Q.Two sources of sound are separated by a distance 4 m. They both emit sound with the same amplitude and frequency (330 Hz), but they are 180º out of phase. At what points between the two sources, will the sound intensity be maximum? (Take velocity of sound to be 330 m/s)

When two waves of the SAME frequency $\omega$ but generally different amplitudes $A_1$, $A_2$ and a fixed phase difference $\phi$ superpose at a point, $y_1=A_1\sin\omega t$ and $y_2=A_2\sin(\omega t+\phi)$, the principle of superposition combines them into a single resultant simple-harmonic oscillation of the SAME frequency, $y=A\sin(\omega t+\theta)$, whose amplitude works out (by expanding the compound-angle term and then squaring-and-adding to eliminate the new phase $\theta$) to $A=\sqrt{A_1^2+A_2^2+2A_1A_2\cos\phi}$.

This single formula captures BOTH extreme interference cases seen with wave pulses: when $\phi=0$ (the waves are exactly in phase), $A=A_1+A_2$, the LARGEST possible resultant amplitude -- constructive interference; when $\phi=\pi$ (exactly out of phase), $A=|A_1-A_2|$, the SMALLEST possible resultant amplitude (zero, if $A_1=A_2$) -- destructive interference. Since intensity is proportional to the square of amplitude, the corresponding maximum and minimum intensities are $I_{max}\propto(A_1+A_2)^2$ and $I_{min}\propto(A_1-A_2)^2$: intensity is greatest wherever two interfering waves arrive in phase, and least wherever …