HW2, EGME 511 (Advanced Mechanical Vibration) Date due and handed in March 17,2008

HW2, EGME 511 (Advanced Mechanical Vibration)
Date due and handed in March 17,2008

by Nasser M. Abbasi

June 15, 2014

1 Problem 1
2 Problem 2
3 Problem 3
4 Problem 4
5 Problem 5 (not correct, left here to check something)
6 Problem 5 (again, correct solution)
7 Problem 6
8 Solving problem shown in class for Vibration 431, CSUF, Spring 2009
9 Solving problem shown in class for Vibration 431, CSUF, Spring 2009. Version 2

1 Problem 1

Find the equation of motion for the following system

Solution

Assume initial conditions are $x(0)= x0$ and $x_(0) = 0$ . Assume that $x0$ was positive (i.e. to the right of the static equilibrium position, and also assume that $kx0 > N μstatic$ ). This second requirement is needed to enable the mass to undergo motion by overcoming static friction. The normal force $N$ is given by

N = mg cos𝜃

And the dynamic friction force $f c$ due to the dynamic friction is deﬁned as follows

( | |||| − μN x_> 0 ||{ fc = ||| 0 x_= 0 ||| |( μN x_< 0

But since $N = mg cos𝜃$ , then the above becomes

( || ||| − μmg cos𝜃 _x> 0 ||{ fc = 0 _x= 0 ||| |||| ( μmg cos𝜃 _x< 0

(1)

Where $μ$ is the coeﬃcient of dynamic friction. Now we can obtain the Lagrangian

Hence

L = 1m _x2− 1kx2 2 2

and

Then the EQM is

Where $fc$ is given by (1). Since $fc$ sign depends in the mass is moving to the left or to the right, we will generate 2 equation of motions, one for each case.

When mass is moving to the left, EQM 1 is

m¨x+ kx = μmg cos𝜃

(2)

When mass is moving to the right, EQM 2 is

mx¨+ kx = − μmg cos𝜃

(3)

So, for the ﬁrst move, starting from $x 0$ and moving to the left, we have

x= xh+ xp

Guess $x = X p$ , hence $ω2X = μg cos𝜃 n$ or $X = μgcos𝜃- ω2n$ , and $x = Acosω t+ Bsin ω t h n n$ , therefore, the solution to EQM 1 is

μg cos𝜃 x(t)= A cosωnt+ Bsinωnt+ ---ω2--- n

$x(0)= x = A + μgcos𝜃 0 ω2n$ hence $A= x − μgcos𝜃- 0 ω2n$ , then

( ) μg-cos𝜃- μg-cos𝜃- x(t)= x0− ω2 cosωnt +B sin ωnt+ ω2 n n

and

Hence $B = 0$ , then EQM is (for $0 < t < πωn-$ )

( ) μg-cos𝜃- μg-cos𝜃- xleft(t)= x0− ω2 cosωnt+ ω2 n n

(4)

The mass will move according to the above equation (4) until the velocity is zero, then it will turn and start moving to the right. To ﬁnd the time this happens:

( ) μg-cos𝜃- _x(t)= − ωn x0− ω2n sin ωnt

Now solve for $t$ when $x_(t)= 0$ , i.e.,

( μgcos𝜃 ) 0 = − ωn x0− ----2--- sin ωnt ω n

(5)

Hence $ωnt = nπ$ , where $n = 0,±1, ±2,⋅⋅⋅$ The case for $n= 0$ do not apply since this implies $t = 0$ , then consider the next time this can happen, which is $n= 1$ , which implies

t = -π- 1 ωn

(6)

Now we need to determine $x(t)$ at this time $t1$ since this will become the initial $x$ for the second equation of motion going to the right in the second leg of the journey. Using (4) and (6) we obtain

Notice that in the above equation, $x0$ is a positive number, since we assumed that the initial conditions $x0$ was to the right of the static equilibrium position, and we are assume the right of the static equilibrium position to be positive. This also implied that $( π-) x ωn$ will be negative number (which is what we expect, as the mass will by the end of its ﬁrst trip be on the left of the static equilibrium position).

