Laplace Transforms Homework: Initial Displacements & Velocities

[jake96]

Homework Statement

Homework Equations

Laplace Transforms

The Attempt at a Solution

Using basic physics knowledge I got
m₁a₁=-k₁x₁+k₂(x₂-x₁)
and
m₂a₂=-k₃x₂-k₂(x₂-x₁)

Sub in values and use laplace transforms and rearrange partial fraction and I found that

By doing this I am assuming that the x^II and x^III will equal 0 when t=0 because this is not stated at all in the question. do you believe this is correct?


x₁=-(1/3)cos2t-(2/3)sin2t+(4/3)cost+(8/3)sint
x₂=-(2/3)cos2t+(1/3)sin2t+(8/3)cost-(4/3)sint

Finding the initial displacements by subbing in t=0 for both _{x_1} and _{x_2} comes out with what is written in the question 1 and 2, respectively.

However, when I try to sub t=0 into the differentials of the 2 above equations. I believe I should receive the initial velocities stated in the question. however I do not receive these results.

I receive 4/3 for x₁ and -2/3 for x₂
the difference between these values and the actual values appears to differ once I differentiate the sin2t and it is multiplied by 2

Does anyone know if I should receive the values listed in the question using this methods and have just made a calculation error in my working earlier on, or should I have done something differently.

Also, the next part of the questions asks to use a matrix and eigenvalues/eigenvectors to solve it. any pointers to help me get started

Thanks very much

 

[Chestermiller]

It doesn't seem like you used Laplace Transforms to solve this. What is the Laplace Transform of x₁''(t) in terms of s, x₁(0), and x₁'(0)?

Chet

 

[jake96]

I got (s³+2s²+5s+10)/((s²+4)(s²+1))

I found this by rearranging my equation earlier to find x₂ and x^II₂ in terms of x₁

This came out as x^IV₁+5x^II₁+4x=0, then used laplace transform to find what I had above.

Does this seem correct?

 

[jake96]

I then split the equation into partial fractions and it worked out as the sin and cos equations earlier after I used the inverse laplace

 

[Chestermiller]

I then split the equation into partial fractions and it worked out as the sin and cos equations earlier after I used the inverse laplace

Try it by first taking the Laplace Transform of each equation, and then doing the algebra.

Chet

 

[jake96]

ok, ill post how I go soon

 

[jake96]

ok, I just did it for x₂ quickly and got (2s³-s²+7s-1)/((s²+4)(s²+1))
which resulted in (1/3)cos2t-sin2t+(5/3)cost

this also gives me the displacement as 2 which is correct, however the velocity is -2 instead of -1

ill check with x₁ now

 

[jake96]

again with x₁, displacement came out as the same (1) but velocity was doubled at 4 instead of 2.
I've noticed the displacement always equals what is infront of the s³. looks like this is because the x(0)= whatever is the only factor to affect the s³ value causing the displacement to be equal each time

 

[Chestermiller]

again with x₁, displacement came out as the same (1) but velocity was doubled at 4 instead of 2.
I've noticed the displacement always equals what is infront of the s³. looks like this is because the x(0)= whatever is the only factor to affect the s³ value causing the displacement to be equal each time

It seems to me that this is just a matter of getting the algebra correct, and then properly inverting the transforms.

Chet

 

[vela]

It would help if you at least showed the initial equations you get from taking the Laplace transform. For all we know, you made an error there and propagated it the rest of the way.

 

[jake96]

I started with x₁^II+3x₁-2x₂=0
and x₂^II+2x₂-x₁=0

 

[jake96]

I rearranged those to get x₂^IV+5x₂^''+4x₂=0

from laplace I got this for x₂
s⁴X₂-s³x₂(0)-s²x₂^'(0)-sx₂^''(0)-x₂^'''+5(s²X₂-sx₂(0)-x₂^'(0))+4X₂

From there I subbed in x₂(0)=2 and x₂^'(0)=-1, and assumed that the second and third derivatives equal 0

Then simplified that to X₂= (2s³-s²+10s-5)/((s²+4)(s²+1))

Used partial fractions and inverse laplace to get what I had for x₂ in the original question

 

[Chestermiller]

I started with x₁^II+3x₁-2x₂=0
and x₂^II+2x₂-x₁=0

Good. Now let's see the laplace transform of each equation.

Chet

 

[jake96]

Good. Now let's see the laplace transform of each equation.

Chet

Then doing the laplace before algebra I got
s²X₁-s-2+3X₁-2X₂=0
s²X₂-2s+1+2X₂-X₁=0

these included the inclusion of the initial x and x^' values

then I simplified both equations to find X₂ which I found as
(2s³-s²+7s-1)/((s²+4)*(s²+1))

 

[jake96]

Then doing the laplace before algebra I got
s²X₁-s-2+3X₁-2X₂=0
s²X₂-2s+1+2X₂-X₁=0

these included the inclusion of the initial x and x^' values

then I simplified both equations to find X₂ which I found as
(2s³-s²+7s-1)/((s²+4)*(s²+1))

that gave me x₂=(1/3)cos2t-sin2t+(5/3)cost

 

[vela]

I got a factor of 1/2 on the sin 2t term.

