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)

When $t =2 s , y =\frac{\sqrt{3}}{2} r$
Now, $y=r \sin \omega t$
$\therefore \quad \frac{\sqrt{3}}{2} r = r \sin \omega \times 2 \quad$ or $\quad \sin 2 \omega=\frac{\sqrt{3}}{2}$
or $\quad 2 \omega=\frac{\pi}{3}$
Or $\omega=\frac{\pi}{6}$
Hence, $T =\frac{2 \pi}{\omega}=\frac{2 \pi}{\pi / 6}=12 s$