Now we can use right equation of motion (EQM 2) to solve for the mass moving to the right. Notice that the initial conditions for this motion are $2μgcos𝜃- x1 = ω2n − x0$ and $π- t1 = ωn$

The equation of motion is now

With the general solution

μg cos𝜃 x(t)= A cosωnt+ Bsinωnt− ----2--- ωn

(7)

At $t = π,x(t)= 2μgcos𝜃− x ωn ω2n 0$ , hence from the above

Hence (7) becomes

( 3μgcos 𝜃) μg cos𝜃 x(t)= x0− ----2---- cosωnt+ Bsinωnt− ----2--- ωn ωn

And

( 3μgcos𝜃 ) _x(t) = − ωn x0− ----2---- sinωnt +ωnB cosωnt ω n

But $_x(t) = 0$ at $-π t = ωn$ , hence the above becomes

Hence $B = 0$ , then the EQM for the right move is, for $-π < t < 2π ωn ωn$

|-------------------------------------| | ( ) | | xright(t)= x0− 3μgωc2os𝜃-cosωnt − μgcωo2s𝜃- | | n n | ---------------------------------------

This diagram below summarize this

Now, we would like to have one equation to express the motion with for any time instance when the mass is moving to the left, or to the right. Looking at the above 2 equation of motion, we see immediately that we can write the equation of motion as follows

|----------------------------------------------| | ( ) | | xn(t) = x0− (2n−1)μ2gcos𝜃 cosωnt + (− 1)n+1 μgcos2𝜃| | ωn ωn | ------------------------------------------------

Where $n$ above is the number of the trip. So, the ﬁrst trip, going from $x0$ and moving to the left, will have $n = 1$ , and then second trip, moving from $x1$ and going to the right will have $n= 2$ , and so on. As for the time during which trip travels, this is found by the following equation

(n−-1)π-< tn < nπ ωn ωn

What the above is saying is that for ﬁrst trip ( $n = 1$ ), we have

π 0< t < --- ωn

And for the second trip, we have

π-- 2π- ωn < t < ωn

etc...

Now that we have one equation, and we have the time during which each equation is valid, we can now plot the equation of motion vs. time. The following is a plot for some values for $k,g,m$ . Please see the appendix for the Matlab code which generated this simulation.

Observation found on this problem: Changing the angle of inclination $𝜃$ causes no change in results. In other words, the same oscillation will occur for ﬂat plane ( $𝜃 = 0$ ) or for $𝜃 = 450$ or any other angle. The reason is because $x0$ , the initial position, is measured from the static equilibrium position, and this static equilibrium position will be diﬀerent as the angle changes, but the eﬀect of the angle change is already accounted for by this change and will not be reﬂected in the actual displacement $x(t)$ .

2 Problem 2

Given $⌊ ⌋ ⌊ ⌋ ⌊ ⌋ |1 0| | 3 − 1| |0| || || ¨X+ || || X = || || ⌈ ⌉ ⌈ ⌉ ⌈ ⌉ 0 4 − 1 1 0$ , $m :kg,k :N∕m$ , use modal analysis to calculate the solution of this given $⌊ ⌋ ⌊ ⌋ |0| | 0| X (0) = || || mm, X˙(0)= || ||mm ∕ sec ⌈ ⌉ ⌈ ⌉ 1 0$ also calculate the eigenvalues of the system and the normalized eigenvectors.

Answer

Since this is a 2 ODE's that are coupled, we use modal analysis to de-couple the system ﬁrst in order to obtain 2 separate ODE's which we can then solve easily.

Let

$⌊ ⌋ ||1 0|| M = | | ⌈0 4⌉$ and let $⌊ ⌋ || 3 − 1|| K = | | ⌈ − 1 1 ⌉$ , then the above system becomes

M X¨+ KX = 0

(1)

Let $1 X = M −2q$ , then $1 ¨X = M −2 ¨q$ and the above equation becomes

1 1 MM − 2q ¨+ KM −2q = 0

premultiply by $1 M −2$ we obtain

Where $~ − 12 − 12 K = M KM$

Let $iωt q = ve$ , then $2 iωt ¨q= − ω ve$ and (2) becomes

Let $2 λ = ω$ then we have

( ) K~− λI v = 0

(3)

For $v⁄= 0$ , we requires that $||K~− λI||= 0$ But

Hence

|------------------------------------------| | { √--- √--} | | λ1,2 = 13−8137,13+8137- = {0.16191,3.0881} | --------------------------------------------

From (3) we then have

When $λ = λ1 = 0.1619$ we obtain

Hence

Let $a = 1$ , then $b= −2.8381= 5.6762 −0.5$ , hence the second eigenvector is

⌊ ⌋ | 1 | v1 = || || ⌈ ⌉ 5.6762

$---------- ∥v1∥= √1 + 5.67622 = 5.7636$ , hence normalized $v1$ is

When $λ = λ2 = 3.0881$ we obtain

Hence

Let $a = 1$ in the ﬁrst equation above, then $b = −0.0881= − 0.1762 0.5$ , hence the ﬁrst eigenvector is

⌊ ⌋ | 1 | v2 = || || ⌈ − 0.1762⌉

$√---------- ∥v2∥= 1 + 0.17622 = 1.0154$ , hence normalized $v2$ is

Then the $P$ matrix

Now let $q = Pr,$ then equation (2) above becomes

Premultiply by $T P$

Let $Λ = P T ~KP$ then the above becomes

I¨r+ Λ r= 0

(4)

Now ﬁnd $Λ1 1Thiscan alsobe found morequicklyby noting thatΛ = diag(λ1,λ2)$

$\text{[math]}$ Hence (4) becomes

⌊ ⌋ | 0.16191 0 | I¨r+ || || r= 0 ⌈ ⌉ 0 3.0881

Which can be written as 2 equations

⌊ ⌋ ⌊ ⌋ ⌊ ⌋ | ¨r1| |0.16191r1| |0| || || + || || = || || ⌈ ⌉ ⌈ ⌉ ⌈ ⌉ ¨r2 3.0881r2 0

With IC given as

⌊ ⌋ |0| X (0)= || || ⌈ ⌉ 1

and $\text{[math]}$

⌊ ⌋ |0| ˙X (0)= || || ⌈ ⌉ 0

Now $X=M − 12q$ and $q=P r$ , hence $X=M − 12Pr$ , then

now need to ﬁnd $r¨(0),$ but since $¨X (0 )= 0$ , then $¨r(0)= 0$ as well.

Now we can solve for $r1(t)$ and $r2(t)$ since we have the IC. From (5) above

At $t = 0,r(0) = 1.9696 1$ , hence $1.9696 = A$ , then

At $t = 0$

_r1(t) = 0= ωn1B

Hence $B = 0$ , then

r1(t)= 1.9696cos ωn1t

But $√ ------- ωn1 = 0.16191 = 0.40238$ , hence

|----------------------------| | r1(t) = 1.9696 cos(0.40238t) | -----------------------------|

Similarly we ﬁnd $r2(t)$

At $t = 0,r2(0) = − 0.34698$ , hence $− 0.34698 = A$ , then

At $t = 0$

_r2(t) = 0= ωn B 2

Hence $B = 0$ , then

r (t)= − 0.34698 cosω t 2 n2

But $ω = √3.0881-= 1.7573 n2$ , hence

|------------------------------| | | | r2(t) = − 0.34698 cos (1.7573t) | -------------------------------

Now that we found the solution in the $r$ space, we switch back to the original $x$ space

1 X (t)=M −2Pr(t)

Then

⌊ ⌋⌊ ⌋ ⌊ ⌋ ||1 0 |||| 0.1735 0.98483 || || 1.9696 cos(0.40238t) || X(t)= |⌈ |⌉|⌈ |⌉ |⌈ |⌉ 0 0.5 0.98484 − 0.17353 − 0.34698 cos(1.7573t)

$\text{[math]}$ $\text{[math]}$ Hence

|-----------------------------------------------------| | ⌊ ⌋ ⌊ ⌋ | | | | || x1(t)|| || 0.34173cos0.40238t− 0.34172 cos1.7573t|| | | |⌈ |⌉ = |⌈ |⌉ | | x2(t) 0.96987 cos 0.40238t+ 0.030106cos1.7573t | | | ------------------------------------------------------

This is a plot of the solutions

Observation on ﬁnal result: Notice that power of the harmonic $ωn = 1.7573$ rad/sec. in the motion $x2(t)$ is small (amplitude is only $0.03$ ) hence the dominant harmonic present in $x2(t)$ is $ωn = 0.40238$ rad/sec. and this reﬂects in the plot where it appears that $x2(t)$ contain one harmonic. In the case of $x1(t)$ we see from the solution that both frequencies contribute equal amount of power, hence the plot for $x1(t)$ reﬂects this.

3 Problem 3

Solution Use as generalized coordinates $𝜃1,𝜃2$ ,. Assume that the spring remain horizontal, and assume that $𝜃2 > 𝜃1$

Hence

1 ( ) 1 ( ) ( 1 ) L = -m1 L_𝜃1 2+ -m2 L_𝜃2 2− m1gL (1− cos𝜃1)+ m2gL(1 − cos 𝜃2) + -k(asin𝜃2− asin𝜃1)2 2 2 2

Now determine the Lagrangian equation

Hence the EQM for $m1$ is

Now apply small angle approximation. $sin 𝜃 ≈ 𝜃$ and $cos𝜃 ≈ 1$ hence

( ) m1L2 ¨𝜃1+ m1gL+ a2k 𝜃1− a2k𝜃2 = 0

(1)

And the EQM for $m 2$ is

Now apply small angle approximation. $sin 𝜃 ≈ 𝜃$ and $cos𝜃 ≈ 1$ hence

Therefore

( ) m2L2 ¨𝜃2+ 𝜃2 m2gL+ a2k − a2k𝜃1 = 0

Now we write the system as $¨ M 𝜃 + K𝜃 = 0$

⌊ ⌋⌊ ⌋ ⌊ ⌋ ⌊ ⌋ ⌊ ⌋ |m1L2 0 || ¨𝜃1| |m1gL + a2k − a2k | |𝜃1| | 0| || |||| || + || || || || = || || ⌈ 2⌉⌈ ¨⌉ ⌈ 2 2⌉ ⌈ ⌉ ⌈ ⌉ 0 m2L 𝜃2 − ak m2gL + ak 𝜃2 0

Substitute numerical values for the above quantities, we obtain

The above can be written as

M ¨𝜃+K 𝜃 = 0

Let $1 𝜃 = M −2q$ , then $1 𝜃¨= M −2 ¨q$ and the above equation becomes

1 1 MM −2𝜃¨+ KM −2𝜃 = 0

premultiply by $1 M −2$ we obtain

Where $− 1 − 1 K~= M 2KM 2$

Let $q = veiωt$ , then $¨q= − ω2veiωt$ and (2) becomes

Let $2 λ = ω$ then we have

( ) K~− λI v = 0

(3)

For $v⁄= 0$ , we requires that $|| ~ || K − λI = 0$ But

Hence

Hence the characteristic equation is

|--------------------------| | 2 | | λ − 39.36 λ + 387.30 = 0 | ---------------------------

Hence

|------------------| | | | λ1,2 = 19.6,19.76 | -------------------

Hence the natural frequencies are

From (3) we then have

When $λ = λ1 = 19.6$ we obtain

Hence

Hence $a = b$ then

⌊ ⌋ ⌊ ⌋ | 1| |0.70711| v1 = √1-|| || = || || 2⌈ ⌉ ⌈ ⌉ 1 0.70711

When $λ = λ = 19.76 2$ we obtain

Hence $a = − b$ , then

⌊ ⌋ ⌊ ⌋ -1- ||− 1|| ||− 0.70711 || v2 = √2-|⌈ |⌉ = |⌈ |⌉ 1 0.70711

Now that we have obtained the eigenvectors of the de-coupled system, we can plot the mode shapes² . I will use a diagram similar to that shown in the textbook Engineering Vibration by Inman on page 313)

$\text{[math]}$

4 Problem 4

M ¨x+K x = 0

Where $⌊ ⌋ ⌊ ⌋ ||27 − 3|| || 9 0 || K = |⌈ |⌉ ,M = |⌈ |⌉ − 3 3 0 1$ Let $− 1 X = M 2q$ , then $− 1 X¨= M 2q ¨$ and the above equation becomes

− 1 − 1 MM 2q ¨+ KM 2q = 0

premultiply by $− 1 M 2$ we obtain

Where $~ − 1 − 1 K = M 2KM 2$

Let $iωt q = ve$ , then $2 iωt ¨q= − ω ve$ and (2) becomes

Let $λ = ω2$ then we have

(~ ) K − λI v = 0

(3)

For $v⁄= 0$ , we requires that $| | |K~− λI|= 0$ But

Hence

Hence the characteristic equation is

|----------------| | 2 | | λ − 6λ + 8= 0 | ------------------

Hence

|--------------| | | | λ1,2 = {2,4} | ---------------

Then the natural frequencies are

{√ -- } ωn = 2,2

From (3) we then have

When $λ = λ1 = 2$ we obtain

Hence

Then $a= b$ , hence

⌊ ⌋ ⌊ ⌋ | 1| |0.70711| v1 = √1-|| || = || || 2⌈ ⌉ ⌈ ⌉ 1 0.70711

When $λ = λ = 4 2$ we obtain

Hence $a = − b$ , then

⌊ ⌋ ⌊ ⌋ -1- || 1 || || 0.70711 || v2 = √2-|⌈ |⌉ = |⌈ |⌉ − 1 − 0.70711

Then the matrix

Now let $q = Pr,$ then equation (2) above becomes

Premultiply by $PT$

Let $T Λ = P ~KP$ then the above becomes

I¨r+ Λ r= 0

(4)

Now ﬁnd $Λ$

Hence (4) becomes

⌊ ⌋ ||2 0|| Ir¨+ | | r= 0 ⌈0 4⌉

Which can be written as 2 equations

⌊ ⌋ ⌊ ⌋ ⌊ ⌋ || ¨r1|| ||2r1|| ||0|| |⌈ |⌉ + |⌈ |⌉ = |⌈ |⌉ ¨r2 4r2 0

With IC given as $⌊ ⌋ ⌊ ⌋ | 13| |0| X(0)= 1√2|| ||, ˙X(0)= || || ⌈ ⌉ ⌈ ⌉ 1 0$ , but

$X =M − 12q$ and $q =Pr$ , hence $X =M − 12Pr$ , then

And since $˙ X (0)= 0$ , then $˙r(0) = 0$ , now we have found IC for $r (t)$ we can solve the ODEs

$r1(0) = 1$ hence $A1 = 1$ , and $B1 = 0$ , similarly, $A2 = 0$ , and $B1 = 0$ , hence

But

X (t)=M − 12Pr(t)

Then

Hence

|----------------------------| | ⌊ ⌋ ⌊ ⌋ | | ( √--) | | || x1(t)|| ||31√2 cos 2t || | | |⌈ |⌉ = |⌈ ( )|⌉ | | x2(t) 1√- cos√2t- | | 2 | -----------------------------

Here is a plot of the solution

5 Problem 5 (not correct, left here to check something)

$m=3,c= 6,k = 12$ , hence $∘ k- ∘ 12 ωn = m = 3 = 2$ rad/sec and $-c -c-- --6-- 1 ξ = ccr = 2ωnm = 2×2×3 = 2$ , hence the system is underdamped and $∘ ------ ∘ -----2 12 √ -- ωd = ωn 1 − ξ = 2 1 − 2 = 3$ rad/sec

Let the response to $3δ (t)$ be $xp1(t)$ and let the response to $δ (t− 1)$ be $xp2(t)$ hence the response of the system becomes

x(t)= x (t)+ x (t)− x (t) h p1 p2

(1)

Where

xh = e−ξωnt(Acosωdt+ B sin ωdt)

(2)

And

3 −ξωnt xp1(t) = mω--e sin ωdt d

(3)

and

--1- −ξωn(t−1) xp2(t)= m ωde sin ωd(t− 1)Φ (t− 1)

Hence, substitute (2),(3) into (1)

Now using IC to ﬁnd $A,B$ . Note, we use only $x(t)= xh(t) +xp1(t)$ for the purpose of ﬁnding $A,B$ from I.C's since the response to the delayed impulse is not active at $t = 0$ . We ﬁnd

x (0 )= -1--= A 100

And for the derivative

Hence

B = ---1√--- 100 3

Therefore the solution is, by substituting values found for $A,B$ into the general solution from above equation (4), we obtain

−t( √ -- √ --) √ -- ( √ -- ) x(t)= -e-- cos 3t+ ---1√--sin 3t + √1-e−tsin 3t− -1√--e−(t−1)sin 3(t− 1)Φ (t− 1) 100 100 3 3 3 3

(5)

The following is a plot of the solution for up to $t = 6$

6 Problem 5 (again, correct solution)

Let the response to $3δ (t)$ be $xp1(t)$ and let the response to $δ (t− 1)$ be $xp2(t)$ hence the response of the system becomes

x(t)= x (t)+ x (t)− x (t) h p1 p2

(1)

Where

xh = e−ξωnt(Acosωdt+ B sin ωdt)

(2)

And

3 −ξωnt xp1(t) = mω--e sin ωdt d

(3)

and

--1- −ξωn(t−1) xp2(t)= m ωde sin ωd(t− 1)Φ (t− 1)

