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Chapter 7

Vibration

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Vibration : (A) Free vibration of particle : ■ ■

free-vibration : absence of any imposed external forces. (a) Undamped free vibration :

k-spring constant [N/m] The equation of motion :

k m x x=0

F = −kx = mx i.e. mx + kx = 0 ( SHM-simple harmonic motion)

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x + ω x = 0 2 n

or ■

where

k ωn = m

[ rad / s ]

= natural angular frequency of the

■

vibration ■

The solution :

x = A cos(ω nt ) + B sin(ω nt ) = C sin(ω nt + ψ )

■ ■

C-amplitude ψ -phase angle

}

They are determined by initial conditions.

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x

τ

x0

C 0

t

− t' ■

xo = C sin ψ

■

∴

0 = C sin (-ω nt' + ψ )

■

⇒

-ω nt' + ψ = 0

■

∴

ψ = ω nt’

■

τ =2π /ω n=period=1/fn

■

fn-natural frequency (Hz) = ω n/ 2π 4 Jump to first page

■

Simple pendulum : (small oscillation)

sin θ ~ θ θ l mg sin θ

+x m

mg

F = −mg sin θ ~ − mgθ = −mgx / l = mx g ∴ x + x = 0 l g ∴ ωn = l

Exercises on Oscillatory Motion :

(1), (3),5 (4), (6), (7), (9) Jump to first page

(b) Damped free vibration : ( s

The equation of motion :

p r

)

in g

k

m

c (dashpot )

x=0

x

Then or

∑ F = −kx − cx = mx c = viscous damping constant (viscous damping coefficient) =[Ns/m]

mx + cx + kx = 0

x + 2ζ ωn x + ω x = 0 2 n

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c ζ ( zeta ) ≡ 2mω n

So

= viscous damping factor (damping ratio) - (dimensionless) Let the solution : x = A eλ t Then

∴ or

λ + 2ζ ωn λ + ω = 0 2

[ λ = ω [−ζ −

2 n

] −1]

λ1 = ωn −ζ + ζ 2 −1 2

n

ζ

2

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The general solution :

x = A1e λ1t + A2 e λ2t

There are three cases : (1) ζ > 1 (overdamping) : λ 1, λ 2 are distinct real ”-" numbers. x decays to zero. No oscillation. (2) ζ = 1 (critical damping) : λ 1 = λ 2 = -ω

n

; real ”-" numbers.

x decays to zero very fast. 8 No oscillation. Jump to first page

x ζ >1 ζ =1

t ζ <1

(3) ζ < 1 (underdamping) :

x =  A1e + A2 e  iω d t − iω d t −ζ ωn t = A1e + A2 e e i 1−ζ 2 ω n t

−i 1−ζ 2 ω n t

[

= C sin ( ω d t + ψ ) e

[

= Ce 9

−ζ ωn t

]

 × e −ζ ωnt 

−ζ ωn t

] sin(ω t +ψ ) d

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ω d = ω n 1 − ζ 2 = damped natural frequency

where

x x1

t1

2π τd = ωd

τd

x2 t2

t

−ζ ωn t1

x1 Ce ζ ωnτ d = −ζ ωn (t1 +τ d ) = e x2 Ce 10

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The logarithmic decrement δ :

 x1  δ ≡ n  = ζ ωnτ d  x2  2π = ζ ωn ωn 1− ζ 2 =

2π ζ 1−ζ

2

 c   ; ζ < 1 ;  ζ = 2mω n  

This is the way to determine the viscous damping constant c. 11 Jump to first page

(B) Forced vibration of particles : F = FO sin ω t

m

x

p r

in g

x=0

)

(dashpot )

k ( s

c

The equation of motion :

∑ F = −kx − cx + Fo sin ω t = mx

∴ i.e. 12

mx + cx + kx = Fo sin ω t Fo 2 x + 2ζ ωn x + ω n x = sin ω t m Jump to first page

Usually, the applied force is coming from the base of the system. neutral position

m

(dashpot)

x

k

xB

( s p r i n g

c

xB = b sin ω t

)

x=0

∴

mx + cx + k ( x − xB ) = 0

i.e.

kb x + 2ζ ωn x + ω x = sin ω t m 2 n

where Fo = kb 13 Jump to first page

Solution of Damped forced vibration : Fo 2 Equation of motion : x + 2ζ ωn x + ω n x = sin ω t

m

Solution is the sum of a complementary solution and a particular solution

x = xc + x p xc = Ce −ζ ωnt sin(ω d t + ψ ) dies down exponentially and is not important Let : x p = X sin(ωt + φ ) then V (t ) = Xω cos(ωt + φ ) = Vo cos(ωt + φ ) 14 Jump to first page

The solution is :

X =

Fo / k

{ [1 − (ω / ω ) ]

2 2

n

+ [ 2ζ ω/ ω n ]

