A) \[\frac{1}{1.57}sec\]
B) 1.57 sec
C) 2 sec
D) 4 sec
Correct Answer: D
Solution :
Given max velocity \[\omega a=1\]and maximum acceleration \[{{\omega }^{2}}a=1.57\] \[\therefore \frac{{{\omega }^{2}}a}{\omega a}=1.57\Rightarrow \omega =1.57\] \[\Rightarrow \frac{2\pi }{T}=1.57\Rightarrow T=4\]You need to login to perform this action.
You will be redirected in
3 sec