Oscillations Ques 44
44. A point performs simple harmonic oscillation of period $T$ and the equation of motion is given by $x=a \sin (\omega t+\pi / 6)$. After the elapse of what fraction of the time period the velocity of the point will be equal to halfof its maximum velocity?
[2008]
(a) $T / 8$
(b) $T / 6$
(c) $T / 3$
(d) $T / 12$
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Answer:
Correct Answer: 44.(d)
Solution:
- (d) We have $x=a \sin (\omega t+\frac{\pi}{6})$
$\therefore$ Velocity, $v=\frac{dx}{dt}=a \omega \cos (\omega t+\frac{\pi}{6})$
Maximum velocity $=a \omega$
According to question,
$\frac{a \omega}{2}=a \omega \cos (\omega t+\frac{\pi}{6})$
or, $\cos (\omega t+\frac{\pi}{6})=\frac{1}{2}=\cos 60^{\circ}$ or $\cos \frac{p}{3}$
$\Rightarrow wt+\frac{p}{6}=\frac{p}{3}$
$wt=\frac{p}{3}-\frac{p}{6}$ or, $wt=\frac{p}{6}$ or,
$\frac{2 p}{T} \cdot t=\frac{p}{6} \Rightarrow t=\frac{T}{12}$
BETA
