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Lecture 11

Lecture 10 | Lecture 12


By Giuseppe Tenti and annotated by Douglas Wilhelm Harder.

Free Oscillations

If the forcing function tex:$$f(t)$$ is zero, the the differential equation becomes homogeneous:

tex:$$\ddot{\phi} + 2\zeta\omega\dot{\phi} + \omega^2\phi = 0$$

From our previous observations, we note that the characteristic equation is tex:$$m^2 + 2\zeta\omega m + \omega^2 = 0$$ and the roots of the characteristic are:

  • tex:$$-\omega \left( \zeta \mp \sqrt{ \zeta^2 - 1 } \right )$$ if tex:$$\zeta > 1$$
  • tex:$$-\omega, -\omega$$ if tex:$$\zeta = 1$$
  • tex:$$-\omega \left( \zeta \mp j\sqrt{ 1 - \zeta^2 } \right )$$ if tex:$$\zeta < 1$$

and the response to the homogeneous system are

  • tex:$$k_1 e^{-\omega \left( \zeta + \sqrt{ \zeta^2 - 1 } \right )t} + k_2 e^{-\omega \left( \zeta - \sqrt{ \zeta^2 - 1 } \right )t}$$,
  • tex:$$k_1 e^{-\omega t}                                           + k_2 te^{-\omega t}$$, and
  • tex:$$e^{-\omega\zeta t} \left ( k_1 \cos \left ( \omega \sqrt{ 1 - \zeta^2 } t \right ) + k_2 j\sin \left ( \omega \sqrt{ 1 - \zeta^2 } t \right ) + \right )$$

respectively.

We will now discuss the physical interpretation of these three classes of responses.

Over Damping

If the friction is stronger than the elastic force, that is tex:$$\zeta > 1$$, the response quickly converges to zero. The response drops monotonically to zero without any oscillations. Figure 1 shows such an example.


Figure 1. An over damped system.

Critically Damping

If the effect of the friction equals the elastic force, the response decays to zero as quickly as is possible without oscillating. In this case, it is called critically damped and is shown in Figure 2.


Figure 2. A critically damped system.

Under Damping

Finally, if the elastic force is greater than the effect of the friction (tex:$$\zeta < 1$$), the response decays to zero but oscillates around that response. This is called underdamping and is shown in Figure 3.


Figure 3. An under damped system.

Mechanical Oscillator

The differential equation

tex:$$m\ddot{x} + \gamma \dot{x} + kx = 0$$

may be written in the form

tex:$$\ddot{x} + \frac{\gamma}{m} \dot{x} + \frac{k}{m}x = 0$$

from which we may determine that:

tex:$$\omega = \sqrt{\frac{k}{m}}$$ and tex:$$\zeta = \frac{\gamma}{2\sqrt{km}}$$.

Suppose that the mass is tex:$$m = 250 {\rm g} = 0.25 {\rm kg}$$ and the spring constant is tex:$$k = 1.44 {\rm kg}/{\rm s}^2$$. Immediately, we note that the natural frequency is tex:$$\omega = 2.4$$. If there was no friction (tex:$$\gamma = 0$$), the mass would naturally oscillate with a period of tex:$$\frac{2\pi}{\omega} \approx 2.618 {\rm s}$$.

and therefore the damping parameter is tex:$$\zeta = \frac{\gamma}{2\sqrt{km}} = \frac{\gamma}{1.2}$$.

In each of these cases, we will move the object to a distance of tex:$$x(0) = 2 {\rm m}$$ and we will release the object, and therefore the initial velocity is tex:$$\dot{x}(0) = 0 {\rm m/s}$$

Over Damping

Suppose the friction coefficient is greater than 1.2, for example, tex:$$\gamma = 2$$, in which case we have that tex:$$\zeta \approx 1.667$$ and therefore we see the response shown in Figures 4 and 5. This is the response tex:$$x(t) = 2.25 e^{-0.8t} - 0.25 e^{-7.2t}$$.


Figure 4. The response to an over damped system.


Figure 5. The movement of the mass following the response in Figure 4.

Note, with sufficiently strong initial conditions (e.g., tex:$$x(0) = 1$$ and tex:$$\dot{x}(0) = -40$$), it may happen that the response may initially change sign, but after this point, it will decay monotonically to zero.

Critically Damped

Suppose the friction coefficient equals 1.2, that is, tex:$$\gamma = 2$$, in which case we have tex:$$\zeta = 1$$ and therefore we see the response shown in Figures 6 and 7. This is the response tex:$$x(t) = 2 e^{-2.4t} + 4.8 te^{-2.4t}$$.


Figure 6. The response to a critically damped system.


Figure 7. The movement of the mass following the response in Figure 6.

Note, with sufficiently strong initial conditions (e.g., tex:$$x(0) = 1$$ and tex:$$\dot{x}(0) = -25$$), it may happen that the response may initially change sign, but after this point, it will decay monotonically to zero.

Under Damped

Suppose the friction coefficient equals 0.3, that is, tex:$$\gamma = 0.3$$, in which case we have tex:$$\zeta = 0.25$$ and therefore we see the response shown in Figures 8 and 9. This is the response tex:$$x(t) \approx e^{-0.6t} \left (0.5164 \sin(2.3238t) + 2 \cos(2.3238 t) \right )$$.


Figure 8. The response to an under damped system.


Figure 9. The movement of the mass following the response in Figure 8.

We can also write this as tex:$$x(t) \approx 2.0656 e^{-0.6t}\cos(2.3238t - 0.2527)$$.