To ﬁnd $A,B$ use only $xh(t).$ At $t = 0$ . We ﬁnd

x (0 )= -1--= A 100

And for the derivative

Hence

Therefore the solution is, by substituting values found for $A,B$ into the general solution from above equation (4), we obtain

e−t( √ -- 101 √ --) 1 √ -- ( 1 √ -- ) x(t)= ---- cos 3t+ ----√--sin 3t + √--e−tsin 3t− -√--e−(t−1)sin 3(t− 1)Φ (t− 1) 100 100 3 3 3 3

(5)

The following is a plot of the solution for up to $t = 6$

7 Problem 6

Let the response by $x(t)$ . Hence $x(t)= x (t)+ x (t) h p$ , where $x (t) p$ is the particular solution, which is the response due the the above forcing function. Using convolution

∫t xp(t)= f(τ)h (t− τ)d τ 0

Where $h (t)$ is the unit impulse response of a second order underdamped system which is

--1- −ξωnt h(t)= m ωde sinωdt

hence

Using $1 sin Asin B= 2[cos (A − B)− cos(A +B )]$ then

1 sin(τ)sin(ωd(t− τ))= --[cos(τ − ωd(t− τ))− cos(τ+ ωd(t− τ))] 2

Then the integral becomes

( ) −ξωnt ∫t ∫t xp (t)= F0e-----( eξωnτcos(τ− ωd (t− τ))dτ − eξωnτcos(τ+ ωd (t− τ))dτ) 2mωd 0 0

Consider the ﬁrst integral $I 1$ where

∫t I = eξωnτcos(τ− ω (t− τ)) dτ 1 d 0

Integrate by parts, where $∫ ∫ udv = uv− vdu$ , Let $ξωnτ eξωnτ- dv = e → v= ξωn$ and let $u = cos(τ− ωd (t− τ))→ du = − (1 + ωd)sin (τ − ωd(t− τ))$ , hence

Integrate by parts again the last integral above, where $∫ ∫ udv= uv− vdu$ , Let $ξω τ eξωnτ dv= e n → v= -ξωn$ and let $u = sin(τ − ωd(t− τ))→ du = (1+ ωd )cos(τ − ωd(t− τ))$ , hence

Substitute (2) into (1) we obtain

Hence

Now consider the second integral $I2$ where

∫t ξωnτ I2 = e cos(τ+ ωd (t− τ)) dτ 0

Integrate by parts, where $∫ ∫ udv = uv− vdu$ , Let $ξωnτ eξωnτ- dv = e → v= ξωn$ and let $u = cos(τ+ ωd (t− τ))→ du = − (1 − ωd)sin (τ + ωd(t− τ))$ , hence

Integrate by parts again the last integral above, where $∫ ∫ udv= uv− vdu$ , Let $ξω τ eξωnτ dv= e n → v= -ξωn$ and let $u = sin(τ + ωd(t− τ))→ du = (1− ωd )cos(τ + ωd(t− τ))$ , hence

Substitute (4) into (3) we obtain

Hence

Using the above expressions for $I ,I 1 2$ , we ﬁnd (and multiplying the solution by $(Φ(t)− Φ (t− π ))$ since the force is only active from $t = 0$ to $t = π$ , we obtain

Hence $xp(t)= (Φ (t)− Φ (t− π))$

$[ ( ξωnt ξωnt ξωnt ξωnt )] F0e−ξωnt ξωn[cos(t)e--−cos(ωdt)]2+(1+ωd)[si2n(t)e--+sin(ωdt)]− ξωn[cos(t)e −cos(ωdt)]+(1−ωd)[sin(t)e +sin(ωdt) 2mωd (ξωn)+(1+ωd) (ξωn)+(1−ωd)$

And

xh(t)= e−ξωnt(Acosωdt + Bsinωdt)

Hence the overall solution is

x (t)= e−ξωnt(A cosωdt+ B sinωdt)+ xp(t)

The above solution is a bit long due to integration by parts. I will not solve the same problem using Laplace transformation method. The diﬀerential equation is

x¨(t)+ 2ξω _x(t)+ ω2 x(t) = f(t) n n

Take Laplace transform, we obtain (assuming $x(0)= x0$ and $_x(0)= v0$ )

Now we ﬁnd Laplace transform of $f(t)$

Integration by parts gives

[1 + e−πs] F (s)= F0 -----2- 1 + s

(8)

