Mechanical Engineering - Theory of machines - Discussion
Discussion Forum : Theory of machines - Section 1 (Q.No. 3)
3.
In a vibrating system, if the actual damping coefficient is 40 N/m/s and critical damping coefficient is 420 N/m/s, then logarithmic decrement is equal to
Discussion:
27 comments Page 1 of 3.
BHEEMESH said:
1 decade ago
Logarithmic Decrement = 2*(Pi)*Zeta/sqrt(1-zeta^2).
Where zeta = Actual damping coefficient/Critical damping coefficient.
Where zeta = Actual damping coefficient/Critical damping coefficient.
Yas said:
1 decade ago
2*pi*(40/440)/sqrt(1-(40/440)^2.
The SARK said:
1 decade ago
2*pi*(40/420)*(1-(40/420)^2) = 0.59.
Gaurav said:
1 decade ago
Solution without calculator.
Zeta = 40/420 = 0.1 (approx.).
Log decrements = 2*pie*0.1/(1-0.1^2) = 2*pie*0.1 = 0.63 (approx..).
Zeta = 40/420 = 0.1 (approx.).
Log decrements = 2*pie*0.1/(1-0.1^2) = 2*pie*0.1 = 0.63 (approx..).
Sarvesh jaiswal said:
1 decade ago
Ld = 2*3.14*0.1/(1-0.1^2).
(1)
Murtadha abass said:
1 decade ago
2*pi*Zeta = 2*3.14*0.09 = 0.565.
Ankit said:
10 years ago
2xpixZeta/sqrt(1-Zeta^2).
Swagat said:
9 years ago
Thanks for all you explanation.
Pradeep said:
9 years ago
What is zeta?
Peviks said:
9 years ago
Thank you all, it great to understand the solution.
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