Physics - SHM Question with Solution | TestHub

x-direction: $x=5\cos 2\omega t+10{\sin }^2\omega t=5\cos 2\omega t+5(1-\cos 2\omega t)=5$ (constant).

$y=10\cos \omega t$ and $z=20{\cos }^2\omega t=10+10\cos \omega t=10+y$.

$y$ and $z$ oscillate in phase with equal amplitude 10, so the motion is SHM along the line $z=y+10$ with amplitude $\sqrt{10^2+10^2}=10\sqrt{2}$ - (A)

Since $x$ is constant, the motion lies in a plane parallel to the Y-Z plane - (D)