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 $A_1 > A_2$]

If $y_1 = 5 \text{ mm} \sin(\pi t)$ is the equation of oscillation of source $S_1$ and $y_2 = 5 \text{ mm} \sin(\pi t + \pi/6)$ is that of $S_2$,and it takes $1 \text{ s}$ and $0.5 \text{ s}$ for the transverse waves to reach point $A$ from sources $S_1$ and $S_2$ respectively,then the resulting amplitude at point $A$ is .... $\text{mm}$.

Two waves of amplitudes $A_0$ and $x A_0$ pass through a region. If $x > 1$,the difference between the maximum and minimum resultant amplitude possible is

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

The amplitude of the resultant wave produced by the superposition of two waves $y_1 = A_1 \sin(wt - \beta_1)$ and $y_2 = A_2 \sin(wt - \beta_2)$ is:

Two waves of equal amplitude and frequency interfere with each other. The ratio of intensity when the two waves arrive in phase to that when they arrive $90^{\circ}$ out of phase is: