Two mutually perpendicular simple harmonic vibrations have the same amplitude,frequency,and phase. When they superimpose,the resultant form of vibration will be:

The equation of $SHM$ is given as:
$x = 3 \sin(20\pi t) + 4 \cos(20\pi t)$,
where $x$ is in $cm$ and $t$ is in $seconds$. The amplitude is ..... $cm$.

The $S.H.M.$ of a particle is given by the equation $y = 3\sin \omega t + 4\cos \omega t$. The amplitude 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

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.

Two particles are executing $S.H.M.$ with the same amplitude of $20 \, cm$ and the same period along the same line about the same equilibrium position. The maximum distance between the two is $20 \, cm$. Their phase difference in radians is equal to