Substitute (8) into (7) we obtain

Hence

Now we can use inverse Laplace transform on the above. It is easier to do partial fraction decomposition and use tables. I used CAS to do this and this is the result. I plot the solution $x(t)$ . I used the following values to be able to obtain a plot $ξ = 0.5,ωn = 2,F0 = 10,x0 = 1,v0 = 0$

8 Solving problem shown in class for Vibration 431, CSUF, Spring 2009

Problem

Solve $x¨+2 _x+ 4x= δ (t)− δ (t− 4 )$ with the IC's $x (0) = 1mm, _x(0)= − 1mm$

Answer

$m = 1,c= 2,k = 4$ , hence $∘ -- √ -- ωn = km = 4 = 2$ rad/sec and $ξ = cccr = 2cωnm-= 2×22×1 = 12$ , hence the system is underdamped and $∘ ------ ∘ ----12 √ -- ωd = ωn 1 − ξ2 = 2 1 − 2 = 3$ rad/sec

Let the response to $δ (t)$ be $x1(t)$ and let the response to $δ(t− 4)$ be $x2(t)$ hence the response of the system becomes

x(t) = xh(t)+ x1(t)− x2(t)

(1)

Where

−ξωnt xh = e (Acosωdt+ B sin ωdt)

(1)

And

-1--− ξωnt x1(t)= mωd e sinωdt

(3)

and

x2(t)= -1--e−ξωn(t−4)sinωd (t− 4)Φ(t− 4) mωd

Hence, substitute (2),(3) ,(4) into (1)

x(t)= e−ξωnt(Acosωdt + Bsin ωdt)+ -1--e−ξωntsin ωdt− -1--e−ξωn(t−4)sin (ωd (t− 4))Φ(t− 4) mωd m ωd

(4)

Now using IC to ﬁnd $A,B$

x(0)= 1 = A

and

At $t = 0, _x(0)= − 1,$ Hence the above becomes (terms with $δ (t− 4)$ and $Φ (t− 4)$ vanish at $t = 0$ by deﬁnition)

Hence (1) becomes

( ) −ξωnt -1- -1-- −ξωnt -1-- −ξωn(t−4) x(t)= e cosωdt− √3-sinωdt + mωd e sin ωdt− m ωde sin (ωd (t− 4)) Φ(t− 4)

If we substitute the numerical values for the problem parameters, the above becomes

Compare the above with the solution given in class, which is

|--------------------------------------------------------| | ( √ -- √--) (√ -- ) | x(t) = | e−t cos 3t+ 1√-sin 3t − 1√-e−(t−4)sin 3(t− 4) Φ(t− 4) | | 3 3 | ----------------------------------------------------------

9 Solving problem shown in class for Vibration 431, CSUF, Spring 2009. Version 2

Problem

Solve $x¨+2 _x+ 4x= δ (t)− δ (t− 4 )$ with the IC's $x (0) = 1mm, _x(0)= − 1mm$

Answer

Let the response to $δ (t)$ be $xp1(t)$ and let the response to $δ (t− 4)$ be $xp2(t)$ hence the response of the system becomes

x(t)= xh(t)+ xp1(t)− xp2(t)

(1)

Where

−ξωnt xh = e (Acosωdt+ B sin ωdt)

(1)

And

-1-- −ξωnt xp1(t) = mωd e sin ωdt

(3)

and

1 xp2(t)= ----e−ξωn(t−4)sin ωd(t− 4)Φ (t− 4) m ωd

Hence, substitute (2),(3) ,(4) into (1)

x(t)= e−ξωnt(Acosωdt + Bsin ωdt)+ -1--e−ξωntsin ωdt− -1--e−ξωn(t−4)sin (ωd (t− 4))Φ(t− 4) mωd m ωd

(4)

Now using IC to ﬁnd $A,B$

x(0)= 1

Hence

|--------| | A = 1 | ---------|

Now take the derivative of the above and evaluate at zero to ﬁnd $B$ . In doing so, we need to consider only the $xh$ . The reason is that the particular solution $xp2(t)$ of the delayed pulse (the second pulse) will have no eﬀect at $t = 0$ and the ﬁrst pulse particular solution $xp1(t)$ will also have no contribution, since its response is assume to occur at $0+$ , i.e. an inﬁtismal time after $t = 0$ . Therefore, since we intend to evaluate $_x(t)$ at $t = 0$ , we only need to take $xh$ derivative at this point