Question
The maximum speed and acceleration of a particle executing simple harmonic motion are $10\ cm/s$ and $50\ cm/s^2.$ Find the position$(s)$ of the particle when the speed is $8\ cm/s.$
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Answer
$\text{v}_{\text{max}}=10\ \text{cm/sec}.$
$\Rightarrow\text{r}\omega=10$
$\Rightarrow\omega^2=\frac{100}{\text{r}^2}\ ...(1)$
$\text{A}_{\text{max}}=\omega^2\text{r}=50\ \text{cm/sec}$
$\Rightarrow\omega^2=\frac{50}{\text{y}}=\frac{50}{\text{r}}\ ...(2)$
$\therefore\frac{100}{\text{r}^2}=\frac{50}{\text{r}}$
$\Rightarrow\text{r}=2\ \text{cm}.$
$\therefore\omega=\sqrt{\frac{100}{\text{r}^2}}=5\sec^2$
Again, to find out the positions where the speed is $8\ \text{m/sec,}$
$\text{v}^2=\omega^2(\text{r}^2-\text{y}^2)$
$\Rightarrow64=25(4-\text{y}^2)$
$\Rightarrow4-\text{y}^2=\frac{64}{25}$
$\Rightarrow\text{y}^2=1.44$
$\Rightarrow\text{y}=\sqrt{1.44}$
$\Rightarrow\text{y}=\pm1.2\ \text{cm}$ from mean position.
$\Rightarrow\text{r}\omega=10$
$\Rightarrow\omega^2=\frac{100}{\text{r}^2}\ ...(1)$
$\text{A}_{\text{max}}=\omega^2\text{r}=50\ \text{cm/sec}$
$\Rightarrow\omega^2=\frac{50}{\text{y}}=\frac{50}{\text{r}}\ ...(2)$
$\therefore\frac{100}{\text{r}^2}=\frac{50}{\text{r}}$
$\Rightarrow\text{r}=2\ \text{cm}.$
$\therefore\omega=\sqrt{\frac{100}{\text{r}^2}}=5\sec^2$
Again, to find out the positions where the speed is $8\ \text{m/sec,}$
$\text{v}^2=\omega^2(\text{r}^2-\text{y}^2)$
$\Rightarrow64=25(4-\text{y}^2)$
$\Rightarrow4-\text{y}^2=\frac{64}{25}$
$\Rightarrow\text{y}^2=1.44$
$\Rightarrow\text{y}=\sqrt{1.44}$
$\Rightarrow\text{y}=\pm1.2\ \text{cm}$ from mean position.
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