A particle of mass m is moving along trajectory given by

$\begin{array}{l}\mathrm{x}={\mathrm{x}}_0+\mathrm{acos}{\mathrm{\omega }}_1\mathrm{t}\\ \mathrm{y}={\mathrm{y}}_0+\mathrm{bsin}{\mathrm{\omega }}_2\mathrm{t}\end{array}$

The torque, acting on the particle about the origin, at t = 0 is

$\overrightarrow{\mathrm{r}}=\left(x_0+a\cos {\omega }_{1}t\right)\hat{i}+\left({\mathrm{y}}_0+\mathrm{bsin}{\mathrm{\omega }}_2\mathrm{t}\right)\hat{j}$

$\mathrm{x}={\mathrm{x}}_0+\mathrm{acos}{\omega }_1\mathrm{t}$

$\Rightarrow {\mathrm{v}}_{\mathrm{x}}=-{\mathrm{a\omega }}_1{\mathrm{sin\omega }}_1\mathrm{t}$

$\Rightarrow a_x=-a{\omega }_1^2\cos {\omega }_1\mathrm{t}$

similarly, $a_x=-b{\omega }_2^2\sin {\omega }_2\mathrm{t}$

$\mathrm{at} \mathrm{t}=0:{\mathrm{a}}_{\mathrm{x}}=-a{\omega }_1^2; {\mathrm{a}}_{\mathrm{y}} =0$

$\Rightarrow {\overrightarrow{\mathrm{a}}}_{t=0}=-{\mathrm{a\omega }}_1^2 \hat{\mathrm{i}}$ ;   ${\overrightarrow{\mathrm{r}}}_{t=0}=\left(x_0+a\right)\hat{i}+\left({\mathrm{y}}_0\right)\hat{j}$

$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}=\mathrm{m}(\overrightarrow{\mathrm{r}}\times \overrightarrow{a})$

$\Rightarrow \overrightarrow{\tau }=\mathrm{m}\left\{\left[\left(x_0+a\right)\hat{i}+\left({\mathrm{y}}_0\right)\hat{j}\right]\times -a{\omega }_1^2\hat{i}\right\}$

$={\mathrm{my}}_{0}a{\omega }_1^2\hat{k}$