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(B) The given wave equation is $y=0.03 \sin (\pi[2 t-0.01 x])$.Expanding the equation,we get $y=0.03 \sin (2\pi t - 0.01\pi x)$.Comparing this with th

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$\frac{\pi}{4}$

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(B) The given wave equation is $y=0.03 \sin (\pi[2 t-0.01 x])$.Expanding the equation,we get $y=0.03 \sin (2\pi t - 0.01\pi x)$.Comparing this with the standard wave equation $y=A \sin (\omega t - kx)$,we identify the wave number $k = 0.01\pi \ rad/m$.The path difference between the two particles is given as $\Delta x = 25 \ m$.The relationship between phase difference $\Delta \phi$ and path difference $\Delta x$ is $\Delta \phi = k \cdot \Delta x$.Substituting the values,$\Delta \phi = (0.01\pi) 	imes 25 = 0.25\pi$.Converting $0.25\pi$ to a fraction,we get $\Delta \phi = \frac{1}{4}\pi = \frac{\pi}{4} \ rad$.

What is the phase difference between two particles $25 \ m$ apart in a wave represented by the equation $y=0.03 \sin (\pi[2 t-0.01 x])$ traveling in a medium?

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