Solving Undamped Vibration Problems, Exams of Mechanics

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MECH 375 MECHANICAL VIBRATIONS FINAL CRASH PART 2-

Harmonically Excited Vibrations Concordia University

Harmonically Excited Vibrations

General Systems

A system with an applied force is system with a

forced vibration, or harmonically excited vibration.

The general differential equation describing this

system is:

𝑚 + 𝑐 + 𝑘 = 𝐹 sin(𝜔𝑡

Note that ω is the input frequency.

For undamped systems, if the input frequency

matches the natural frequency of the system, the

system’s oscillations will continue to grow. This

would result in catastrophic failure of the system.

The system does not require initial conditions in

order to be moving. If there were no initial

conditions, the system would start at the steady-

state solution.

Parts of the Solution:

The total response of the solution has two parts:

 Homogenous solution

 Particular/steady-state solution

1. A spring-mass system, resting on an inclined plane, is subjected to a harmonic force as shown. Find

the response of the system by assuming zero initial conditions.

Given:

m = 100 kg

k = 4000 N/m

x o

ẋ o

Equation of Motion :

Method 1:

Homogeneous Solution:

o

cos 𝜔𝑡

o

cos 𝜔𝑡

Particular Solution:

For an undamped system, the solution is:

(𝑡) = 𝐴 cos 𝜔

𝑛

𝑡 + 𝐴 sin 𝜔

𝑛

Let:

o

cos 𝜔𝑡

= X cos 𝜔𝑡

= 𝜔X sin 𝜔𝑡

2

X cos 𝜔𝑡

𝑛

2

X cos 𝜔𝑡 + 4000X cos 𝜔𝑡 = 𝐹

o

cos 𝜔𝑡

𝑛

2

)X = 𝐹

o

X =

o

𝑛

2

ℎ(𝑡) = 𝐴 cos √40𝑡 + 𝐵 sin √40𝑡

𝑝

o

2

cos 𝜔𝑡

Total Solution:

ℎ

𝑝

𝐴 cos √40𝑡 + 𝐵 sin √40𝑡 +

o

cos 𝜔𝑡

2

−√ 40 𝐴 sin √40𝑡 + √40𝐵 cos √40𝑡 −

o

sin 𝜔𝑡

2

Initial Conditions:

1. t = 0, x = 0

o

2

o

2

o

o

(𝑡) = − cos √40𝑡 + cos 𝜔𝑡

2. t = 0, ẋ = 0 4000 − 100 𝜔

2

4000 − 100 𝜔

2

𝑜

x(𝑡

2

(Cos𝑚𝑡 − Cos√40𝑡)

Method 1:

o

cos 𝜔𝑡

Laplace Transform:

2

X − 𝑠

− (0)) + 4000X =

o

2

2

2

• 4000)X =

o

2

2

o

X =

2

• 𝜔

2

)(100𝑠

2

Partial Fractions:

2

• 4000

2

• 𝜔

2

Numerator:

3

2

2

2

3

2

3

2

2

2

2

o

2

• 4000𝐷 = 0

When solving the system of equations:

o

2

o

2

2. An aircraft engine has a rotating unbalanced mass m at radius r. If the wing can be modeled as a

cantilever beam of uniform cross section a X b, as shown, determine the maximum deflection of the

engine at a speed of N rpm. Assume damping and effect of the wing between the engine and free end to

be negligible.

Equivalent Stiffness:

e𝑞

3

3

3

e𝑞

3

12

3

e𝑞

3

Equivalent Mass:

e𝑞

Unbalanced Mass:

o

2

Equation of Motion :

o

sin 𝜔𝑡

Let:

= X sin 𝜔𝑡

= 𝜔X cos 𝜔𝑡

2

X sin 𝜔𝑡

2

𝑀X sin 𝜔𝑡 + 𝑘X sin 𝜔𝑡 = 𝐹

o

sin 𝜔𝑡 2 𝜋𝑁

2

2

𝑀X + 𝑘X = 𝐹

3

(140)(60)

2

2

o

)X = 𝐹

o

X =

3

2

3

2

o

X =

2

3

2

𝑚𝑟𝑁

2

X =

3

2

2

3. The landing gear of an airplane can be idealized as the spring-mass-damper system shown. If the

runway surface is described y(t) = y o

cos(ωt), determine the values of k and c that limit the amplitude of

vibration of the airplane (x) to 0.1 m. Assume m = 2000 kg, y o

= 0.2 m, and ω = 157.08 rad/s.

