7. Direct Solution of Resonance Condition

my + ky Fcoswt ;

"

2

F

y" + w²y = coswt, forcing function at frequency of w.

m

y = y + y₂ = C₁sinwt + C₂coswt + Y #(30)

Р

Use Variation of Parameters to obtain Y.

Formula Solution (See B/DP for derivation)

Y

//*

= natural frequency = w#(27)

==

y₁ S

y₂coswtdt

Wy₁v₂

+

S

Y₂

y,coswtdt

-#(47)

11/₂2

- sinut coswtcoswtdt

wy,₂

W|y₁v₂| = |V₁ V₂ V₁¹ V₁₂ | = v₁y₁₂ — y₂y₁#(49)

+ coswt f.

sinwtcoswtdt

w|y

-#(48)

= (sinwt) (- wsinwt) - coswt (wcoswt)

=- wwt + wt] = - w # (50)

Note: Wy₁, y₂ #0 for y, y to be independent solutions (not proportional to each other.)

(29)/nENGG 614 2023

Ninth Class 10-31-23

Revised of 10-30-23, 8:30pm

24/4/1

Y

F/m

But I sinwtcoswtdt →

Y

=

sinut fwtdt

W

sinut

W

sinut

6

S

But sin2wt=2sinotcoswt

+

udu

W

Ftsinuat

2mw

+

Ftsinut

2mw

costat

W

2

cos wt =

1+cos2wat

2

sin2wt

4ய

2w

But

2)dt.

4 ய

f sinwtcoswtdt # (51)

utorint

2²

1+cos2wat

2

cosut

(sinust) (sin2ust)

2

So

-;fsinwtcoswtdt =

m

cont

cosut

W

-f sinwtcoswtdt#(52)

So

y(t) = C₁y₁ +₂₂

- sinwtcoswtdt

+

wit

2w

costsin wat

2w

Ftsinust

2m

-#(54)

Equation 54 is identical to equation (38) as before for final value of y(t) when 0→w. But we still

need to evaluate, C₁, C₂ for rest conditions, i.e. for y₁ = y₁ = 0

Ftsinat

2mw

772

#(53)

-#(55)

Page 17 of 22/nENGG 614 2023

Ninth Class 10-31-23

Revised of 10-30-23, 8:30pm

Let y(0) = y(0) = 0

y'(t) =

y(t) = C,sinat + C₂coswt +

y(0) = 0 = C₂.0 + C₂1 +

C₂ = 0#(57)

wC coswt - C₂wsinwt +

0 = wC₁ - 0 +

2mw

F

2mw

y(t) =

C₁ = 0#(59)

Ftsinuat

2mw

So the complete solution for system initially at rest is:

Ftsinust

2mw

-#(56)

F.0.0

2mw

-(0 + w.0.1)

-(sinwt + wtcoswt)

As before by undetermined coefficients and use of L'Hopital's Rule (see equation (38) page 13.)

Note: If you generate arbitrary constants of integration when evaluating the integrals in the

Variation of Parameters formula, you will get the complete general solution; i.e.,

C₁y₁ + C₂y₂ + Y, not just the Y. But we already know y, so no need to repeat it.

8. Damped Free Vibration, 80

my + 8y + ky = 0, y(0) = yo

↓

initial displacement

y + (y + (y = 0#(4)

y (0) = y,# (3)

↓

initial velocity

Page 18 of 22/nDimensions of

4km

8

(MT²) (M)

(+)**

Case I: If 4km/8² <1, discriminant is positive and < 1, so both roots are negative.

Case II: If 4km/8² = 1, discriminant is zero and roots 2m

y(t)=r₁c₁e¹² +r₂C₂¹²²

y₁ = C₁ + C₁₂; C₁ = y₁ - C₂

1

y₁ = r₁₂C₁ + r₂C₂

11

=r₁(y₁ - C₂) +r₂C₂

--√√

2m

=(√₂-T₁) C₂+T₁Y₁

+r

0

=

Case III: If 4km/8² > 1, discriminant is negative and roots are complex conjugates

Case I: roots, real & unequal, both negative

=

= dimensionless

1

4km

7₂-7₂

Particular Solution in terms of ICs on y and y':

y₁(t₂-₁)-v₁+ry_ryo-yo

#(63)

-#(64)