Two waves each of intensity $I$ interfere with a phase difference of $120^o$. The resultant intensity of the waves will be:

Two waves are propagating to point $P$ along a straight line,produced by two sources $A$ and $B$ of simple harmonic motion and equal frequency. The amplitude of every wave at $P$ is $a$. The phase of $A$ is ahead by $\frac{\pi}{3}$ compared to $B$,and the distance $AP$ is greater than $BP$ by $50 \ cm$. If the wavelength is $1 \ m$,what will be the resultant amplitude at point $P$?

Two loudspeakers $L_1$ and $L_2$,driven by a common oscillator and amplifier,are arranged as shown. The frequency of the oscillator is gradually increased from zero,and the detector at $D$ records a series of maxima and minima. If the speed of sound is $330 \, m/s$,then the frequency at which the first maximum is observed is .... $Hz$.

There is a destructive interference between two waves of wavelength $\lambda$ coming from two different paths at a point. To get maximum sound or constructive interference at that point,the path of one wave is to be increased by

Four independent waves are represented by the following equations. Which of these equations represent the phenomenon of interference?
$Y_1 = a_1 \sin \omega t$
$Y_2 = a_2 \sin \omega t$
$Y_3 = a_3 \cos \omega t$
$Y_4 = a_4 \sin (\omega t + \frac{\pi}{3})$

Two sound waves having intensities $I$ and $4I$ interfere to produce an interference pattern. The phase difference between the waves is $\pi / 2$ at point $A$ and $\pi$ at point $B$. Then the difference between the resultant intensities at $A$ and $B$ is (in $I$)