What is the phase difference between two particles \(25 \mathrm{~m}\) apart in a wave represented by…
What is the phase difference between two particles \(25 \mathrm{~m}\) apart in a wave represented by equation \(y=0.03 \sin (\pi[2 t-0.01 x]) \mathrm{s}\) travelling in a medium?
\(\frac{\pi}{8}\)
\(\frac{\pi}{4}\)
\(\frac{\pi}{2}\)
\(\pi\)
Solution
Equation of wave is given as
\(y=0.03 \sin (\pi[2 t-0.01 x])\)
Path difference, \(\Delta x=25 \mathrm{~m}\)
Comparing the wave equation by
\(\begin{aligned}
y & =A \sin (\omega t-k x) \text {, we get } k=0.01 \pi \\
\Rightarrow \frac{2 \pi}{\lambda} & =0.01 \pi \Rightarrow \lambda=\frac{2}{0.01} \Rightarrow \lambda=2 \times 10^2 \mathrm{~m}
\end{aligned}\)
\(\therefore\) Phase difference,
\(\Delta \phi=\frac{2 \pi}{\lambda} \times \Delta x=\frac{2 \pi}{2 \times 10^2} \times 25=\frac{\pi}{4}\)