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System with mass 20 kg and equivalent stiffness of 2 kN/m is modified by attaching the viscous damper which is parallel with spring. Damping factor of system is 0.32. Modified system will have maximum magnification factor when frequency of exciting force is equal to k times of natural frequency of system. The value of k is_________ . (up to two decimal places)
Correct answer is '0.89'. Can you explain this answer?
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System with mass 20 kg and equivalent stiffness of 2 kN/m is modified...
Given information:
- Mass of the system (m) = 20 kg
- Equivalent stiffness of the system (k) = 2 kN/m
- Damping factor (ξ) = 0.32
Calculation:
The natural frequency (ωn) of the undamped system can be calculated using the formula:
ωn = √(k/m)
Substituting the given values:
ωn = √(2 kN/m / 20 kg) = √0.1 rad/s
The damped natural frequency (ωd) can be calculated using the formula:
ωd = ωn √(1 - ξ^2)
Substituting the given damping factor:
ωd = √0.1 rad/s √(1 - 0.32^2) = √0.1 rad/s √0.896 = √0.0896 rad/s
The maximum magnification factor occurs when the frequency of the exciting force (ω) is equal to k times the damped natural frequency (ωd):
ω = k * ωd
Substituting the given values:
ω = k * √0.0896 rad/s
To find the value of k, we need to solve this equation. Rearranging the equation:
k = ω / √0.0896
The maximum magnification factor occurs when k is equal to 1. Therefore, we need to find the value of ω when k = 1:
1 = ω / √0.0896
Squaring both sides:
1 = ω^2 / 0.0896
Rearranging the equation:
ω^2 = 1 * 0.0896
ω = √0.0896 rad/s
Substituting this value of ω in the equation for k:
k = √0.0896 / √0.0896 = 1
Therefore, the value of k is 1.
However, the correct answer given is 0.89. It seems there might be an error in either the given answer or the calculation. Please double-check the values and calculations to ensure accuracy.
- Mass of the system (m) = 20 kg
- Equivalent stiffness of the system (k) = 2 kN/m
- Damping factor (ξ) = 0.32
Calculation:
The natural frequency (ωn) of the undamped system can be calculated using the formula:
ωn = √(k/m)
Substituting the given values:
ωn = √(2 kN/m / 20 kg) = √0.1 rad/s
The damped natural frequency (ωd) can be calculated using the formula:
ωd = ωn √(1 - ξ^2)
Substituting the given damping factor:
ωd = √0.1 rad/s √(1 - 0.32^2) = √0.1 rad/s √0.896 = √0.0896 rad/s
The maximum magnification factor occurs when the frequency of the exciting force (ω) is equal to k times the damped natural frequency (ωd):
ω = k * ωd
Substituting the given values:
ω = k * √0.0896 rad/s
To find the value of k, we need to solve this equation. Rearranging the equation:
k = ω / √0.0896
The maximum magnification factor occurs when k is equal to 1. Therefore, we need to find the value of ω when k = 1:
1 = ω / √0.0896
Squaring both sides:
1 = ω^2 / 0.0896
Rearranging the equation:
ω^2 = 1 * 0.0896
ω = √0.0896 rad/s
Substituting this value of ω in the equation for k:
k = √0.0896 / √0.0896 = 1
Therefore, the value of k is 1.
However, the correct answer given is 0.89. It seems there might be an error in either the given answer or the calculation. Please double-check the values and calculations to ensure accuracy.
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Community Answer
System with mass 20 kg and equivalent stiffness of 2 kN/m is modified...
According to question, ω/ωn = k
For MF to be maximum, (1 - k2)2 + (2ξk)2 should be minimum.
So, for the denominator to be minimum,
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System with mass 20 kg and equivalent stiffness of 2 kN/m is modified by attaching the viscous damper which is parallel with spring. Damping factor of system is 0.32. Modified system will have maximum magnification factor when frequency of exciting force is equal to k times of natural frequency of system. The value of k is_________ . (up to two decimal places)Correct answer is '0.89'. Can you explain this answer?
Question Description
System with mass 20 kg and equivalent stiffness of 2 kN/m is modified by attaching the viscous damper which is parallel with spring. Damping factor of system is 0.32. Modified system will have maximum magnification factor when frequency of exciting force is equal to k times of natural frequency of system. The value of k is_________ . (up to two decimal places)Correct answer is '0.89'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about System with mass 20 kg and equivalent stiffness of 2 kN/m is modified by attaching the viscous damper which is parallel with spring. Damping factor of system is 0.32. Modified system will have maximum magnification factor when frequency of exciting force is equal to k times of natural frequency of system. The value of k is_________ . (up to two decimal places)Correct answer is '0.89'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for System with mass 20 kg and equivalent stiffness of 2 kN/m is modified by attaching the viscous damper which is parallel with spring. Damping factor of system is 0.32. Modified system will have maximum magnification factor when frequency of exciting force is equal to k times of natural frequency of system. The value of k is_________ . (up to two decimal places)Correct answer is '0.89'. Can you explain this answer?.
System with mass 20 kg and equivalent stiffness of 2 kN/m is modified by attaching the viscous damper which is parallel with spring. Damping factor of system is 0.32. Modified system will have maximum magnification factor when frequency of exciting force is equal to k times of natural frequency of system. The value of k is_________ . (up to two decimal places)Correct answer is '0.89'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about System with mass 20 kg and equivalent stiffness of 2 kN/m is modified by attaching the viscous damper which is parallel with spring. Damping factor of system is 0.32. Modified system will have maximum magnification factor when frequency of exciting force is equal to k times of natural frequency of system. The value of k is_________ . (up to two decimal places)Correct answer is '0.89'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for System with mass 20 kg and equivalent stiffness of 2 kN/m is modified by attaching the viscous damper which is parallel with spring. Damping factor of system is 0.32. Modified system will have maximum magnification factor when frequency of exciting force is equal to k times of natural frequency of system. The value of k is_________ . (up to two decimal places)Correct answer is '0.89'. Can you explain this answer?.
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