Oscillations Ques 11
11. The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is
[2007]
(a) $\pi$
(b) $0.707 $ $\pi$
(c) zero
(d) $0.5 $ $\pi$
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Answer:
Correct Answer: 11.(d)
Solution:
- (d) Let $y=A \sin \omega t$
$v _{\text{inst }}=\frac{d y}{d t}=A \omega \cos \omega t=A \omega \sin (\omega t+\pi / 2)$
Acceleration $=-A \omega^{2} \sin \omega t$
$=A \omega^{2} \sin (\pi+\omega t)$
$\therefore \phi=\frac{\pi}{2}=0.5$ $ \pi$
BETA
