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

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

$A$ particle is subjected to two mutually perpendicular simple harmonic motions such that its $x$ and $y$ coordinates are given by:
$x = 2 \sin \omega t$
$y = 2 \sin \left( \omega t + \frac{\pi}{4} \right)$
The path of the particle will be:

Two particles $P$ and $Q$ describe $SHM$ of same amplitude $a$ and same frequency $f$ along the same straight line. The maximum distance between the two particles is $a \sqrt{2}$. The phase difference between the particles is

$A$ vibratory motion is represented by $x = 2A \cos \omega t + A \cos \left( \omega t + \frac{\pi}{2} \right) + A \cos ( \omega t + \pi ) + \frac{A}{2} \cos \left( \omega t + \frac{3\pi}{2} \right)$. The resultant amplitude of the motion is

Two $SHM$ are represented by equations,$y_1 = 6\cos \left( {6\pi t + \frac{\pi }{6}} \right)$ and $y_2 = 3\left( {\sqrt 3 \sin 3\pi t + \cos 3\pi t} \right)$. Which of the following statements is true?