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 are represented by $y_1 = a \sin(\omega t + \frac{\pi}{6})$ and $y_2 = a \cos(\omega t)$. What will be their resultant amplitude?

Two identical coherent sound sources $R$ and $S$ with frequency $f$ are $5 \,m$ apart. An observer standing equidistant from the sources at a perpendicular distance of $12 \,m$ from the line $RS$ hears maximum sound intensity. When the observer moves parallel to the line $RS$ to a position directly in front of one of the sources,the sound intensity is a minimum. $A$ possible value of $f$ is close to ............ $Hz$ (the speed of sound is $330 \,m/s$).

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 $mm$)

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:

When two sound waves with a phase difference of $\pi / 2$,and each having amplitude $A$ and frequency $\omega$,are superimposed on each other,then the maximum amplitude and frequency of the resultant wave is: