Question

Three component sinusoidal waves progressing in the same direction along the same path have the same period, but their amplitudes are $A$, $\frac{A}{2}$ and $\frac{A}{3}$ respectively. The phase of the variation at any position $x$ on their path at time $t=0$ are $0$, $-\frac{\pi }{2}$ and $-\pi$ respectively. Find the amplitude and phase of the resultant wave.

• A. $\frac{5}{6}A$, $-ta{n}^{-1}\left(\frac{3}{4}\right)$
• B. $\frac{7}{6}A$, $-ta{n}^{-1}\left(\frac{3}{4}\right)$
• C. $\frac{5}{6}A$, $-ta{n}^{-1}\left(\frac{1}{4}\right)$
• D. $\frac{7}{6}A$, $-ta{n}^{-1}\left(\frac{1}{4}\right)$

Answer 1

The correct option is A $\frac{5}{6}A$, $-ta{n}^{-1}\left(\frac{3}{4}\right)$

The waves with opposite phases superimpose to give a resultant wave of amplitude, $A-\frac{A}{3}=\frac{2A}{3}$ with phase angle $0$ at time $t=0$.

This superimposes with wave of amplitude $\frac{A}{2}$ in with phase $-\frac{\pi }{2}$ at $t=0$.

Hence, the resulting wave has amplitude $\sqrt{(\frac{2A}{3}{)}^2+(\frac{A}{2}{)}^2}=\frac{5}{6}A$

The phase of the resulting wave is $ta{n}^{-1}\frac{-\frac{A}{2}}{\frac{2A}{3}}$$=-ta{n}^{-1}\frac{3}{4}$