Textbook 'The Physics of Waves': Linearity of Forced Oscillator

[brettng]

Homework Statement

Prove, just using linearity, without using the explicit solution, that the steady state solution to the complex equation of motion for forced oscillator must be proportional to the real force $F_0$.

Relevant Equations

$\frac {d^2z\left( t \right)} {dt^2} + \Gamma \frac {dz\left( t \right)} {dt} + \omega^2_0 z\left( t \right) = \frac{\mathcal {F}\left( t \right)}{m}$

$\mathcal {F}\left( t \right)=F_0 e^{-i \omega_d t}$

Reference textbook “The Physics of Waves” in MIT website:
https://ocw.mit.edu/courses/8-03sc-...es-fall-2016/resources/mit8_03scf16_textbook/

Chapter 2 - Problem 2.2 [Page 51] (see attached file)

Question: In the content of Page 43 (see attached file), it also states that the amplitude of complex displacement is proportional to the amplitude of the driving force, and this is expected from linearity.

However, it means the linearity of what? I guess it is linearity of solution $x_1(t)+x_2(t)$. Do I misunderstand it?

Also, grateful if you could give me some hints on Problem 2.2 please.

 

[pasmith]

The left hand side of (2.16) is an operator $$L(z) = z'' + \Gamma z' + \omega_0^2 z.$$ This operator is linear, in that $L(kz) = kL(z)$ for any constant $k$ and any function $z$. I can't comment further on 2.2 without essentially answering it for you.

 

[brettng]

The left hand side of (2.16) is an operator $$L(z) = z'' + \Gamma z' + \omega_0^2 z.$$ This operator is linear, in that $L(kz) = kL(z)$ for any constant $k$ and any function $z$. I can't comment further on 2.2 without essentially answering it for you.

May I confirm if I can have the logical flow as follows:

1) Set $z(t) = F_0 u(t)$

2) The equation of motion becomes $$F_0 \left( u'' + \Gamma u’ + \omega_0^2 u\right) = \frac {F_0 e^{-i \omega_d t}} {m}$$

$$\left( u'' + \Gamma u’ + \omega_0^2 u\right) = \frac {e^{-i \omega_d t}} {m}$$

3) Now, we can conclude that any constant $C$ multiplies the $e^{-i \omega_d t}$ on the R.H.S. of the 2nd equation in 2) above, the solution $z(t)$ would be proportional to the constant $C$.

4) Therefore, the steady state solution $z(t)$ must be proportional to $F_0$ in this case (i.e. answer of Problem 2.2).

5) A step more: I can also conclude that the steady state solution must be proportional to $\frac 1 {m}$.

Does my logic correct?

 

[pasmith]

Yes. In fact dimensional analysis requires that $$z(t) = \frac{F_0}{m\omega_0^2} Z\left( \omega_0 t, \frac{\omega_d}{\omega_0}, \frac{\Gamma}{\omega_0} \right)$$ where we find $$Z(\tau, \alpha, \beta) = \frac{e^{-i\alpha \tau}}{1 - i\alpha\beta - \alpha^2 }.$$

 

[brettng]

Yes. In fact dimensional analysis requires that $$z(t) = \frac{F_0}{m\omega_0^2} Z\left( \omega_0 t, \frac{\omega_d}{\omega_0}, \frac{\Gamma}{\omega_0} \right)$$ where we find $$Z(\tau, \alpha, \beta) = \frac{e^{-i\alpha \tau}}{1 - i\alpha\beta - \alpha^2 }.$$

So, your demonstration is based on dimensional analysis; while my logical flow follows linearity.

Am I understand correctly?

 

[Steve4Physics]

Since the question has essentially already been answered, here's what looks like (IMO) a simple/direct approach.

You are being asked to show that the steady-state solution of equation 2.16 is proportional to $F_0$. That means if $F_0$ is multiplied by some constant, say $k$, then the new solution is $k z(t)$.

And to do this, you are being asked to use the fact that $\frac d{dt}$ and $\frac {d^2}{dt^2}$ are linear operators.

Start by multiplying both sides of equation 2.16 by $k$.

Then use the fact that since $\frac d{dt}$ is a linear operator, $k\frac d{dt} z(t) = \frac d{dt} (k z(t))$. Similarly for $\frac {d^2}{dt^2}$.

 

[brettng]

Thank you so much for your help!!