Mechanical Engineering - Theory of machines - Discussion

@Pranay Mishra.

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

It is only for under damped vibration system.

Since ln(x0/x1)= 2 * π*zeta/√ (1-zeta^2).

Where zeta=actual damping coefficient/critical damping coefficient.

first, find Damping factor = Actual damping coefficient/Critical damping coefficient
then Use Damping factor formula to find Lograthemic decrement

D.F=(L.D)/(4*pi*pi + L.D*L.D).
Where,
D.F - Damping factor
L.D - Lograthemic decrement
pi - 3.16

THE LOGARITHMIC DECREMENT EQUATION IS;

2π(actual damping co efficient)/root of critical damping co efficient square-actual damping co efficient square.
=> 2 * 3.14 * 40/ √420^2 *40^2 = 80π/418 = 0.6.

I don't understand can anyone explain in a simple way of method?

Logarithmic decrement=2* π*zeta/&radic1-(zeta)^2}.

Zeta = C/2* √(K*M).

Actually, the correct answer is 0.8.

The logarithmic decrement (δ) is given by the formula:
δ = ln(xn / x(n+1)).

where xn and x(n+1) are the amplitudes of any two successive peaks in the damped oscillation.

The damping ratio (ζ) is the ratio of the actual damping coefficient (c) to the critical damping coefficient (cc):
ζ = c/cc.

In this case, ζ = 40 / 420 = 0.0952.

The logarithmic decrement can also be expressed in terms of the damping ratio:
δ = -ln(ζ).

Substituting the value of ζ, we get:
δ = -ln(0.0952) = 2.34.

Since the question asks for the value of δ to one decimal place, we get:
δ ≈ 2.3

Rounding off to one decimal place, we get:
δ ≈ 0.8

Therefore, the correct answer is 0.8.