Harmonic oscillator, friction, impulse

[covers]

Homework Statement

A linear harmonic oscillator is subject to friction(stokes). The Oscillator gets an Impulse at the time t = 0 at the rest position. What is the equation of motion for the time interval 0 - t0?

Homework Equations

Friction force: Fr = -a * x'(t); a = constant (stokes friction)
Impulse: F(t) = m *v0 / t0 (for 0 <= t <= t0)

k = spring constant
m = mass

x(t = 0) = 0 (rest position at the time t = 0)
x'(t = 0) = 0 (velocity at the time t = 0 is zero)

The Attempt at a Solution

I have a little trouble with the impulse here. My attempt at a solution was this:
m*x'' = -k*x - a*x' + m*v0/t0 ==> x'' + (a/m)*x' + (k/m)*x = v0 / t0

But my book claims that the equation really goes like this:
x'' + (a/m)*x' + (k/m)*x = v0 / t0 * (1/m)

Does somebody know why I am wrong and the book is right? I don't understand it, because I thought I needed a force(respectively acceleration) in above equation of motion. How does v0 / t0 * (1/m) fit into the equation, doesn't it have the wrong units?

 

[AlephZero]

I don't understand your notation. You say the Impulse is at time t = 0 but then you give it as a change of momentum divided by the whole time of the motion (t0).

I would have interpreted the "impulse at t = 0" to mean the initial velocity was v0 (not 0) and the equation for time 0 <= t <= t0 was just mx" + ax' + kx = 0.

But I learned this stuff 30 years ago, so the standard terminology might have changed since then.

If v0 and t0 are a velocity and a time then you are correct, your units are consistent and the book apparently is not.

 

[covers]

Thanks for helping!

I don't understand your notation. You say the Impulse is at time t = 0 but then you give it as a change of momentum divided by the whole time of the motion (t0).

As far as I understand it, the impulse lasts per definition for a certain time interval. In my example here, it goes from t = 0 to t = t0. In my book it is stated as like that: F(t) = m*v0/t0 (for 0 <= t <= t0).

I would have interpreted the "impulse at t = 0" to mean the initial velocity was v0 (not 0) and the equation for time 0 <= t <= t0 was just mx" + ax' + kx = 0.

The velocity at t = 0 is clearly stated as 0. Maybe it would be more correct to write the impulse like that: F(t) = m*v0/t0 (for 0 < t <= t0). But I'm not sure about that. Also, x'(t = 0) = 0 is used to determine initial conditions later in the exercise.

If v0 and t0 are a velocity and a time then you are correct, your units are consistent and the book apparently is not.

Yes, v0 and t0 are velocity and time (sorry if I was not precise enough). The Problem is just, if that's an error in the book its hard to believe! Because as the exercise goes on, this has a lot of consequences, and the term 'v0 / t0 * (1/m)' is used throughout the whole exercise...

 

[AlephZero]

OK, so he's assuming the impulse comes from a constant force for the time interval 0 <= t <= t0, and if it was applied to a free mass m it would produce velocity v0 at time t0.

In which case, sorry, I don't understand why the 1/m is there.

 

[covers]

Thanks anyway. Maybe it's really an error.

I take the opportunity to ask another question about this exercise.

At one point, I have to decide whether 'a / (2*m)' is bigger or smaller or equal to 'sqrt(k/m)'. This is relevant when deciding if the oscillator is overdamped, underdamped or critically damped. Is there a way to decide with the information given what kind of oscillator it is? The book is assuming that the oscillator is underdamped, but it does not explain why...

 

[AlephZero]

The solutions of the equation of motion mx" + ax' + kx = 0 (with no external forces) are exponentials. The exponents are either real (overdamped case) or complex (underdamped, oscillating solutions).

A trial solution x = e^st gives the auxiliary equation ms^2 + as + k = 0 to determine s. The condition for real or imaginary roots of this quadratic equation is that a^2 - 4mk is greater or less than 0. For the critically damped case a^2 - 4mk = 0, there is only one repeated root for s, and the two independent solutions are e^st and t.e^st.

Look for a text on solutions of second order linear differential equations if you need more on this.

Beware - some texts on vibration write this equation in the form mx" + 2ax' + kx. Take care with the 2's.