Two simple harmonic motions are represented by the equations $y_{1} = 10 \sin(3 \pi t + \frac{\pi}{3})$ and $y_{2} = 5(\sin 3 \pi t + \sqrt{3} \cos 3 \pi t)$. The ratio of the amplitude of $y_{1}$ to $y_{2}$ is $x : 1$. The value of $x$ 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:

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

Two simple harmonic motions are represented as $y_1 = 10 \sin \omega t$ and $y_2 = 10 \sin \omega t + 5 \cos \omega t$. The ratio of the amplitudes of $y_1$ and $y_2$ is

Two particles execute simple harmonic motion along the same straight line with the same amplitude and same frequency. The two particles pass one another when moving in opposite directions each time at a distance of $\frac{1}{\sqrt{2}}$ times the amplitude from their common mean position. The phase difference between the two particles is (in $^{\circ}$)

Two particles undergo $SHM$ along parallel lines with the same time period $(T)$ and equal amplitudes $(A)$. At a particular instant,one particle is at its extreme position while the other is at its mean position. They move in the same direction. They will cross each other after a further time: