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 $I$)

When two sound waves having amplitude $3$ and $5$ units are superimposed,then the ratio of maximum to minimum intensity of the wave produced is:

When two progressive waves $y_1=4 \sin (2 x-6 t)$ and $y_2=3 \sin \left(2 x-6 t-\frac{\pi}{2}\right)$ are superimposed,the amplitude of the resultant wave is

Two coherent sources of sound,$S_{1}$ and $S_{2}$,produce sound waves of the same wavelength,$\lambda = 1\, m$,in phase. $S_{1}$ and $S_{2}$ are placed $1.5\, m$ apart (see figure). $A$ listener,located at $L$,directly in front of $S_{2}$,finds that the intensity is at a minimum when he is $2\, m$ away from $S_{2}$. The listener moves away from $S_{1}$,keeping his distance from $S_{2}$ fixed. The adjacent maximum of intensity is observed when the listener is at a distance $d$ from $S_{1}$. Then,$d$ is $......\, m$.

Obtain the resultant wave of more than two wave functions by representing the superposition principle mathematically.

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: