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
0.2
0.4
0.6
0.8
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
27 comments Page 2 of 3.

Mithun Kumar said:   7 years ago
Since ln(x0/x1)= 2 * π*zeta/√ (1-zeta^2).

Where zeta=actual damping coefficient/critical damping coefficient.

PRINCE KUMAR said:   8 years ago
@Pranay Mishra.

Logarithmic decrement is the rate through which amplitude of free damped vibration is decreases.

It is only for under damped vibration system.

Nagamalleswarao said:   9 years ago
Could you please confirm the zeta. Is this zeta is a damping factor or damping coefficient, or both are same?

Mohan said:   9 years ago
Good explanation. Thank you all.

Kareena shah said:   9 years ago
Perfect answer @Jithin.

BHEEMESH said:   1 decade ago
Logarithmic Decrement = 2*(Pi)*Zeta/sqrt(1-zeta^2).

Where zeta = Actual damping coefficient/Critical damping coefficient.

Vicky pooraneeswar said:   9 years ago
Can anyone please explain in a simple way?

Ramu said:   9 years ago
Zeta is damping factor or damping ratio.

Rohit said:   9 years ago
May be Zeta is the damping coefficient.

Peviks said:   9 years ago
Thank you all, it great to understand the solution.


Post your comments here:

Your comments will be displayed after verification.