Given:

m = 2000 kg

y o

= 0.2 m

ω = 157.08 rad/s

X = 0.1 m

Equation of Motion :

Let:

= Xe

−i𝜔𝑡

= i𝜔Xe

−i𝜔𝑡

2

Xe

−i𝜔𝑡

𝑦 = 𝑌e

−i𝜔𝑡

𝑦 = i𝜔𝑌e

−i𝜔𝑡

2

𝑌e

−i𝜔𝑡

2

Xe

−i𝜔𝑡

• i𝑐𝜔Xe

−i𝜔𝑡

• 𝑘Xe

−i𝜔𝑡

= i𝑐𝜔𝑌e

−i𝜔𝑡

• 𝑘𝑌e

−i𝜔𝑡

2

X + i𝑐𝜔X + 𝑘X = i𝑐𝜔𝑌 + 𝑘𝑌

[(𝑘 − 𝑀𝜔

2

) + i𝑐𝜔]X = (𝑘 + i𝑐𝜔)𝑌

X 𝑘 + i𝑐𝜔

2

) + i 𝑐𝜔

X √𝑘

2

2

2

)

2

2

Assume large stiffness k to reduce oscillations. Let k = 5 MN/m.

√(5 × 10

6

2

2

√(5 × 10

6

− 2000(157.08)

2

)

2

2

2

1.4583Θe

−i𝜔𝑡

• i62.5𝜔Θe

−i𝜔𝑡

• 3125Θe

−i𝜔𝑡

= 100e

−i𝜔𝑡

2

1.4583Θ + i62.5𝜔Θ + 3125Θ = 100

[(3125 − 1.4583(104.72)

2

) + i62.5(104.72)]Θ = 100

−12867+ i

2

2

Θ = 6.927 × 10

− 3

𝜑 = 0 − t a n

− 1

(

𝑝

6. 9 27 × 10

− 3

C os

ƒ

ƒ

ƒ

5. A rigid plate, weighing 100 lb, is hinged along an edge (P) and is supported on a dashpot with c = 1 lb-

sec/in, at the opposite edge (Q), as shown. A small fan weighing 50 lb and rotating at 750 rpm is

mounted on the plate through a spring with k = 200 lb/in. If the center of gravity of the fan is located at

0.1 in, from its axis of rotation, find the steady-state motion of the edge Q and the force transmitted to

the point S.

Given:

W

p

= 100 lb

C = 1 lb-s/in

W

f

= 50 lb

ω = 750 rpm = 78.54 rad/s

k = 200 lb/in

e = 0.1 in

Equation of Motion : fan

ƒ

• 𝑘 = 𝑚e𝜔

2

sin 𝜔𝑡

FBD: Plate

Let:

= X sin 𝜔𝑡

= 𝜔X cos 𝜔𝑡

2

X sin 𝜔𝑡

From Parallel-Axis Theorem for plate and fan:

ƒ

2

X sin 𝜔𝑡 + 𝑘X sin 𝜔𝑡 = 𝑚

ƒ

e𝜔

2

sin 𝜔𝑡

ƒ

2

) X = 𝑚

ƒ

e𝜔

2

𝑝

2

• 𝑚

𝑝

ƒ

2

ƒ

e𝜔

2

X =

2

2

𝑝

ƒ

2

2

X =

2

X = 0.1334 i𝑘

(𝑡) = 0.1334sin 78.54𝑡

2

2

6. An electric motor of mass 68 kg is mounted on an isolator block of mass 1200 kg, as shown. The

natural frequency of the total assembly is 160 Hz and its damping ratio is 0.1. An unbalance in the motor

armature causes a harmonic force F(t) = 100 sin 31.4t. Determine the vibration response of the block.

Two-Degrees-of-Freedom Systems

General System

This system can be analyzed as:

 Free or forced.

 Damped or undamped.

As previously, they can be solved as

ODEs or using Laplace.

Free-Vibration of an Undamped System

Consider the above system, where f(t) = c 1

= c 2

= 0. This system would be considered undamped with

free vibration. The system would therefore need initial conditions in order to have motion.

The systems set of input-output equations are:

1 1

1

2

1

2 2

2 2

2 2

2 1

Mode Shapes

When the natural frequencies of the system are found (usually there are two), the relative amplitudes

can be found using the transfer function. By setting one of the variables to 1 and solving for the other,

the mode shape can then be sketched for each value of ω.

2

Method of Solving (Laplace Transform):

1. Apply the Laplace transform for both equations, solving for a total response.
2. Put the set of equations in matrix form, ie,

1

2

1

2

2

[

2

] [

X

1

] = [

1

1o

1o

]

2

2

2

X

2

2o

2o

Apply Cramer’s Rule to solve for X 1

and X 2

, ie,

1

2

1

2

2

2

2

2

• 𝐾

1

1o

1o

2

1

2

2o

2o

2

2

2

1

2

1

2

1

1o

1o

2

2

1

X

1

, X

2

2

2

2o

2o

Solve for x 1

(t) and x 2

(t) by applying inverse Laplace on X 1

and X 2

For mode shapes, divide the coefficient in front of the sine/cosine function in x 1

(t) to its

corresponding sine/cosine function in x 2

(t). This will solve for X 1

assuming X 2

For More General Solutions:

 For free vibration of damped systems, use the substitution of = Xe

i𝜔𝑡

 For forced vibration of either damped or undamped, use Cramer’s Rule. Note that simplifying

the determinant of the denominator may be difficult due to higher order polynomials.

1. A two-mass system consists of a piston mass m, connected by two elastic springs, that moves inside a

tube as shown. A pendulum of length l and end mass m 2

is connected to the piston as shown. (a) Derive

the equation of motion of the system in terms of x 1

(t) and θ(t). (b) Derive the equations of motion of the

system in terms of x 1

(t) and x 2

(t). (c) Find the natural frequencies of vibration of the system.

FBD:

Slider:

1 1

1

FBD:

Pendulum:

O

2 1

2

2

Note that since θ is really

small,

sin 𝜃 ≈ 𝜃

cos 𝜃 ≈ 1

2

Also, since this is a 2 DOF

system,

2

1

2

1

a) x 1

and θ:

S

o

1

1

2

1

− 𝑇 sin 𝜃 = 0

2 1

𝑙 cos 𝜃 + 𝑚

2

2

2

𝑔𝑙 sin 𝜃 = 0

1

1

2

1

2

2 1

2

2

[

] [

1

] + [

k 1

• k 2

2

x

1

] =

] [

2 1

2

2

2

−

1 2

2

2

2

2

2

2

2

2

2

2

1

2

2

2

2

2

2

2

2

4

2

2

1

2

2

1

2

2

4

2

2

2

2

2

2 2 1 2

1 2

2

4

− 𝑚

2

2

(𝑘

1

2

2

2

2

• 𝑚

2

1

2

2

4

− [𝑚𝑚

2

2

1

2

2

𝑔)]𝜔

2

• 𝑚

2

1

2

From quadratic equation:

2

2

2

1

2

2

𝑔) ± √[𝑚𝑚

2

2

1

2

2

𝑔)]

2

2

𝑙)[𝑚

2

1

2

)]

2

2

2

1

2

2

2

[𝑚𝑔 + (𝑘

1

2

2

𝑔]

2

2

𝑔𝑙(𝑘

1

2

2

2 2

2

𝜔

2

𝑚𝑚 2 𝑔 + 𝑚 2

( 𝑘 1 𝑙 + 𝑘 2 𝑙 + 𝑚 2 𝑔

) ± 𝑚 2 √

( 𝑚𝑔

)

2

• 𝑙

2 ( 𝑘 1 + 𝑘 2

)

2

• (𝑚 2 𝑔

)

2

• 2𝑚𝑔𝑙

( 𝑘 1 + 𝑘 2

) + 2 𝑚 2 𝑔𝑙

( 𝑘 1 + 𝑘 2

) + 2 𝑚𝑚 2 𝑔

2

− 4 𝑚𝑔𝑙

( 𝑘 1 + 𝑘 2

)

=

2 𝑚𝑚 2 𝑙

2

1

2

2

2

2

( 𝑘 1

2

2

2

2

2

1

2

2

2

1

2

2

2

)𝑔 + (k

1

• k

2

2

2

2

• (k

1

• k

2

)𝑙[2(𝑚

2

− 𝑚)𝑔 + (k

1

• k

2

)𝑙]

Note that if applied for part a, it will give the same result.

2

[ ]

Find the natural frequencies of the system shown with m 1

= m, m 2

= 2m, k 1

= k, and k 2

= 2k. Determine

the response of the system when k = 1000 N/m, m = 20 kg, and the initial values of the displacements of

the masses m 1

and m 2

are 1 and -1, respectively. Also, determine the natural modes of the system.

Given:

m 1

= m

m 2

= 2m

k 1

= k

k 2

= 2k

k = 1000 N/m

m = 20 kg

x 1o

x 2o

Equations of Motion :

1 1

1

2

1

2 2

2 2

2 2

2 1

1

1

2

2

2

1

1

1

2

2

2

1

Let:

= Xe

−i𝜔𝑡

2

X

1

e

−i𝜔𝑡

• 3000X

1

e

−i𝜔𝑡

− 2000X

2

e

−i𝜔𝑡

= 0

= i𝜔Xe

−i𝜔𝑡

2

X

2

e

−i𝜔𝑡

+ 2000X

2

e

−i𝜔𝑡

− 2000X

1

e

−i𝜔𝑡

2

Xe

−i𝜔𝑡

(3000 − 20 𝜔

2

)X

1

− 2000X

2

2

)X 2

− 2000X

1

[

2

−2000 ] [

X

1

] =

X

2