Damped spring differental equation

[jblakes]

Afternoon All

I have a math question I don't actually have a clue what to do. Can some help me out.

A mass M is suspended vertically by a damped spring of length L and stiffness k such that the distance x between the centre of the mass and the top of the springis given by

M (d^2 x)/(dt^2 )=Mg-k(x-L)-c dx/dt



Given that M=0.2 kg, k=25 N m^(-1), c=4.5, L=0.40 m, x(0)=0.8 m and dx/dt (0)=0 m s^(-1).

Use the particular integral approach to determine how x varies as a function of t.

State which parameter(s) control(s) the level of damping in the system.

Now consider the ODE describing a forced horizontal mass-spring oscillator where the mass slides over a frictionless surface:

M (d^2 x)/(dt^2 )=0.2cos⁡(4t)-kx
Here x measures how far the mass is to the right of the rest position of the system whileM and k are the same as in part a). Given the initial conditions

x(0)=0 m ; dx/dt (0)=-0.5 m s^(-1)
determine how x varies as a function of t and describe this behaviour in physical terms.

Thanks In advance

James

 

[MarkFL]

Hello, James! Welcome to MHB! (Wave)

Let's look at the first ODE:

\(\displaystyle m\frac{d^2x}{dt^2}=mg-k(x-L)-c\d{x}{t}\)

The first thing we want to do is write this equation in standard linear form:

\(\displaystyle \frac{d^2x}{dt^2}+\frac{c}{m}\d{x}{t}+\frac{k}{m}x=g+\frac{kL}{m}\)

Can you determine the roots of the corresponding characteristic (auxiliary) equation?

 

[jblakes]

Hi Mark.

Thanks for you reply. I don't really now what to do. I'm guessing it goes it the form.
y"+22.5y+125=0? Dont know what to do with the g+(kl) /m side of the equation

Regards
James.

 

[MarkFL]

Well, if you are going to replace the parameters with the given values so that your solution applies only to the given system, then the corresponding homogeneous equation would be:

\(\displaystyle x''+22.5x'+125x=0\)

And so the characteristic equation is:

\(\displaystyle r^2+22.5r+125=0\)

So, we first want to determine the roots of this equation so we can determine the homogeneous solution, and then we will use the fact that expression on the RHS of the ODE in linear form is a constant to determine the form the particular solution must take, since we want this particular solution to be linearly independent from all of the terms in the homogeneous solution.

So, what are the roots of the characteristic equation?

 

[jblakes]

Well I can't complete the square so assume I use quadratic equations so I get -10 and -12.5. Is that correct? Where do I go from here?

 

[MarkFL]

Well I can't complete the square so assume I use quadratic equations so I get -10 and -12.5. Is that correct? Where do I go from here?

You could complete the square as well as factor, but you have the correct roots:

\(\displaystyle r=-\frac{25}{2},\,-10\)

And so from these roots, we know the homogeneous solution must therefore be:

\(\displaystyle x_h(t)=c_1e^{-\frac{25}{2}t}+c_2e^{-10t}\)

Now, if we are going to use the method of undetermined coefficients to determine the form of the particular solution $x_p(t)$, what must this form be?

 

[jblakes]

Looking at the undetermined coefficient table then xp(t)=A. I'm guessing that's not what you mean?

 

[MarkFL]

Looking at the undetermined coefficient table then xp(t)=A. I'm guessing that's not what you mean?

Yes, that's exactly what I meant...since none of the characteristic roots are zero (which would mean one of the terms in the homogeneous solution would be a constant), a constant is therefore linearly independent. And so we may state:

\(\displaystyle x_p(t)=A\)

Substituting this solution into our ODE, what must the particular solution then be?

