When a 2.50 kg object is hung vertically on a certain light spring described by hookes law the spring stretches 2.76 cm.
1 answer:
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Answer: (a). ΓL = 0.246 < 75°
(b). S = 1.7
(c). Zin = (30-j)λ
(d). jreal = Arc Po = 0.105λ
(e). jmax = jreal = 0.105λ
Explanation:
attached is a document to help in understanding.
So we will begin with a step by step analysis of the problem.
from the diagram we have that ZL = (50 + j25) Ω.
where ZL = ZL / Z₀ = 50 + j25 / 50 = 1 + j0.5
so we mark this on the chart as point 'P'
(a) ΓL = mP/m 'P' < Θ L = 1.7/6.9 < 75°
ΓL = 0.246 < 75°
(b) This s-circle 's' is given thus s = r = 1.7 on the RHS of the chart
S = 1.7
(c) we are to calculate the input impedance;
ζin = Q = 0.6 - j0.02
therefore Zin = Z₀ζin = 50(0.6 - j0.02) = (30-j)λ
Zin = (30-j)λ
(d) here we are taking R as the diameter opposite of Q on the s=circle
so R = γin = 1.7 + j0.02
yin = yo (γin) = (1.7+j0.02) / 50 = (34 + j0.4)ms
yin = (34 + j0.4)ms
(e) move from 'p' on s-circle to 'o'
where maximum impedance = Znxl = Zos
which gives jreal = Arc Po = 0.105λ
(f) jmax = jreal = 0.105λ
cheers i hope this helps
