# Second order differential equations on forced vibrations

## Homework Statement

A spring-mass system has a spring constant of 3 N/m. A mass of 2 kg is attached to the spring, and the motion takes place in a viscous fluid that offers a resistance numerically equal to the magnitude of the instantaneous velocity. If the system is driven by an externam force of 3 cos(3t) - 2 sin(3t) N, determine the steady-state response.

## Homework Equations

From the problem:
m = 2, k = 3, gamma = y' ?
2y" + y'$$^{2}$$ + 3y = 3 cos(3t) - 2 sin(3t), or in terms of delta and omega_0
y" + 2 (y'/4) y' + 3/2 y = 3/2 cos(3t) - sin(3t)

## The Attempt at a Solution

Don't know where to go from there. I need to find A, omega and gamma, but I don't know how.

Tom Mattson
Staff Emeritus
Gold Member
Well for one thing, the y' in your differential equation should not be squared. The equation should be linear.

Does that help?

HallsofIvy
Homework Helper

## Homework Statement

A spring-mass system has a spring constant of 3 N/m. A mass of 2 kg is attached to the spring, and the motion takes place in a viscous fluid that offers a resistance numerically equal to the magnitude of the instantaneous velocity.
You may have interpreted this as saying that the coefficient is equal tot he magnitude of the instaneous velocity so the force is (-y')y'. But it says the force itself is: you want only -y'. Tom Mattson is correct.

try a solution of the form Acos(3t-delta) + Btcos(3t-delta). do you know why?

ice109 said:
try a solution of the form Acos(3t-delta) + Btcos(3t-delta). do you know why?

Not exactly. It looks very similar to the guess for undamped free vibrations, y = A cos(omega_0 t) + B sin(omega_0 t), and I can see where the -delta comes from (the damping part), but why 3t? Isn't 3 = omega, not omega_0? My omega_0 = $$\sqrt{3/2}$$, I think.

HallsofIvy said:
you want only -y'. Tom Mattson is correct.

So my equation is 2y" + y' + 3y = etc.

HallsofIvy
Homework Helper
Not exactly. It looks very similar to the guess for undamped free vibrations, y = A cos(omega_0 t) + B sin(omega_0 t), and I can see where the -delta comes from (the damping part), but why 3t? Isn't 3 = omega, not omega_0? My omega_0 = $$\sqrt{3/2}$$, I think.

So my equation is 2y" + y' + 3y = etc.
Yes, that is correct.

try a solution of the form Acos(3t-delta) + Btcos(3t-delta). do you know why?
I certainly don't see any reason to include the "delta" or "t". Acos(3t)+ Bsin(3t) should be sufficient. Solutions to the homogeneous equation are nothing like cos(3t) or sin(3t).

I certainly don't see any reason to include the "delta" or "t". Acos(3t)+ Bsin(3t) should be sufficient. Solutions to the homogeneous equation are nothing like cos(3t) or sin(3t).

Acos(3t-d) is compact notation for Acos(d)cos(3t)+Asin(d)sin(3t). and the reason for both is that i was under the impression that since the solution for an inhomogeneous part of something like y'=cos(t) is Acos(t-d) then the solution for y'=cos(t) + sin(t) is the sum of the solutions for cos(t) and sin(t) but those are the same hence the t multiplying out in front.