When two progressive waves $y_1=4 \sin (2 x-6 t)$ and $y_2=3 \sin \left(2 x-6 t-\frac{\pi}{2}\right)$ are superimposed,the amplitude of the resultant wave is

There is a destructive interference between two waves of wavelength $\lambda$ coming from two different paths at a point. To get maximum sound or constructive interference at that point,the path of one wave is to be increased by

Two waves originating from $S_1$ and $S_2$,having zero initial phase difference and common wavelength $\lambda$,will show completely destructive interference at $P$,if $(S_1 P - S_2 P)$ is

Two waves $Y_1 = A_1 \sin(\omega t - \beta_1)$ and $Y_2 = A_2 \sin(\omega t - \beta_2)$ superimpose to form a resultant wave whose amplitude is

Two sound waves (expressed in $CGS$ units) given by $y_1 = 0.3 \sin \frac{2\pi}{\lambda}(vt - x)$ and $y_2 = 0.4 \sin \frac{2\pi}{\lambda}(vt - x + \theta)$ interfere. The resultant amplitude at a place where phase difference is $\pi/2$ will be .... $cm$.

Two waves each of intensity $I$ interfere with a phase difference of $120^o$. The resultant intensity of the waves will be: