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

Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to $A$ and $T$,respectively. At time $t=0$,one particle has displacement $A$ while the other one has displacement $\frac{-A}{2}$ and they are moving towards each other. If they cross each other at time $t$,then $t$ is

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

Two simple harmonic motions are represented by $y_1 = 5[\sin 2 \pi t + \sqrt{3} \cos 2 \pi t]$ and $y_2 = 5 \sin [2 \pi t + \frac{\pi}{4}]$. The ratio of their amplitudes is