Question

Position vector of a particle as a function of time is given by,
$\overrightarrow{R}=\left(asin\omega t\right)\hat{i}+\left(acos\omega t\right)\hat{j}+\left(bsin{\omega }_{0}t\right)\hat{k}$.The particle appears to be performing simple harmonic motion along $z$ direction, to an observer moving in $xy$ plane. The distance travelled by the observer while he sees the particle completing one oscillation is $2\pi ax\left[\frac{\omega }{{\omega }_0}\right].$
The value of $x$ is

Answer 1

The particle appears to be performing simple harmonic motion along $Z$ direction, this observer must move in $xy$ plane with his position vector changing with time as,
$\overrightarrow{R_o}=\left(asin\omega t\right)\hat{i}+\left(acos\omega t\right)\hat{j}$
This is a circle with radius $a$. Angular speed of the observer is $\omega$ and his linear speed is $v=a\omega$.
Position of particle relative to the observer $\overrightarrow{R}-\overrightarrow{R_o}=\left(bsin{\omega }_{0}t\right)\hat{k}$

He finds that the particle oscillates along $z$ direction with angular frequency = ${\omega }_o$.
Time period of oscillation of the particle is,
$T_o=\frac{2\pi }{{\omega }_o}.$

In this time observer travels a distance
$\therefore s=v{T}_o=2\pi a\left[\frac{\omega }{{\omega }_o}\right]$