Two simple harmonic motions are represented by the equations $x_{1}=5 \sin \left(2 \pi t+\frac{\pi}{4}\right)$ and $x_{2}=5 \sqrt{2}(\sin 2 \pi t+\cos 2 \pi t)$. The ratio of the amplitude of $x_{1}$ and $x_{2}$ is

Two particles $p$ and $q$ perform $SHM$ with the same amplitude $a$ and the same frequency $f$ along a straight line. The maximum distance between the two particles is $a\sqrt{2}$. The initial phase difference between the particles is:

Two particles are executing $SHM$ in a straight line. The amplitude $A$ and time period $T$ of both particles are equal. At time $t = 0$,one particle is at displacement $x_1 = +A$ and the other is at $x_2 = -A/2$,and they are approaching each other. The time after which they will cross each other is:

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

Two particles are oscillating in $SHM$ along two very close parallel paths such that they have the same mean position. The equations of $SHM$ for the two particles are $x_1 = A \sin \omega t$ and $x_2 = A \sin(\omega t + \phi)$ respectively. If the maximum distance between them is $\frac{6A}{5}$,then $\phi$ is equal to ..... $^o$.

Two simple harmonic motions $y_1 = A \sin \omega t$ and $y_2 = A \cos \omega t$ are superimposed on a particle of mass $m.$ The total mechanical energy of the particle is: