Three waves, $y_1 = A \sin (kx - \omega t)$ $y_2 = A \sin (kx - \omega t + \phi)$ and $y_3 = A \sin (kx -…
Three waves, $y_1 = A \sin (kx - \omega t)$
$y_2 = A \sin (kx - \omega t + \phi)$
and $y_3 = A \sin (kx - \omega t + 2\phi)$
are superimposed, so that $y_1 + y_2 + y_3 = 0$ at all positions. Then, the value of $\phi$ is
$\frac{2\pi}{3}$
$\frac{\pi}{3}$
$\frac{4\pi}{3}$
$\frac{\pi}{2}$
Solution
Phase, $\phi = \frac{2\pi}{3} = 120^\circ$
$\Rightarrow 2\phi = 240^\circ \quad (\because y_1 + y_2 + y_3 = 0)$
The diagram shows three amplitude vectors $A_1$, $A_2$, and $A_3$ arranged symmetrically with phase angles $\phi = 120^\circ$ and $2\phi = 240^\circ$ relative to $A_1$.