2.7 Logarithmic Decrement
47
or
ζ =
1
2π
ln
x 1
x 2
(2.62)
If a system has 5% of critical damping (ζ = 0.05), the logarithmic decrement will
be 0.1π which indicates that successive peaks would be e
0.1 π or 1.37. Inverting this
quantity, it can be said that each and every peak would have a value of 0.73 times
that of the preceding. This facilitates to a great extent the visualisation of the effect
of damping.
Example 2.16 For a system having mass 10 kg and spring constant 12 kN/m, the
ampli-tude decreases to 0.2 of the initial value after six consecutive cycles. Find the
damping constant of the damper.
The ratio of two successive amplitudes will remain the same.
x 0
x 1
=
x 1
x 2
=
x 2
x 3
=
x 3
x 4
=
x 4
x 5
=
x 5
x 6
= e
δ
Therefore,
x 0
x 1
·
x 1
x 2
·
x 2
x 3
·
x 3
x 4
·
x 4
x 5
·
x 5
x 6
=
x 0
x 6
= e
6 δ
=
1
0.2
= 5
δ =
1
6
ln 5 = 0.268
therefore
2πζ
1 − ζ 2
= 0.268
or
ζ = 0.0427
The damping constant is
c = 2ζ
√
km = 2 × (0.0427)
√
12000 × 10 = 29.58 Ns/m
Based on the above example, a general statement can be made. If the amplitude
is reduced by a factor N after n cycles, then
x 0
x n
= N = e
n δ
or
47
or
ζ =
1
2π
ln
x 1
x 2
(2.62)
If a system has 5% of critical damping (ζ = 0.05), the logarithmic decrement will
be 0.1π which indicates that successive peaks would be e
0.1 π or 1.37. Inverting this
quantity, it can be said that each and every peak would have a value of 0.73 times
that of the preceding. This facilitates to a great extent the visualisation of the effect
of damping.
Example 2.16 For a system having mass 10 kg and spring constant 12 kN/m, the
ampli-tude decreases to 0.2 of the initial value after six consecutive cycles. Find the
damping constant of the damper.
The ratio of two successive amplitudes will remain the same.
x 0
x 1
=
x 1
x 2
=
x 2
x 3
=
x 3
x 4
=
x 4
x 5
=
x 5
x 6
= e
δ
Therefore,
x 0
x 1
·
x 1
x 2
·
x 2
x 3
·
x 3
x 4
·
x 4
x 5
·
x 5
x 6
=
x 0
x 6
= e
6 δ
=
1
0.2
= 5
δ =
1
6
ln 5 = 0.268
therefore
2πζ
1 − ζ 2
= 0.268
or
ζ = 0.0427
The damping constant is
c = 2ζ
√
km = 2 × (0.0427)
√
12000 × 10 = 29.58 Ns/m
Based on the above example, a general statement can be made. If the amplitude
is reduced by a factor N after n cycles, then
x 0
x n
= N = e
n δ
or
