What is the position of the mass over time in a damped spring system?

[Shackleford]

A 1-kg mass, when attached to a large spring, stretches the spring a distance of 4.9 m.

(a) Calculate the spring constant.

my'' + uy' + ky = F(t)

F(t) = gravity = mg = 1 kg (9.8 m/s^2) = 9.8 N

m(0) + u(0) + ky = 9.8 N

y = 2 N/m

(b) The system is placed in a viscous medium that supplies a damping constant u = 3 kg/s. The system is allowed to come to rest. Then the mass is displaced 1 m in the downward direction and given a sharp tap, imparting an instantaneous velocity of 1 m/s in the downward direction. Find the position of the mass as a function of time and plot the solution.

my'' + uy' + ky = F(t)

m = 1
u = 3
y'(0) = -1
y(0) = -1
F(t) = 9.8

It says it was displaced downward and given a velocity in the downward direction.

y'' + 3y' + 2y = -9.8

 

[HallsofIvy]

The mass is displaced downward and given an upward initial velocity. While you are free to choose any direction to be positive, you must be consistent. If you choose upward to be positive then your initial position if negative: y(0)= -1, y'(0)= 1, and F= -9.8. Also the resistance is always opposite the direction of motion and so the coefficient of y' is negative. the Most importantly, y and y' are NOT constants- those are only their values at t= 0.

Now, your differential equation is y"- 4y'+ 2y= -9.8 with initial conditions y(0)= -1, y'(0)= 1. You can solve that by finding its characteristic equation and characteristic roots.

 

[Shackleford]

The mass is displaced downward and given and upward initial velocity. While you are free to choose any direction to be positive, you must be consistent. If you choose upward to be positive then your initial position if negative: y(0)= -1, y'(0)= 1, and F= -9.8. Also the resistance is always opposite the direction of motion and so the coefficient of y' is negative. the Most importantly, y and y' are NOT constants- those are only their values at t= 0.

Now, your differential equation is y"- 4y'+ 2y= -9.8 with initial conditions y(0)= -1, y'(0)= 1. You can solve that by finding its characteristic equation and characteristic roots.

Which is exactly what we've been doing.

It says it was displaced downward and given a velocity in the downward direction.

m = 1
u = 3
y'(0) = -1
y(0) = -1
F(t) = -9.8

y'' +3y' + 2y = -9.8

I'm not sure how to get the roots with an inhomogeneous equation. Typically, we'd just divide by e^lambda(t) or whatever with zero on the right and it would reduce to the simple lambda polynomial easily solved by the quadratic formula. If I don't use gravitation as the external force, I wouldn't have this issue.

 

[Mark44]

You need to look at this as two problems: the homogeneous problem y'' + 3y' + 2y = 0, and the nonhomogeneous problem y'' + 3y' + 2y = -9.8.

It sounds like you know how to get the characteristic equation for the homogeneous problem, and solve it to get your two solutions that are exponential functions, and from that you should get y_h = c₁ e^r₁t + c₂ e^r₂t, where r₁ and r₂ are the (known) roots of the characteristic equation.

For the nonhomogeneous problem, since the forcing function on the right side is a constant, try a particular solution of the form y_p = A. Substitute this back into your nonhomogeneous equation to see what value you need for A.

Your general solution will be the solution to the homogeneous problem plus the particular solution. That is, y_g = y_h + y_p.

 

[Shackleford]

You need to look at this as two problems: the homogeneous problem y'' + 3y' + 2y = 0, and the nonhomogeneous problem y'' + 3y' + 2y = -9.8.

It sounds like you know how to get the characteristic equation for the homogeneous problem, and solve it to get your two solutions that are exponential functions, and from that you should get y_h = c₁ e^r₁t + c₂ e^r₂t, where r₁ and r₂ are the (known) roots of the characteristic equation.

For the nonhomogeneous problem, since the forcing function on the right side is a constant, try a particular solution of the form y_p = A. Substitute this back into your nonhomogeneous equation to see what value you need for A.

Your general solution will be the solution to the homogeneous problem plus the particular solution. That is, y_g = y_h + y_p.

Ah. We haven't gotten to those yet. Since we haven't, for this problem should I use -9.8 as the external force or not?

 

[Mark44]

I'm not sure how to get the roots with an inhomogeneous equation. Typically, we'd just divide by e^lambda(t) or whatever with zero on the right and it would reduce to the simple lambda polynomial easily solved by the quadratic formula. If I don't use gravitation as the external force, I wouldn't have this issue.

My explanation covered what you needed to do to solve your nonhomogeneous differential equation. Take another look at what I said in my previous post.

Ah. We haven't gotten to those yet. Since we haven't, for this problem should I use -9.8 as the external force or not?

Of course you need to use this constant. Again, see what I wrote in my previous post.

 

[Shackleford]

My explanation covered what you needed to do to solve your nonhomogeneous differential equation. Take another look at what I said in my previous post.

Of course you need to use this constant. Again, see what I wrote in my previous post.

I'm just guessing here - I need to find a particular solution that's a constant to equal the forcing term value of -9.8. How about A = -4.9?

 

[Mark44]

That works. Do you also have the solution to the homogeneous equation?

 

[Shackleford]

That works. Do you also have the solution to the homogeneous equation?

Let's see, the characteristic equation would be R^2 + 3R + 2.

(R+2)(R+1)

R = -1, -2.

y(t) = C1e^(-t) + C2e^(-2t)

y'(t) = -C1e^(-t) + -2C2e^(-2t)

y(0) = -1 = C1e^(-t) + C2e^(-2t) = C1 + C2 = -1

y'(t) = -1 = -C1e^(-t) + -2C2e^(-2t) = -C1 - 2C2 = -1

 

[Mark44]

Let's see, the characteristic equation would be R^2 + 3R + 2.

An equation has an = sign, so the char. equation would be R^2 + 3R + 2 = 0, from which you can say that (R+2)(R+1) = 0.

(R+2)(R+1)

R = -1, -2.

y(t) = C1e^(-t) + C2e^(-2)

Make that y(t) = C1e^(-t) + C2e^(-2t). Same fix needed below.

y'(t) = -C1e^(-t) + -2C2e^(-2)

y(0) = -1 = C1e^(-t) + C2e^(-2) = C1 + C2 = -1

y'(t) = -1 = -C1e^(-t) + -2C2e^(-2) = -C1 - 2C2 = -1

So, now solve for C1 and C2. Then your general solution will be the above, plus your particular solution, and you're home.

 

[Shackleford]

An equation has an = sign, so the char. equation would be R^2 + 3R + 2 = 0, from which you can say that (R+2)(R+1) = 0.

Make that y(t) = C1e^(-t) + C2e^(-2t). Same fix needed below.

So, now solve for C1 and C2. Then your general solution will be the above, plus your particular solution, and you're home.

Yes. I know an equation has an equal sign. I was just doing a little bit of shorthand. Thanks for your help so far. And, oops, I forgot the "t"s.

 

[Mark44]

Sure, you're welcome. Glad to do it.