2ζ ω/ ω n tan φ = 2 1 − (ω / ωn ) X M= = Fo / k 1 − (ω / ω

{[

n

)

]

2 2

2

}

1/ 2

1 + [ 2ζ ω/ ω n ]

2

}

1/ 2

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ζ =0 ζ = 0.1

M

ζ = 0.2

ζ =1

1 1

ω / ωn

d ω 2 2 2ζ ω 2 2 {[1 − ( ) ] + ( ) } =0 dω ωn ωn

ω resonance = 1 - 2ζ 2 ω n 16 Jump to first page

φ π

ζ =0 ζ = 0.2

ζ =1

π/2

0

1

(resonance )

ω / ωn

(1) ω is small, tanφ > 0, φ → 0+, xp in phase with driving force (2) ω is large, tanφ < 0, φ → 0-, φ = π , xp leads the driving force by 90o (3) ω → ω n-, tanφ →+∞, φ → π /2(-) ω → ω n+, tanφ →-∞, φ → π /2(+) 17 Jump to first page

Electric circuit analogy : C

L

R

q Lq + Rq + = E C q = i (current ) q-charge

E = EO sin ω t

L-inductionce C-capacitance R-resistance

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Example

y = sin θ

d l O

θ

τ o = (ky )  = − I oθ

k

y

m

∴

I o = md

(

2

)

2   md θ + k sin θ = 0 2

For small oscillation :

∴

2 k  sin θ ~ θ ⇒ θ + θ =0 2 md  k ω  ω= ⇒ f = = d m 2π 2πd

k m

#

19

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Example

R

R−r

θ r

( R − r )(1 − cos θ )

α ω

vG

G

f

Find the period. RN

mg sin θ

Rolling without slipping Method I : By conservation of energy :

1 2 1 2 mvG + I Gω + mg ( R − r )[1 − cos θ ] =constant 2 2 But 20

vG ω= r

(Rolling without slipping) Jump to first page

and

∴

vG = ( R − r )θ

2 1 1  1 2  ( R − r )    m ( R − r )θ +  mr   θ 2 22  r  + mg ( R − r ) [1 − cosθ ] = constant

[

]

2

Differentiate both side w.r.t. t :

1 2      m( R − r ) θθ + m( R − r ) θθ + mg ( R − r )θ sin θ = 0 2 2 g   ∴ θ+ sin θ = 0 3 R−r 2

For small oscillation : sinθ ~ θ

3( R − r ) T = 2π 2g 21

# Jump to first page

Method II

τ G = I Gα

1 2  1 2   [ R − r ]  rf =  mr ω =  mr   θ 2  2  r  1 f = m( R − r ) θ 2 ∑ Ft = m(aG )t

∴ ∴ Also

− f − mg sin θ = m( R − r )θ

∴

1   m( R − r )θ + m( R − r )θ + mg sin θ = 0 2

i.e.

∴ 22

θ +

2g sin θ = 0 3( R − r ) Jump to first page

Example

A

p

m k

c

m=45 kg k = 35 kN/m c = 1250 N.s/m p = 4000 sin (30 t) Pa A= 50 x 10-3 m2.

Determine the steady-state displacement as a function of time and the max. force transmitted to the base. 23 Jump to first page

k 35 × 10 ωn = = = 27.9 rad / s m 45 c 1250 ζ = = = 0.498 (underdamped ) 2mω n 2 × 45 × 27.9 3

The steady-state amplitude : Fo / k X= 2 2 2 1 − ( ω / ω n ) + [ 2ζ ω/ ω n ]

{[

=

]

}

1/ 2

4000 × 50 × 10 −3 / 35 × 103

{ [1 − ( 30 / 27.9) ]

2 2

= 0.00528 m

+ [ 2 × 0.498 × 30 / 2.79]

2

}

1/ 2

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The phase angle φ :

 2ζ ω/ ω n  φ = tan  2 1 − ( ω / ω n )  −1

 2 × 0.498 × 30 / 27.9  = tan   2  1 − ( 30 / 27.9 )  = 1.716 rad −1

∴

The steady state solution :

xP = X sin(ω t − φ ) = 0.00528 sin(30t − 1.716) m

#

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The force transmitted to the base :

Ftr = kx p + cx p

= kX sin (ω t − φ ) + cωX cos (ω t − φ ) For max Ftr :

dFtr =0 ⇒ d (ω t − φ )

k ω t − φ = tan cω 3 35 × 10 ω t − φ = tan −1 = 0.75 rad = 43o 1250 × 30

∴

( Ftr ) max = 35 ×103 × 0.00528 sin 43o

∴ tan θ = sin −1

θ

−1

= cos

−1

−1

1+θ 2 1

+ 1250 × 30 × 0.00528 cos 43o = 271 N

#

26

1+θ 2

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