The equations of two waves acting in perpendicular directions are given as $x=a \cos (\omega t+\delta)$ and $y=a \cos (\omega t+\alpha)$,where $\delta=\alpha+\frac{\pi}{2}$. The resultant wave represents:

Three simple harmonic motions of equal amplitudes $A$ and equal time periods in the same direction combine. The phase of the second motion is $60^{\circ}$ ahead of the first and the phase of the third motion is $60^{\circ}$ ahead of the second. Find the amplitude of the resultant motion.

$A$ particle is subjected to two simple harmonic motions in the same direction having equal amplitudes and equal frequency. If the resultant amplitude is equal to the amplitude of the individual motion, the phase difference $(\delta)$ between the two motions is

$A$ point mass is subjected to two simultaneous sinusoidal displacements in $x$-direction,$x_1(t) = A \sin \omega t$ and $x_2(t) = A \sin \left(\omega t + \frac{2 \pi}{3}\right)$. Adding a third sinusoidal displacement $x_3(t) = B \sin (\omega t + \phi)$ brings the mass to a complete rest. The values of $B$ and $\phi$ are

Two particles are executing $SHM$ of the same amplitude $A$ and frequency $\omega$ along the $x$-axis. Their mean positions are separated by $X_0$ (where $X_0 > A$). If the maximum separation between them is $X_0 + 2A$,then the phase difference between their motion is:

For a periodic motion represented by the equation $Y = \sin \omega t + \cos \omega t$,the amplitude of the motion is: