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
Answer:
displacement power factor is 0.959087
Explanation:
given data
THD = 88%
true power factor = 0.72
solution
we get here total harmonic distribution THD is express as here
THD =
..............1
her g is distortion factor
so put here value and we will get g that is
0.88² =
solve it we get
g = 0.750714
and
displacement power factor is express as
DPF =
.................2
put here value and we will get
DPF =
DPF = 0.959087
Answer:
30
Explanation:
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