A linear harmonic oscillator having a mass of 5.4kg is set into motion with viscous damping. if the frequency is 7Hz and if two successive amplitudes a full cycle apart are measured to be 4.65mm and 4.30mm, compute the viscous damping coefficient. Note: Give your answer to 3 decimal places x (t) + Other Parameters: Logarithmic Decrement (6): Answer: 8 = In= In 1 = 5wnTa = 5 wn 2n 2n c %3D Xn+1 wa 2m t [s] Damping Ratio (5): % = Ce %3D 2vkm 2man V(2m) +8 Frequency of damped vibration (wa): wa - Undamped Forced Vibration: Frequency Ratio (r): r- Xp - x sin wt
A linear harmonic oscillator having a mass of 5.4kg is set into motion with viscous damping. if the frequency is 7Hz and if two successive amplitudes a full cycle apart are measured to be 4.65mm and 4.30mm, compute the viscous damping coefficient. Note: Give your answer to 3 decimal places x (t) + Other Parameters: Logarithmic Decrement (6): Answer: 8 = In= In 1 = 5wnTa = 5 wn 2n 2n c %3D Xn+1 wa 2m t [s] Damping Ratio (5): % = Ce %3D 2vkm 2man V(2m) +8 Frequency of damped vibration (wa): wa - Undamped Forced Vibration: Frequency Ratio (r): r- Xp - x sin wt
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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![A linear harmonic oscillator having a mass of 5.4kg is set into motion with viscous damping. if the frequency is 7Hz and if two successive
amplitudes a full cycle apart are measured to be 4.65mm and 4.30mm, compute the viscous damping coefficient.
Note: Give your answer to 3 decimal places
Other Parameters:
x (t)
Logarithmic Decrement (6):
Answer:
x2
2nC
2n C
2T
8 = In1 = In1 = 5 wnTd = 5 wn
%3D
X3
X2
Xn+1
uz Pm
Damping Ratio (5): 5 =
%3D
t [s]
(2T)2+8
Cc
2vkm
2mwn
Wn
Frequency of damped vibration (wa): wa =
Undamped Forced Vibration:
Fo/k
x sin ot
Frequency Ratio (r): r =
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Transcribed Image Text:A linear harmonic oscillator having a mass of 5.4kg is set into motion with viscous damping. if the frequency is 7Hz and if two successive
amplitudes a full cycle apart are measured to be 4.65mm and 4.30mm, compute the viscous damping coefficient.
Note: Give your answer to 3 decimal places
Other Parameters:
x (t)
Logarithmic Decrement (6):
Answer:
x2
2nC
2n C
2T
8 = In1 = In1 = 5 wnTd = 5 wn
%3D
X3
X2
Xn+1
uz Pm
Damping Ratio (5): 5 =
%3D
t [s]
(2T)2+8
Cc
2vkm
2mwn
Wn
Frequency of damped vibration (wa): wa =
Undamped Forced Vibration:
Fo/k
x sin ot
Frequency Ratio (r): r =
Next page
jous page
E//learnonline.gmit.ie/mod/quiz/attempt.php?attempt3D167036&cmid%3D155306&page%3D2...
P Type here to search
3| 5
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