Question

The shortest distance travelled by a particle (performing S.H.M.) from mean position in $2$ seconds is equal to $\frac{\sqrt{3}}{2}$ of its amplitude. Find its period.

• A. $11s$
• B. $12s$
• C. $13s$
• D. $14s$

Answer 1

Answer: (B)

$\text{When} \text{t} = 2 \text{s} \text{,} \text{y} = \frac{\sqrt{3}}{2} \text{r}$
Now, y = r sin ωt
$\therefore \frac{\sqrt{3}}{2} \text{r} = \text{r} \text{sin} \omega \times 2 \text{or} \text{sin} 2 \omega = \frac{\sqrt{3}}{2}$
$\text{or} 2 \omega = \frac{\text{\pi }}{3} \text{or} \omega = \frac{\text{\pi }}{6}$
$\text{Hence,} \text{T} = \frac{2 \text{\pi }}{\text{ω}} = \frac{2 \text{\pi }}{\text{\pi } /6} = 1 2 \text{s}$