Simple Harmonic Motion — Physics STD 11 Science — Question

$\Rightarrow\text{r}\omega=10$

$\Rightarrow\omega^2=\frac{100}{\text{r}^2}\ ...(1)$

$\text{A}_{\text{max}}=\omega^2\text{r}=50\text{cm/sec}$

$\Rightarrow\omega^2=\frac{50}{\text{y}}=\frac{50}{\text{r}}\ ...(2)$

$\therefore\frac{100}{\text{r}^2}=\frac{50}{\text{r}}$

$\Rightarrow\text{r}=2\text{cm}.$

$\therefore\omega=\sqrt{\frac{100}{\text{r}^2}}=5\sec^2$

Again, to find out the positions where the speed is 8m/sec,

$\text{v}^2=\omega^2(\text{r}^2-\text{y}^2)$

$\Rightarrow64=25(4-\text{y}^2)$

$\Rightarrow4-\text{y}^2=\frac{64}{25}$

$\Rightarrow\text{y}^2=1.44$

$\Rightarrow\text{y}=\sqrt{1.44}$

$\Rightarrow\text{y}=\pm1.2\text{cm}$ from mean position.