Equation of a progressive wave is given by

$y=0.2\mathrm{cos}\pi \left(0.04t+.02x-\frac{\pi }{6}\right)$

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/2

(1) 4 cm

(2) 8 cm

(3) 25 cm

(4) 12.5 cm

(3) Comparing with $y=a\mathrm{cos}\left(\omega t+kx-\varphi \right)$,

We get $k=\frac{2\pi }{\lambda }=0.02\mathrm{\pi }\text{\hspace{0.17em}}⇒\text{\hspace{0.17em}}\lambda =100\text{\hspace{0.17em}}cm$

Also, it is given that phase difference between particles $\Delta \varphi =\frac{\pi }{2}.$

Hence path difference between them $\Delta =\frac{\lambda }{2\pi }×\Delta \varphi =\frac{\lambda }{2\pi }×\frac{\pi }{2}=\frac{\lambda }{4}=\frac{100}{4}=25\text{\hspace{0.17em}}cm$

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