3.2 Response of Damped Systems to Harmonic Loading
67
=
1
(1 − 0.32 2 ) 2 + (2 × 0.32 × 0.099) 2
= 1.111
Example 3.3 An automobile whose weight is 150 N is mounted on four identical
springs. Due to its weight, it sags 0.23 m. Each shock absorber has a damping
coefficient of 0.4 N for a velocity of 3 cm per second. The car is placed on a platform
which moves vertically at resonant speed, having an amplitude of 1 cm. Find the
amplitude of vibration of the car.
The centre of gravity is assumed to be at the centre of the wheelbase.
The natural frequency of the car as per Eq. (2.20) is
p =
9.81
0.23
= 6.53 rad/s
The damping coefficient for the four shock absorbers is
c =
0.4 × 4
0.03
=
160
3
Ns/m
The platform vibrates with an amplitude x 0 . Therefore, spring extension is x −
x 0 sin ω t, where ω is the frequency of oscillation of the platform. The equation of
motion is
m ¨
x + c ( ˙
x − x 0 ω cos ω t) + k (x − x 0 sin ω t) = 0
or
m ¨
x + c ˙
x + kx = cx 0 ω cos ω t + kx 0 sin ω t
=
(cx 0 ω ) 2 + (kx 0 ) 2 sin (ω t + α)
Therefore, equivalent external force is
F =
(cx 0 ω ) 2 + (kx 0 ) 2
At resonance,
ω = p
1 − 2ζ 2
ζ =
160
3 × 2 × 150 × 6.53
= 0.027
ω = 6.53
1 − 2 (0.027) 2 = 6.53
x 0 = 0.01 m and k =
150
0.23