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

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:

The equations of two waves are given as
$\begin{aligned}
& y_1=a \sin \left(\omega t+\phi_1\right) \\
& y_2=a \sin \left(\omega t+\phi_2\right)
\end{aligned}$
If the amplitude and time period of the resultant wave are the same as those of the individual waves,then $(\phi_1-\phi_2)$ is

Out of the given four waves $(1), (2), (3)$ and $(4)$:
$y = a \sin(kx + \omega t)$ ......$(1)$
$y = a \sin(\omega t - kx)$ ......$(2)$
$y = a \cos(\omega t - kx)$ ......$(3)$
$y = a \cos(kx + \omega t)$ ......$(4)$
emitted by four different sources $S_1, S_2, S_3$ and $S_4$ respectively,interference phenomena would be observed in space under appropriate conditions when:

Two waves of amplitudes $A_0$ and $x A_0$ pass through a region. If $x > 1$,the difference between the maximum and minimum resultant amplitude possible is

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: