A bird is flying directly toward a stationary bird-watcher and emits a frequency of 1420 Hz. The bird-watcher, however, hears a
1 answer:
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Answer:
a) the maximum shear stress τ the bar is 16T
/πd³
b) the angle of twist between the ends of the bar is 16tL² / πGd⁴
Explanation:
Given the data in the question, as illustrated in the image below;
d is the diameter of the prismatic bar of length AB
t is the intensity of distributed torque
(a) Determine the maximum shear stress tmax in the bar
Maximum Applied torque T_max = tL
we know that;
shear stress τ = 16T/πd³
where d is the diameter
so
τ = 16T
/πd³
Therefore, the maximum shear stress τ the bar is 16T
/πd³
(b) Determine the angle of twist between the ends of the bar.
let theta () be the angle of twist
polar moment of inertia = πd⁴/32
now from the second image;
lets length dx which is at distance of "x" from "B"
Torque distance x
T(x) = tx
Elemental angle twist = d = T(x)dx / G
so
d = tx.dx / G(πd⁴/32)
d = 32tx.dx / πGd⁴
so total angle of twist will be;
=
=
32tx.dx / πGd⁴
= 32t / πGd⁴
= 32t / πGd⁴ [ L²/2]
= 16tL² / πGd⁴
Therefore, the angle of twist between the ends of the bar is 16tL² / πGd⁴
Answer:
0.72 Hz minimum frequency
Explanation:
When the damping is negligible,Amplitude is given as
here = (6.30)/(0.135) = 46.67 N/m kg
= 1.70/(0.135)(0.480) = 26.2 N/m kg
From the above equation , rearranging for ω,
⇒ ω² =46.67 ± 26.2 = 72.87 or 20.47
⇒ ω = 8.53 or 4.52 rad/s
Frequency = f
ω=2 π f
⇒ f = ω / 2π = 8.53 /6.28 or 4.52 / 6.28 = 1.36 Hz or 0.72 Hz
The lower frequency is 0.72 Hz and higher is 1.36 Hz