An equation of a simple harmonic progressive wave is given by $y=A \sin (100 \pi t-3 x)$. The distance…
An equation of a simple harmonic progressive wave is given by $y=A \sin (100 \pi t-3 x)$. The distance between two particles having a phase difference of $\left(\frac{\pi}{3}\right)$ in meter is
$\frac{\pi}{6}$
$\frac{\pi}{9}$
$\frac{\pi}{18}$
$\frac{\pi}{3}$
Solution
The correct option is (B).
Concept: The phase difference given $\phi=\left(\frac{\pi}{3}\right)$, so we can obtain the distance between two particles by using the path difference:
$\mathrm{k}\left(\mathrm{x}_2-\mathrm{x}_1\right)=\phi$
Where, $\mathrm{k}$ is the propagation constant.
The general equation of the wave is given by $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx})$ on comparing with the given equation, $\mathrm{k}=3$ and $\omega=100 \pi$.
Therefore, the distance between two particles is given by,
$\left(\mathrm{x}_2-\mathrm{x}_1\right)=\frac{\phi}{\mathrm{k}}=\frac{\pi}{9} \mathrm{~m}$