0
$\begingroup$

I'm given a mass-and-spring system with a couple second-order differential equations describing the behavior of the system. Those are:

$$x^{\prime \prime} +50x - \frac{25}{2}y = 0$$ and $$y^{\prime \prime} + 50y -50x = 0$$

We set up an operational determinant and calculate the general solutions as follows, as well as make sure we have four independent arbitrary constants

$$x(t) = a_1 \cos (5t) + a_2 \sin (5t) + a_3 \cos (5t \sqrt{3}) + a_4 \sin (5t \sqrt {3})$$ $$y(t) = 2a_1 \cos (5t) + 2a_2 \sin (5t) - 2a_3 \cos (5t \sqrt{3}) - 2a_4 \sin (5t \sqrt {3})$$

Using a bit of trigonometry, we can rewrite the equations a bit in terms of two particular solutions $(x_1, y_1)$ and $(x_2, y_2)$ of the system for

$$x_1 (t) = a_1 \cos (5t) + a_2 \sin (5t) = A \cos(5t - \alpha)$$ $$y_1 (t) = 2a_1 \cos (5t) + 2a_2 \sin (5t) = 2A \cos(5t - \alpha)$$ and $$x_2 (t) = a_3 \cos (5t \sqrt{3}) + a_4 \sin (5t \sqrt {3}) = B \cos(5t\sqrt{3} - \beta)$$ $$y_2 (t) = -2a_3 \cos (5t \sqrt{3}) - 2a_4 \sin (5t \sqrt {3}) = -2B \cos(5t\sqrt{3} - \beta)$$

Here are my questions (I need a really intuitive answer because I don't really understand this):

#1. How do we know that those two solutions, $(x_1, y_1)$ and $(x_2, y_2)$, are actually solutions? More specifically, what is the significance about these solutions? Obviously we can reach the solutions when we set $a_3$ and $a_4$ to zero and then do the same with $a_1$ and $a_2$... but what's the intuition behind this?

**#2. The book describes these two solutions as natural modes of oscillation that exhibit two natural frequencies. What are natural modes of oscillation and how do I find the natural frequencies? Could someone give me a physical description of this answer? **

Thank you in advance.

$\endgroup$

1 Answer 1

0
$\begingroup$

1a) You can check that $(x_1,y_1)$ and $(x_2,y_2)$ are solutions by substituting them in the system of ODEs, as they should solve the system, i.e. they should obey the equations.

1b) and 2. Your solutions are linear combinations of the functions $\cos(5t)$, $\sin(5t)$, $\cos(5\sqrt{3}\,t)$ and $\sin(5\sqrt{3}\,t)$. Note that these functions are of the form $\cos/\sin(\alpha\,t)$, with $\alpha = 5$ or $\alpha = 5 \sqrt{3}$. The function $\cos(\alpha t)$ describes an oscillation with period $\frac{2\pi}{\alpha}$, i.e. with frequency $\frac{\alpha}{2\pi}$, see here. The same holds for $\sin(\alpha t)$. Therefore, you see that the general solution to the system of ODEs combines oscillations with two separate frequencies, $\frac{5}{2\pi}$ and $\frac{5\sqrt{3}}{2\pi}$. By choosing either $a_{1,2}=0$ or $a_{3,4}=0$, you can find two solutions, with $(x_1,y_1)$ only exhibiting oscillations with frequency $\frac{5}{2\pi}$, and $(x_2,y_2)$ only exhibiting oscillations with frequency $\frac{5\sqrt{3}}{2\pi}$.

Now, you can see that the general solution $(x,y)$ is a superposition (i.e. a linear combination) of these two solutions $(x_1,y_1)$ and $(x_2,y_2)$.

$\endgroup$
5
  • $\begingroup$ The answer in the book reads "In the natural mode with frequency $\omega_1 = 5$ the masses move in the same direction, while in the natural mode with frequency $\omega_2 = 5\sqrt{3}$ they move in opposite directions. In each case the amplitude of the motion $m_2$ is twice that of $m_1$." How do we know this? How do I find the frequencies, and from there how can I determine the amplitude and the direction of the movement? $\endgroup$
    – user312437
    Apr 26, 2016 at 0:25
  • $\begingroup$ And what is the natural oscillation? $\endgroup$
    – user312437
    Apr 26, 2016 at 0:25
  • 1
    $\begingroup$ These questions consider the physical interpretation of the solutions you found. I very, very strongly suggest you read the Wikipedia lemma I linked in my answer. $\endgroup$ Apr 26, 2016 at 9:24
  • $\begingroup$ I had only read a small part regarding the frequency, but I will go back and read it in its entirety! Thank you. $\endgroup$
    – user312437
    Apr 26, 2016 at 11:04
  • $\begingroup$ You're welcome! Don't hesitate to ask if some things remain unclear. $\endgroup$ Apr 26, 2016 at 13:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy