5.5 Vibration Absorber
165
Fig. 5.8 Example 5.2
W 1
g
¨
x 1 + k 1 x 1 + k 2 ( x 1 − x 2 ) = meω
2 sin ω t
W 2
g
¨
x 2 + k 2 ( x 2 − x 1 ) = 0
⎫
⎪ ⎪ ⎬
⎪ ⎪ ⎭
(a)
Equation (a) is similar to Eq. (5.20), except that F 1 is replaced by meω
2 . Therefore,
solutions of Eq. (a) are identical to Eq. (5.25), except that F 1 is substituted as (mew
2 )
ω =
2 π × 1800
60
= 188.5 rad/s
For the system to act as vibration absorber, from Eq. (5.26)
ω
2
=
k 2
m 2
Therefore
k 2 = ω
2 m 2 = 188.5
2
×
225
981
= 814,960 N/m
The amplitude of W 2 is given by Eq. (5.28) as
B = −
meω
2
k 2
= −
40
1000
×
188.5
2
814,960
= − 0.0017 m
165
Fig. 5.8 Example 5.2
W 1
g
¨
x 1 + k 1 x 1 + k 2 ( x 1 − x 2 ) = meω
2 sin ω t
W 2
g
¨
x 2 + k 2 ( x 2 − x 1 ) = 0
⎫
⎪ ⎪ ⎬
⎪ ⎪ ⎭
(a)
Equation (a) is similar to Eq. (5.20), except that F 1 is replaced by meω
2 . Therefore,
solutions of Eq. (a) are identical to Eq. (5.25), except that F 1 is substituted as (mew
2 )
ω =
2 π × 1800
60
= 188.5 rad/s
For the system to act as vibration absorber, from Eq. (5.26)
ω
2
=
k 2
m 2
Therefore
k 2 = ω
2 m 2 = 188.5
2
×
225
981
= 814,960 N/m
The amplitude of W 2 is given by Eq. (5.28) as
B = −
meω
2
k 2
= −
40
1000
×
188.5
2
814,960
= − 0.0017 m