A) 0.4 \[-\pi /4\]
B) 2.5 \[-\pi /2\]
C) 2.5 \[+\pi /2\]
D) 2.5 \[-\pi /4\]
Correct Answer: B
Solution :
From the graph shown below. It is clear that phase lead of N over M is \[-\frac{\pi }{2}\]. Since time period (i.e. taken to complete one cycle) = 0.4 sec. Hence frequency \[\nu =\frac{1}{T}=2.5\,Hz\]You need to login to perform this action.
You will be redirected in
3 sec