Oscillations Ques 46
46. 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
[2007]
(a) $T / 8$
(b) $T / 12$
(c) $T / 2$
(d) $T / 4$
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Answer:
Correct Answer: 46.(b)
Solution:
- (b) Displacement from the mean position
$y=a \sin (\frac{2 \pi}{T}) t$
According to problem $y=a / 2$
$a / 2=a \sin (\frac{2 \pi}{T}) t$
$\Rightarrow \frac{\pi}{6}=(\frac{2 \pi}{T}) t \Rightarrow t=T / 12$
This is the minimum time taken by the particle to travel half of the amplitude from the equilibrium position.
BETA