 

[jblakes]

=-3.2e^(-25/2)t+4e^-10t

I hope. I subbed x(0)=0.8 and dx/dt=0

 

[MarkFL]

=-3.2e^(-25/2)t+4e^-10t

I hope. I subbed x(0)=0.8 and dx/dt=0

Your initial condition are to be used to determine the parameters of the homogeneous solution, once we have the complete general solution. What you want to do is begin with:

\(\displaystyle x_p(t)=A\)

And then compute the first and second time derivatives and then substitute those into the ODE. Since the particular solution is a constant, we then find:

\(\displaystyle x_p'(t)=0\)

\(\displaystyle x_p''(t)=0\)

Okay, now we have (assuming \(\displaystyle g=9.8\frac{\text{m}}{\text{s}^2}\)):

\(\displaystyle 0+22.5(0)+125A=\frac{299}{5}\)

Hence:

\(\displaystyle x_p(t)=A=\frac{299}{625}\)

And then by the principle of superposition, we may state:

\(\displaystyle x(t)=x_h(t)+x_p(t)=c_1e^{-\frac{25}{2}t}+c_2e^{-10t}+\frac{299}{625}\)

Now once you differentiate this to get $x'(t)$, you may use the initial values to determine the parameters.

 

[jblakes]

So I'm guessing I do the step I did become
Subbing x(0)=0.8 x'(0)=0
So therefore
0.8=c1+c2+ 299/625
-c1+0.3216=c2
X'(0)=0= (-25/2)c1+(-10)c2
Sub them into I get.
=1.2864e^(-25/2)t-0.9684e^-10t

Thank you for helping I'm literally learning as you post.

 

[MarkFL]

Okay, first I get:

\(\displaystyle x'(t)=-\frac{25}{2}c_1e^{-\frac{25}{2}t}-10c_2e^{-10t}\)

And then using the given initial values, we obtain:

\(\displaystyle x(0)=c_1+c_2+\frac{299}{625}=\frac{4}{5}\implies c_1+c_2=\frac{201}{625}\)

\(\displaystyle x'(0)=-\frac{25}{2}c_1-10c_2=0\implies 5c_1+4c_2=0\)

Solving the resulting 2X2 linear system, we obtain:

\(\displaystyle \left(c_1,c_2\right)=\left(-\frac{804}{625},\frac{201}{125}\right)\)

And so the solution to the given IVP is:

\(\displaystyle x(t)=-\frac{804}{625}e^{-\frac{25}{2}t}+\frac{201}{125}e^{-10t}+\frac{299}{625}=\frac{1}{625}\left(1005e^{-10t}-804e^{-\frac{25}{2}t}+299\right)\)

 

[jblakes]

If I had worked it out correctly I would have got the same. Is that the answer to the question? I think I'm going to have to do some more reading. Do you do the same process for every first order ode?

 

[MarkFL]

If I had worked it out correctly I would have got the same. Is that the answer to the question? I think I'm going to have to do some more reading. Do you do the same process for every first order ode?

Yes, that's the solution to the first given IVP. The process is the same for all second order problems of the same type, however there are other methods that can be used, such as annihilation and variation of parameters.

 

[jblakes]

Thank you. I'm guessing the second part. M(d^2x/dt^2)=0.2cos(4t)-kx.I'm guessing you rearrange to get d^2x/dt^2+k/mx=0.2cos(4t)/m then do the same process?

 

[MarkFL]

Thank you. I'm guessing the second part. M(d^2x/dt^2)=0.2cos(4t)-kx.I'm guessing you rearrange to get d^2x/dt^2+k/mx=0.2cos(4t)/m then do the same process?

Yes, your second order ODE will have the form:

\(\displaystyle \frac{d^2x}{dt^2}+k_1x=k_2\cos(4t)\)

So, first determine the roots of the characteristic equation (it appears they will be imaginary which means the homogeneous solution will be sinusoidal rather than exponential).

Then consult your table to determine the forum of the particular solution, and since the forcing function is a sinusoid, make certain your particular solution is linearly independent from all terms in the homogeneous solution).

Then use the method of undetermined coefficients to find the particular solution, and finally use the initial values to determine the parameters of the homogeneous solution. :)

 

[jblakes]

Ok can you see if this is correct.

y"+0y'+25=0

so roots are r+/-sqrt25
r=0+5i
r=0-5i

into
y(t)= Acos4t+Bsin4ty(t)=0
y'(t)=0

0+0+125A=1cos(4*0)
A=1/125

x(t)=Acos4t+Bsin4t+(1/125)

x'(t) = 4Acos(4t)−4Bsin(4t)

sub initial value into it
x(0)=0
X'(0)=-0.5

X(t)= A+(1/125)
A= -(1/125)

x'(t)= -0.5 = 0.999A-(-0.035)B
sub A = (1/125)
B=-145.145

x(t) = 0.032cos(4t)-580.58sin(4t)-1/125)

i'm guessing i have done something wrong somewhere. Can you point out where?

 

[MarkFL]

Well, substituting for the values given for the constants, the ODE becomes:

\(\displaystyle \frac{d^2x}{dt^2}+125x=\cos(4t)\)

And so the roots of the characteristic equation are:

\(\displaystyle r=\pm5\sqrt{5}i\)

which means the homogeneous solution is:

\(\displaystyle x_h(t)=c_1\cos(5\sqrt{5}t)+c_2\sin(5\sqrt{5}t)\)

You have the correct form for the particular solution:

\(\displaystyle x_p(t)=A\cos(4t)+B\sin(4t)\)

But...what you want to do next is compute the first and second time derivatives of this particular solution, substitute everything into the ODE and equate corresponding coefficients to determine the values of $A$ and $B$...as follows:

\(\displaystyle x_p'(t)=-4A\sin(4t)+4B\cos(4t)\)

\(\displaystyle x_p''(t)=-16A\cos(4t)-16B\sin(4t)\)

Now, make the substitutions:

\(\displaystyle \left(-16A\cos(4t)-16B\sin(4t)\right)+125\left(A\cos(4t)+B\sin(4t)\right)=\cos(4t)\)

Arrange as follows:

\(\displaystyle 109A\cos(4t)+109B\sin(4t)=1\cdot\cos(4t)+0\cdot\sin(4t)\)

Equating coefficients, we find:

\(\displaystyle A=\frac{1}{109},\,B=0\)

Thus, our particular solution is:

\(\displaystyle x_p(t)=\frac{1}{109}\cos(4t)\)

And so, by the principle of superposition, there results:

\(\displaystyle x(t)=x_h(t)+x_p(t)=c_1\cos(5\sqrt{5}t)+c_2\sin(5\sqrt{5}t)+\frac{1}{109}\cos(4t)\)

Next, which I will leave to you to attempt and post your progress, you want to take the time derivative of this solution to the ODE, and use the solution and its derivative to determine the parameters $c_i$.

 

[jblakes]

Hi again thanks for your update.

x(t)= c1sin(5√5t)+c2sin(5√5t)+1/109cos(4t)
x(t)= 0 c1= -1/109

x'(t)= (-4sin(4t)/109)- c1(5√5sin(5√5t))/(2√t)+c2(5√5tcos(5√5t))/(2√t)st

just checking this is ok before i sub c1 and -0.5 into x'(t) to get c2. solution will then be

x(t) with values of c1 and c2 added?

 

[MarkFL]

Okay, we have:

\(\displaystyle x(t)=c_1\cos(5\sqrt{5}t)+c_2\sin(5\sqrt{5}t)+\frac{1}{109}\cos(4t)\)

Hence:

\(\displaystyle x'(t)=-5\sqrt{5}c_1\sin(5\sqrt{5}t)+5\sqrt{5}c_2\cos(5\sqrt{5}t)-\frac{4}{109}\sin(4t)\)

Now, using the given initial values, we obtain:

\(\displaystyle x(0)=c_1+\frac{1}{109}=0\implies c_1=-\frac{1}{109}\)

\(\displaystyle x'(0)=5\sqrt{5}c_2=-\frac{1}{2}\implies c_2=-\frac{1}{10\sqrt{5}}\)

And so the solution to the given IVP is:

\(\displaystyle x(t)=-\frac{1}{109}\cos(5\sqrt{5}t)-\frac{1}{10\sqrt{5}}\sin(5\sqrt{5}t)+\frac{1}{109}\cos(4t)\)