An accelerometer has a damping ratio of 0.5 and a natural frequency of 18,000 Hz. It is used to sense the relative displacement
1 answer:
Answer:
A) i) Dynamic error ≈ 3.1%
ii) phase shift ≈ -12°
B) 79971.89 rad/s
Explanation:
Given data :
Damping ratio = 0.5
natural frequency = 18,000 Hz
<u>a) Calculate the dynamic error and phase shift in accelerometer output at an impart vibration of 4500 Hz</u>
i) Dynamic error
This can be calculated using magnitude ratio formula attached below is the solution
dynamic error ≈ 3.1%
ii) phase shift
This phase shift can be calculated using frequency dependent phase shift formula
phase shift ≈ -12°
<u>B) Determine resonance frequency </u>
Wr = 2 ( 18000
) = 79971.89 rad/s
C) The maximum magnitude ratio that the system can achieve
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Answer:
0.07756 m
Explanation:
Given mass of object =0.20 kg
spring constant = 120 n/m
maximum speed = 1.9 m/sec
We have to find the amplitude of the motion
We know that maximum speed of the object when it is in harmonic motion is given by where A is amplitude and
is angular velocity
Angular velocity is given by where k is spring constant and m is mass
So
Refer to the diagram shown below.
Define the unit vector i to point in the eastern direction, and the unit vector j to point in the northern direction.
The first distance is 184 m west. It is represented by
d₁ = -184 i
The second distance is 220 m at 30° south of east. It is
d₂ = 220(cos 30° i - sin 30° j) = 190.53 i - 110 j
The third distance is 104 m at 80 east of north. It is
d₃ = 104(sin 80° i + cos 80° j) = 102.42 i + 18.06 j
Let the fourth distance be
d₄ = a i + b j
Because the traveler ends back at the original position, the vector sum of the distances is zero. It means that each component of the vector sum is zero.
The x-component yields
-184 + 190.53 + 102.42 + a = 0
a = -108.95
The y-component yields
0 - 110 + 18.06 + b = 0
b = 91.94
The magnitude of the fourth displacement is
√[(-108.95)² + 91.94² ] = 142.56 m
The direction is at an angle θ north of west, given by
θ = tan⁻¹ (91.94/108.95) = 40.2°
Answer:
The fourth displacement has a magnitude of 142.56 m. It is about 40° north of west.
Define the unit vector i to point in the eastern direction, and the unit vector j to point in the northern direction.
The first distance is 184 m west. It is represented by
d₁ = -184 i
The second distance is 220 m at 30° south of east. It is
d₂ = 220(cos 30° i - sin 30° j) = 190.53 i - 110 j
The third distance is 104 m at 80 east of north. It is
d₃ = 104(sin 80° i + cos 80° j) = 102.42 i + 18.06 j
Let the fourth distance be
d₄ = a i + b j
Because the traveler ends back at the original position, the vector sum of the distances is zero. It means that each component of the vector sum is zero.
The x-component yields
-184 + 190.53 + 102.42 + a = 0
a = -108.95
The y-component yields
0 - 110 + 18.06 + b = 0
b = 91.94
The magnitude of the fourth displacement is
√[(-108.95)² + 91.94² ] = 142.56 m
The direction is at an angle θ north of west, given by
θ = tan⁻¹ (91.94/108.95) = 40.2°
Answer:
The fourth displacement has a magnitude of 142.56 m. It is about 40° north of west.