A single-phase transformer circuit feeds a motor and lighting load of 50 kilowatts. At a power factor of .8, the KVA rating of t
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Answer:
1) A=282.6 mm
2)
3)T=0.42 sec
4)f= 2.24 Hz
Explanation:
Given that
V=3.5 m/s at x=150 mm ------------1
V=2.5 m/s at x=225 mm ------------2
Where x measured from mid position.
We know that velocity in simple harmonic given as
Where A is the amplitude and ω is the natural frequency of simple harmonic motion.
From equation 1 and 2
------3
--------4
Now by dividing equation 3 by 4
So A=0.2826 m
A=282.6 mm
Now by putting the values of A in the equation 3
ω=14.609 rad/s
Frequency
ω= 2πf
14.609= 2 x π x f
f= 2.24 Hz
Maximum acceleration
Time period T
T=0.42 sec