Question
The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes $2cm, 1m/s$ and $10m/s^2$ at a certain instant. Find the amplitude and the time period of the motion.
✓
Answer
Given that, at a particular instant, $X = 2cm = 0.02m\ V = 1m/sec A = 10 m/s^2$ We know that $\text{a}=\omega^2\text{x}$
$\Rightarrow\omega=\sqrt{\frac{\text{a}}{\text{x}}}=\sqrt{\frac{10}{0.02}}=\sqrt{500}=10\sqrt{5}$
$\text{T}=\frac{2\pi}{\omega}=\frac{2\pi}{10\sqrt{5}}=\frac{2\times3.14}{10\times2.236}=0.28\text{ seconds.}$
Again, amplitude r is given by $\text{v}=\omega\Big(\sqrt{\text{r}^2-\text{x}^2}\Big)$
$\Rightarrow\text{v}^2=\omega^2(\text{r}^2-\text{x}^2)$
$1=500(\text{r}^2-0.0004)$
$\Rightarrow\text{r}=0.0489\approx0.049\text{m}$
$\therefore\text{r}=4.9\text{cm.}$
$\Rightarrow\omega=\sqrt{\frac{\text{a}}{\text{x}}}=\sqrt{\frac{10}{0.02}}=\sqrt{500}=10\sqrt{5}$
$\text{T}=\frac{2\pi}{\omega}=\frac{2\pi}{10\sqrt{5}}=\frac{2\times3.14}{10\times2.236}=0.28\text{ seconds.}$
Again, amplitude r is given by $\text{v}=\omega\Big(\sqrt{\text{r}^2-\text{x}^2}\Big)$
$\Rightarrow\text{v}^2=\omega^2(\text{r}^2-\text{x}^2)$
$1=500(\text{r}^2-0.0004)$
$\Rightarrow\text{r}=0.0489\approx0.049\text{m}$
$\therefore\text{r}=4.9\text{cm.}$
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