.02\pi} = 25\text{ cm}\).
Equation of a progressive wave is given by \( y = 0.2 \cos \pi(0.04t + 0.02x - \pi/6) \). The distance is expressed in cm and time in second. What will be the minimum distance between two particles having the phase difference of \( \pi/2 \)?
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