Two identical waves $1$ and $2$,each of intensity $I_0$,are superimposed. The resulting intensity is:

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 $P$?

If two waves represented by $y_1 = 4 \sin \omega t$ and $y_2 = 3 \sin (\omega t + \frac{\pi}{3})$ interfere at a point,then the amplitude of the resulting wave will be about:

Two waves have intensities $x$ and $y$. If the time difference between them is $3T/2$,what is the resultant intensity?

Two waves are superimposed whose ratio of intensities is $9: 1$. The ratio of maximum and minimum intensity is

Three harmonic waves having equal frequency $v$ and same intensity $I_{0}$,have phase angles $0, \frac{\pi}{4}$ and $-\frac{\pi}{4}$ respectively. When they are superimposed,the intensity of the resultant wave is close to