Two waves of the same frequency and same amplitude $a$ are superimposed. If the resultant amplitude is also $a$,what is the phase difference between the two waves?

Two sources of sound $A$ and $B$ produce waves of $350 Hz$,and they vibrate in the same phase. $A$ particle $P$ is vibrating under the influence of these two waves. If the amplitudes at point $P$ produced by the two waves are $0.3 mm$ and $0.4 mm$,what will be the resultant amplitude of the point $P$ when $AP - BP = 25 cm$ and the velocity of sound is $350 m/s$?

If two waves having amplitudes $2A$ and $A$ and same frequency and velocity,propagate in the same direction in the same phase,the resulting amplitude will be

When two light waves each of amplitude $A$ and having a phase difference of $\frac{\pi}{2}$ are superimposed,the amplitude of the resultant wave is:

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 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: