$x=10 \cos \left[2 \pi t+\frac{\pi}{2}\right]$
At $t=\frac{1}{6} s$
$x=10 \cos \left[\frac{\pi}{2}+\frac{\pi}{3}\right]$
$x=-10 \sin \frac{\pi}{3}$
$x=-5 \sqrt{3}$
$v=\omega \sqrt{A^2-x^2}$
$v=2 \pi \sqrt{100-75}$
$v=10 \pi$
$v=31.4 \,cm / s$



$x = 3\,sin\, 20\pi t + 4\, cos\, 20\pi t$ ,
where $x$ is in $cms$ and $t$ is in $seconds$ . The amplitude is ..... $cm$
${y_1} = 8\,\cos\, \omega t;\,{y_2} = 4\,\cos \,\left( {\omega t + \frac{\pi }{2}} \right)$ ;
${y_3} = 2\cos \,\left( {\omega t + \pi } \right);\,{y_4} = \,\cos \,\left( {\omega t + \frac{{3\pi }}{2}} \right)$ ,
are superposed on each other. The resulting amplitude and phase are respectively;

