Kinematics Question 375
Question: The acceleration of a particle, starting from rest, varies with time according to the relation $ a=-s{{\omega }^{2}}\sin \omega t $ . The displacement of this particle at a time t will be
Options:
A) $ \text{s sin }\omega \text{ t} $
B) $ \text{s }\omega \text{ cos }\omega \text{ t} $
C) $ \text{s }\omega \text{ sin}\omega t $
D) $ -\frac{1}{2}( s{{\omega }^{2}}\text{ sin }\omega t ){{t}^{2}} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ a=\frac{d^{2}x}{dt^{2}}=-s\omega \sin \omega t. $
On integrating, $ \frac{dx}{dt}=s{{\omega }^{2}}\frac{\cos \omega t}{\omega }=s\sin \omega t $
Again on integrating, we get $ x=s\omega \frac{\sin \omega t}{\omega }=s\sin \omega t $
