Oscillations - Result Question 57
62. A simple harmonic oscillator has an amplitude A and time period $T$. The time required by it to travel from $x=A$ to $x=A / 2$ is
[1992]
(a) $T / 6$
(b) $T / 4$
(c) $T / 3$
(d) $T / 2$
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Solution:
- (a) For S.H.M., $x=A \sin (\frac{2 \pi}{T} t)$
When $x=A, A=A \sin (\frac{2 \pi}{T} . t)$
$\therefore \quad \sin (\frac{2 \pi}{T} \cdot t)=1$
$\Rightarrow \sin (\frac{2 \pi}{T} \cdot t)=\sin (\frac{\pi}{2}) \Rightarrow t=(T / 4)$
When $x=\frac{A}{2}, \frac{A}{2}=A \sin (\frac{2 \pi}{T} . t)$
or, $\sin \frac{\pi}{6}=\sin (\frac{2 \pi}{T} t)$ or $t=(T / 12)$
Now, time taken to travel from $x=A$ to $x=A / 2$ is $(T / 4-T / 12)=T / 6$
