- #1

zenterix

- 515

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- Homework Statement
- Consider the undamped harmonic oscillator with Heaviside unit step function ##u(t)## as input.

$$m\ddot{x}+kx=u(t)\tag{1}$$

and initial conditions

$$x(0^-)=0\tag{2}$$

- Relevant Equations
- $$\dot{x}(0^-)=0\tag{3}$$

I'd like to go through the derivation of the response function (ie, the so-called 2nd order unit step response).

This question is based on the calculations in these notes on 2nd order unit step response.

The scenario modeled here is an undamped spring-mass system that is at rest until time ##0##, at which point a constant force starts to act on the mass.

The force is finite and causes a constant acceleration, which goes from being ##0## to suddenly being ##1/m##. Thus, acceleration has a discontinuity at ##0##.

Velocity, on the other hand, starts to increase from 0 in small increments. Position does as well.

Thus, position and velocity are continuous.

In the calculation below I will not assume this continuity. Rather it must come out as a result of the calculations.

Note that

$$u(t)=\begin{cases} 0\ \ \ \ \ \text{for}\ t<0\\ 1\ \ \ \ \ \text{for}\ t>0\end{cases}$$

The general solution of (1) is

$$x(t)=\left (\frac{1}{k}+c_1\cos{\omega_nt}+c_2\sin{\omega_n t}\right )u(t)\tag{4}$$

Note that to find this solution I used initial condition (2).

Then

$$\dot{x}(t)=\left ( -c_1\omega_n\sin{\omega_n t}+c_2\omega_n\cos{\omega_n t} \right ) u(t)+\left (\frac{1}{k}+c_1\right ) \delta(t)\tag{5}$$

$$\ddot{x}(t)=(-c_1\omega_n^2\cos{\omega_n t}-c_2\omega_n^2\sin{\omega_n t})u(t)+c_2\omega_n\delta(t)+\left (\frac{1}{k}+c_1\right )\delta'(t)\tag{6}$$

In order to find the constants ##c_1## and ##c_2##, we can sub in (4) and (6) into the original differential equation.

$$m\ddot{x}+kx=u(t)$$

$$m\left ( (-c_1\omega_n^2\cos{\omega_n t}-c_2\omega_n^2\sin{\omega_n t})u(t)+c_2\omega_n\delta(t)+\left (\frac{1}{k}+c_1\right )\delta'(t) \right ) +k(\left (\frac{1}{k}+c_1\cos{\omega_nt}+c_2\sin{\omega_n t}\right )u(t))=u(t)\tag{7}$$

$$u(t)\left ( m(-c_1\omega_n^2\cos{\omega_n t}-c_2\omega_n^2\sin{\omega_n t}) +k\left (\frac{1}{k}+c_1\cos{\omega_nt}+c_2\sin{\omega_n t}\right ) \right )+mc_2\omega_n\delta(t)+m\left (\frac{1}{k}+c_1\right )\delta'(t)=u(t)\tag{8}$$

Since there are no products of ##\delta(t)## or ##\delta'(t)## on the rhs, we have

$$mc_2\omega_n=0\implies c_2=0$$

$$m\left (\frac{1}{k}+c_1\right )=0\implies c_1=-\frac{1}{k}$$

At this point we've found the constants we're after, but we haven't checked what happens with the factor on the ##u(t)## term in (8). In fact, the notes also do not check.

The check doesn't work for me.

$$m\left (-\left ( -\frac{1}{k}\right )\omega_n^2\cos{\omega_n t})+k\left (\frac{1}{k}+\left ( -\frac{1}{k}\right )\cos{\omega_n t}\right )\right )= 1$$

$$\frac{m\omega_n^2}{k}\cos{\omega_n t}+1-\cos{\omega_n t}=1$$

$$\cos{\omega_n t}\left (\frac{m\omega_n^2}{k}-1\right )=0$$

$$\frac{m\omega_n^2-k}{k}=0$$

I'm not sure what to make of these last equations.

I expected something like ##0=0## or ##1=1##.

**Some Initial Observations**The scenario modeled here is an undamped spring-mass system that is at rest until time ##0##, at which point a constant force starts to act on the mass.

The force is finite and causes a constant acceleration, which goes from being ##0## to suddenly being ##1/m##. Thus, acceleration has a discontinuity at ##0##.

Velocity, on the other hand, starts to increase from 0 in small increments. Position does as well.

Thus, position and velocity are continuous.

In the calculation below I will not assume this continuity. Rather it must come out as a result of the calculations.

Note that

$$u(t)=\begin{cases} 0\ \ \ \ \ \text{for}\ t<0\\ 1\ \ \ \ \ \text{for}\ t>0\end{cases}$$

**Calculation of 2nd Order Unit Step Response**The general solution of (1) is

$$x(t)=\left (\frac{1}{k}+c_1\cos{\omega_nt}+c_2\sin{\omega_n t}\right )u(t)\tag{4}$$

Note that to find this solution I used initial condition (2).

Then

$$\dot{x}(t)=\left ( -c_1\omega_n\sin{\omega_n t}+c_2\omega_n\cos{\omega_n t} \right ) u(t)+\left (\frac{1}{k}+c_1\right ) \delta(t)\tag{5}$$

$$\ddot{x}(t)=(-c_1\omega_n^2\cos{\omega_n t}-c_2\omega_n^2\sin{\omega_n t})u(t)+c_2\omega_n\delta(t)+\left (\frac{1}{k}+c_1\right )\delta'(t)\tag{6}$$

In order to find the constants ##c_1## and ##c_2##, we can sub in (4) and (6) into the original differential equation.

$$m\ddot{x}+kx=u(t)$$

$$m\left ( (-c_1\omega_n^2\cos{\omega_n t}-c_2\omega_n^2\sin{\omega_n t})u(t)+c_2\omega_n\delta(t)+\left (\frac{1}{k}+c_1\right )\delta'(t) \right ) +k(\left (\frac{1}{k}+c_1\cos{\omega_nt}+c_2\sin{\omega_n t}\right )u(t))=u(t)\tag{7}$$

$$u(t)\left ( m(-c_1\omega_n^2\cos{\omega_n t}-c_2\omega_n^2\sin{\omega_n t}) +k\left (\frac{1}{k}+c_1\cos{\omega_nt}+c_2\sin{\omega_n t}\right ) \right )+mc_2\omega_n\delta(t)+m\left (\frac{1}{k}+c_1\right )\delta'(t)=u(t)\tag{8}$$

Since there are no products of ##\delta(t)## or ##\delta'(t)## on the rhs, we have

$$mc_2\omega_n=0\implies c_2=0$$

$$m\left (\frac{1}{k}+c_1\right )=0\implies c_1=-\frac{1}{k}$$

At this point we've found the constants we're after, but we haven't checked what happens with the factor on the ##u(t)## term in (8). In fact, the notes also do not check.

The check doesn't work for me.

$$m\left (-\left ( -\frac{1}{k}\right )\omega_n^2\cos{\omega_n t})+k\left (\frac{1}{k}+\left ( -\frac{1}{k}\right )\cos{\omega_n t}\right )\right )= 1$$

$$\frac{m\omega_n^2}{k}\cos{\omega_n t}+1-\cos{\omega_n t}=1$$

$$\cos{\omega_n t}\left (\frac{m\omega_n^2}{k}-1\right )=0$$

$$\frac{m\omega_n^2-k}{k}=0$$

I'm not sure what to make of these last equations.

I expected something like ##0=0## or ##1=1##.

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