Two sound waves each of intensity $I$ are superimposed. If the phase difference between the waves is $\frac{\pi}{2}$, then the intensity of the resultant wave is

The displacement equations of two interfering waves are given by
$y_1 = 10 \sin \left(\omega t + \frac{\pi}{3}\right) \text{ cm}$
$y_2 = 5[\sin (\omega t) + \sqrt{3} \cos \omega t] \text{ cm}$ respectively.
The amplitude of the resultant wave is $............. \text{ cm}$.

The sources of sound $A$ and $B$ produce a wave of $350 Hz$ in the same phase. $A$ particle $P$ is vibrating under the influence of these two waves. If the amplitudes at $P$ produced by the two waves are $0.3 mm$ and $0.4 mm$,the resultant amplitude of the point $P$ will be,when $AP-BP=25 cm$ and the velocity of sound is $350 m s^{-1}$. (in $mm$)

Two waves of amplitudes $A_1$ and $A_2$ respectively are superimposed. The ratio between the maximum and minimum intensities of the resultant waves is $9 : 4$. The value of $A_2 / A_1$ is [Assume $A_1 > A_2$]

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:

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