 

[Chestermiller]

I got a factor of 1/2 on the sin 2t term.

I got a factor of 1/3 on the sin 2t term.

Chet

 

[jake96]

I got a factor of 1/3 on the sin 2t term.

Chet

Did you get similar factors for the other terms?
Do you know where our work differs to result in this?

 

[vela]

I got a factor of 1/3 on the sin 2t term.

I cranked it out in Mathematica. There was a factor of 1/3 on the cosine term, but on the sine term, it was 1/2.

Did you get similar factors for the other terms?
Do you know where our work differs to result in this?

I got the same $X_2(s)$ as you did, so the problem is in your inversion.

 

[jake96]

From here
(2s³-s²+7s-1)/((s²+4)*(s²+1))

I used partial fraction expransions and found that 2s³-s²+7s-1=(A+C)s³+(B+D)s²+(A+4C)s+(B+4D)

that gave me A+C=2, B+D=-1, A+4C=7 and B+4D=-1

therefore A=(1/3), B=-1, C=(5/3) and D=0

 

[jake96]

I cranked it out in Mathematica. There was a factor of 1/3 on the cosine term, but on the sine term, it was 1/2.


I got the same $X_2(s)$ as you did, so the problem is in your inversion.

the only inversion I used was for s/(s²+a) and 1/(s²+a)
which go to sin and cos

 

[jake96]

From here
(2s³-s²+7s-1)/((s²+4)*(s²+1))

I used partial fraction expransions and found that 2s³-s²+7s-1=(A+C)s³+(B+D)s²+(A+4C)s+(B+4D)

that gave me A+C=2, B+D=-1, A+4C=7 and B+4D=-1

therefore A=(1/3), B=-1, C=(5/3) and D=0

sub those into A(s/(s²+4))+b(1/(s²+4))+C(s/(s²+1))+D(1/(s²+1))
and that dwindles down to (1/3)(s/(s²+4))-(1/(s²+4))+(5/3)(s/(s²+1))

then the inverses I mentioned above comes up with x2=(1/3)cos2t-sin2t+(5/3)cost

Are these Inversions correct?

 

[Chestermiller]

I cranked it out in Mathematica. There was a factor of 1/3 on the cosine term, but on the sine term, it was 1/2.

Yes. I made an algebra error in the last step. When I corrected it, I got your result. I think the mistake that Jake made was inverting $-\frac{1}{s^2+4}$. The inverse of this is $-\frac{1}{2}\sin 2t$, not -sin 2t.

Chet

 

[jake96]

Yes. I made an algebra error in the last step. When I corrected it, I got your result. I think the mistake that Jake made was inverting $-\frac{1}{s^2+4}$. The inverse of this is $-\frac{1}{2}\sin 2t$, not -sin 2t.

Chet

ahhh, thankyou very much. that would fix the problem I've had with all the answers I've received because the sin2t has always been double what it should be.

it appears that the method I use comes up with multiple answers that all equal the correct displacement and velocities.
Is this correct that there will be multiple results?

 

[Chestermiller]

ahhh, thankyou very much. that would fix the problem I've had with all the answers I've received because the sin2t has always been double what it should be.

it appears that the method I use comes up with multiple answers that all equal the correct displacement and velocities.
Is this correct that there will be multiple results?

No. There is only one answer. It's a linear problem.

Chet

 

[jake96]

No. There is only one answer. It's a linear problem.

Chet

so the answers we have been talking about should be the correct ones (laplace then algebra) cos the first time I used algebra before laplace and that resulted in x^''(0) and x^'''(0) values which are not stated

 

[jake96]

Yes. I made an algebra error in the last step. When I corrected it, I got your result. I think the mistake that Jake made was inverting $-\frac{1}{s^2+4}$. The inverse of this is $-\frac{1}{2}\sin 2t$, not -sin 2t.

Chet

I see my error now, I had 1 over instead of 2 over which will cause the factor to be 1/2.

thanks very much

 

[jake96]

Would you have any pointers to help me get started on doing this problem with a matrix using eigenvalues and eigenvectors

I can't seem to see where the matrix will come in for this question. I'm confident I can solve the eigenvalues and vectors once I've gotten out the matrix.

Thanks again

 

[vela]

so the answers we have been talking about should be the correct ones (laplace then algebra) cos the first time I used algebra before laplace and that resulted in x^''(0) and x^'''(0) values which are not stated

Now that you know the correct answers, you can actually calculate those derivatives at t=0, and you'll see that you can't justify setting them to 0 when solving the fourth-order DE using Laplace.

 

[Chestermiller]

OK Jake,

Suppose you had a set of 4 coupled first order linear homogeneous ODEs in 4 unknowns, y₁,...,y₄. Would you know how to obtain the complementary solution to that set of equations using eigenvalues and eigenvectors?

Chet