Q93 / 160 · JIPMER 2009 · Physics
A particle executes simple harmonic oscillation with an amplitude $a$. The period of oscillation is $T$. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is
- A$\frac{T}{4}$
- B$\frac{T}{8}$
- Correct answerC$\frac{T}{12}$
- D$\frac{T}{2}$
Solution
Let displacement equation of particle executing SHM is
$y=a \sin \omega t$
As particle travels half of the amplitude from the equilibrium position, so
$y=\frac{a}{2}$
Therefore,
$\frac{a}{2}=a \sin \omega t$
or
$\sin \omega t=\frac{1}{2}=\sin \frac{\pi}{6}$
or
$\omega t=\frac{\pi}{6}$
$or$
$t=\frac{\pi}{6 \omega}$
or
$t=\frac{\pi}{6\left(\frac{2 \pi}{T}\right)} \quad\left(\text { as } \omega=\frac{2 \pi}{T}\right)$
or $\quad t=\frac{T}{12}$
Hence, the particle travels half of the amplitude from the equilibrium in $\frac{T}{12} \mathrm{~s}$.
Hence, the correct answer is option: 'C'.