A particle is performing U.C.M. along the circumference of circle of diameter $50 \mathrm{~cm}$ with…

A particle is performing U.C.M. along the circumference of circle of diameter $50 \mathrm{~cm}$ with frequency $2 \mathrm{~Hz}$. The acceleration of the particle in $\mathrm{m} / \mathrm{s}^2$ is
  1. $2 \pi^2$
  2. $4 \pi^2$
  3. $8 \pi^2$
  4. $\pi^2$

Solution

$\begin{aligned} & \mathrm{d}=50 \mathrm{~cm} \\ & \therefore \mathrm{r}=25 \times 10^{-2} \mathrm{~m}, \mathrm{f}=2 \mathrm{~Hz} \\ & \mathrm{a}=\mathrm{r} \omega^2=4 \pi^2 \mathrm{f}^2 \mathrm{r}=4 \pi^2 \times 4 \times 25 \times 10^{-2}=4 \pi^2\end{aligned}$ .

Asked in: MHT CET 2021 (24 Sep Shift 1)

Practice more Motion In Two Dimensions questions on Aicharya