Two harmonic travelling waves are described by the equations $y_1 = a \sin (kx - \omega t)$ and $y_2 = a \sin (-kx + \omega t + \phi)$. The amplitude of the superposed wave is:

$A$ narrow tube is bent in the form of a circle of radius $R,$ as shown in the figure. Two small holes $S$ and $D$ are made in the tube at positions right-angled to each other. $A$ source placed at $S$ generates a wave of intensity $I_0$ which is equally divided into two parts: one part travels along the longer path,while the other travels along the shorter path. Both the waves meet at point $D$ where a detector is placed. The maximum value of $\lambda$ to produce a minima at $D$ is given by

The superposition takes place between two waves of frequency $f$ and amplitude $a$. The total intensity is directly proportional to

If two waves represented by $y_1 = 4\sin \omega t$ and $y_2 = 3\sin (\omega t + \pi/3)$ interfere at a point,the amplitude of the resulting wave will be about

$A$ sound wave of wavelength $32 \ cm$ enters the tube at $S$ as shown in the figure. Then the smallest radius $r$ so that a minimum of sound is heard at detector $D$ is ... $cm$.

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?