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:

Out of the given four waves $(1), (2), (3)$ and $(4)$:
$y = a \sin(kx + \omega t)$ ......$(1)$
$y = a \sin(\omega t - kx)$ ......$(2)$
$y = a \cos(\omega t - kx)$ ......$(3)$
$y = a \cos(kx + \omega t)$ ......$(4)$
emitted by four different sources $S_1, S_2, S_3$ and $S_4$ respectively,interference phenomena would be observed in space under appropriate conditions when:

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

Two sound waves each of wavelength $\lambda$ and same amplitude $A$ interfere at point $Q$. If the path difference is $\frac{\lambda}{4}$, the amplitude of the resultant wave at point $Q$ is $[\sin \frac{\pi}{2} = 1, \cos \frac{\pi}{2} = 0]$

The similarity between sound waves and light waves is:

The resultant amplitude due to the superposition of two waves $y_1 = 5 \sin (\omega t - kx)$ and $y_2 = -5 \cos (\omega t - kx - 150^{\circ})$ is: