Textbook 'The Physics of Waves': Varying Force Amplitude

[brettng]

Homework Statement

Problem 2.3 (See screen capture below)

Relevant Equations

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

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 2.3 [Page 52] (see screen capture below)

Question: In Problem 2.3, I have proved the hint equation, and it leads to the complex equation of motion to become $$\frac{d^2 z(t)}{dt^2} + \omega^2_0 z(t) = \frac{f_0}{2m} \left( e^{-i \delta t} + e^{i \delta t} \right) e^{-i \omega_0 t}$$

Then, I try a complex solution as usual, but this time with a time-dependent amplitude, $z(t) = \mathcal{A}(t) e^{-i \omega_0 t}$. This could simplify the complex equation of motion to $$\frac{d^2 \mathcal{A}(t)}{dt^2} - i2 \omega_0 \frac{\mathcal{A}(t)}{dt} = \frac{f_0}{2m} \left( e^{-i \delta t} + e^{i \delta t} \right) = \frac{f_0}{m} \cos (\delta t)$$

At this stage, I try a complex form for the time-dependent amplitude $\mathcal{A}(t) = A\cos (\delta t) + iB \sin (\delta t)$. The complex equation of motion would further be simplified into real and imaginary parts: $$(-A \delta^2 + 2 \omega_0 \delta B) \cos (\delta t) + (-B \delta^2 + 2 \omega_0 A \delta) \sin (\delta t) = \frac {f_0}{m} \cos (\delta t)$$

By setting the following equations:
$$(-A \delta^2 + 2 \omega_0 \delta B) \cos (\delta t) = \frac {f_0}{m} \cos (\delta t)$$

and

$$(-B \delta^2 + 2 \omega_0 A \delta) \sin (\delta t) = 0$$

We can solve that $A = \frac {f_0}{m} \left( \frac {1}{4 \omega^2_0 - \delta^2} \right)$ and $B = \frac {2 \omega_0 f_0}{m \delta} \left( \frac {1}{4 \omega^2_0 - \delta^2} \right)$.

Finally, by using $z(t) = \left[ A\cos (\delta t) + iB \sin (\delta t) \right] \left[ \cos (\omega_0 t) - i \sin (\omega_0 t) \right]$, we can solve the exact real displacement for nonzero $\delta$ :
$$x(t) = \frac {f_0 \cos (\delta t)}{m \left( {4 \omega^2_0 - \delta^2} \right)} \cos (\omega_0 t) + \frac {2 f_0 \omega_0 \sin (\delta t)}{m \delta \left( {4 \omega^2_0 - \delta^2} \right)} \sin (\omega_0 t)$$

Therefore, answering the question,
$$\alpha (t) = \frac {f_0}{m \left( {4 \omega^2_0 - \delta^2} \right)} \cos (\delta t)$$

and

$$\beta (t) = \frac {2 f_0 \omega_0}{m \delta \left( {4 \omega^2_0 - \delta^2} \right)} \sin (\delta t)$$.


Question 1: Do I get the correct answer for this Problem question?

Question 2: What does the meaning of "for $\delta$ nonzero to leading order in $\delta / \omega_0$" as shown in the problem question (see the screen capture)? Because in my approach, I did not use this criteria and I could find the exact solution for nonzero $\delta$.

Question 3: In simple cases (i.e. sinusoidal force with constant amplitude; e.g. $f_0 e^{-i \omega_0 t}$), we could try the solution by directly copying the force complex component (e.g. $A(t) e^{-i \omega_0 t}$) as demonstrated in Chapter 2 of the reference book. In my approach above, it shows that we could not use such technique directly, just because the amplitude is also time-dependent (which makes the time derivative parts complicated). Therefore, we need to use some slightly different approach for this Problem case. Does my view correct?

 

[TSny]

Question 1: Do I get the correct answer for this Problem question?

Your work looks correct. You have found an exact expression for a particular solution of the differential equation. To this particular solution, you can add the general solution of the associated homogeneous differential equation to get the general solution for this problem. It will contain two arbitrary constants that are determined by initial conditions. Since there is no damping in this problem, the solutions to the associated homogeneous differential equation do not decay with time.

Question 2: What does the meaning of "for $\delta$ nonzero to leading order in $\delta / \omega_0$" as shown in the problem question (see the screen capture)? Because in my approach, I did not use this criteria and I could find the exact solution for nonzero $\delta$.

I guess they want you to find an approximate result that holds when the system is near resonance ( $\delta$ small). For example, if $\delta / \omega_0 \ll 1$ you can simplify the denominators in your result. Also, if $\delta \, t \ll 1$, you can simplify some of the trig functions and see how the amplitude of the oscillations grows with time initially.

Question 3: In simple cases (i.e. sinusoidal force with constant amplitude; e.g. $f_0 e^{-i \omega_0 t}$), we could try the solution by directly copying the force complex component (e.g. $A(t) e^{-i \omega_0 t}$) as demonstrated in Chapter 2 of the reference book. In my approach above, it shows that we could not use such technique directly, just because the amplitude is also time-dependent (which makes the time derivative parts complicated). Therefore, we need to use some slightly different approach for this Problem case. Does my view correct?

In this problem, the total driving force is a superposition of two harmonic driving forces: $$\cos \omega_0 t \cos \delta t = \frac 1 2 \textrm{Re} \left( e^{-i(\omega_0 + \delta)t} + e^{-i(\omega_0 - \delta)t}\right).$$ Since your differential equation is linear, you could find the solution by superposition of the solutions for each harmonic driving force acting alone. This way, you can use the text's solution for a harmonic driving force.

 

[brettng]

Thank you for your reply TSny!

Since your differential equation is linear, you could find the solution by superposition of the solutions for each harmonic driving force acting alone. This way, you can use the text's solution for a harmonic driving force.

In this case, do you mean that I could try the superpostion solutions with $Ae^{-i (\omega_0 + \delta) t}$ and $Be^{-i (\omega_0 - \delta) t}$?

 

[TSny]

In this case, do you mean that I could try the superpostion solutions with $Ae^{-i (\omega_0 + \delta) t}$ and $Be^{-i (\omega_0 - \delta) t}$?

Yes. Solve the two differential equations $$\ddot z_1 + \omega_0^2 z_1 = \frac{f_0}{2m}e^{-i\omega_1 t} \,\,\,\,\,\,\,\, \textrm{and} \,\,\,\,\,\,\,\, \ddot z_2 + \omega_0^2 z_2 = \frac{f_0}{2m} e^{-i\omega_2 t},$$ where $\omega_1 = \omega_0 + \delta$ and $\omega_2 = \omega_0 - \delta$.

Then, $x(t) = \textrm{Re}\left[z_1(t) + z_2(t) \right]$.