The equations of motion for two waves traveling in the same direction are given by ${y_1} = A\sin (\omega t - kx)$ and ${y_2} = A\sin (\omega t - kx - \theta )$. The resultant amplitude of the medium particle will be

Two identical sinusoidal waves each of amplitude $10 \,mm$, with a phase difference of $90^{\circ}$ are travelling in the same direction in a string. The amplitude of the resultant wave is

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 .... $\text{mm}$.

The displacement of a particle executing periodic motion is given by :
$y = 4 \cos^2(t/2) \sin(1000t)$
This expression may be considered to be a result of the superposition of

$A$ 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 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?