Effect of damping

Introducing damping, the equation of motion

$$m \ddot{x}+ c \dot{x}+ k x = F \cos(\omega t + \psi)$$ (3.12)

written using the notation introduced earlier is

$$\ddot{\tilde{x}}+ 2 \beta \dot{x}+ \omega_0^2 \tilde{x}= \tilde{f}e^{i \omega t } \,.$$ (3.13)

Here again we separately discuss the complementary functions and the particular integral. The complementary functions are the decaying solutions that arise when there is no external force. These are short lived transients which are not of interest when studying the long time behaviour of the oscillations. These have already been discussed in considerable detail and we do not consider them here. The particular integral is important when studying the long time or steady state response of the oscillator. This solution is

$$\tilde{x}(t)= \frac{\tilde{f}}{(\omega_0^2-\omega^2)+2 i \beta \omega} e^{i \omega t}$$ (3.14)

which may be written as $\tilde{x}(t)=C e^{i \phi} \tilde{f}e^{i \omega t}$ where $\phi$ is the phase of the oscillation relative to the force $\tilde{f}$.

This has an amplitude

$$\mid \tilde{x}\mid = \frac{f}{\sqrt{(\omega_0^2-\omega^2)^2+ 4 \beta^2 \omega^2}}$$ (3.15)

and the phase $\phi$ is

$$\phi = \tan^{-1} \left( \frac{-2 \beta \omega}{\omega_0^2-\omega^2} \right)$$ (3.16)

Figure 3.2: Amplitudes and phases for various damping coefficients as a function of driving frequency

Figure 3.2 shows the amplitude and phase as a function of $\omega$ for different values of the damping coefficient $\beta$. The damping ensures that the amplitude does not blow up at $\omega=\omega_0$ and it is finite for all values of $\omega$. The change in the phase also is more gradual.

The low frequency and high frequency behaviour are exactly the same as the situation without damping. The changes due to damping are mainly in the vicinity of $\omega=\omega_0$. The amplitude is maximum at

$$\omega=\sqrt{\omega_0^2 - 2 \beta^2} \,$$ (3.17)

For mild damping ( $\beta \ll \omega_0$) this is approximately $\omega=\omega_0$.

We next shift our attention to the energy of the oscillator. The average energy $E(\omega)$ is the quantity of interest. Calculating this as a function of $\omega$ we have

$$E(\omega)={{m}f^2\over 4}{{(\omega^2+\omega_0^2)}\over {[(\omega_0^2-\omega^2)^2+4\beta^2\omega^2]}}$$ (3.18)

Figure 3.3: Energy resonance

The response to the external force shows a prominent peak or resonance (Figure 3.3) only when $\beta \ll \omega_0$, the mild damping limit. This is of great utility in modelling the phenomena of resonance which occurs in a large variety of situations. In the weak damping limit $E(\omega)$ peaks at $\omega \approx \omega_0$ and falls rapidly away from the peak. As a consequence we can use

$$(\omega_0^2-\omega^2)^2=(\omega_0+\omega)^2 (\omega_0-\omega)^2 \approx 4 \omega_0^2 (\omega_0-\omega)^2$$ (3.19)

which gives

$$E(\omega) \approx \frac{k}{8} \frac{f^2}{\omega_0^2[(\omega_0-\omega)^2 + \beta^2 ]}$$ (3.20)

in the vicinity of the resonance. This has a maxima at $\omega \approx \omega_0$ and the maximum value is

We next estimate the width of the peak or resonance. This is quantified using the FWHM (Full Width at Half Maxima) defined as FWHM =  $2 \Delta \omega$ where $E(\omega_0+\Delta \omega)=E_{max}/2$ ie. half the maximum value. Using equation (3.20) we see that $\Delta \omega=\beta$ and FWHM $= 2 \beta$. as shown in Figure 3.3. The FWHM quantifies the width of the curve and it records the fact that the width increases with the damping coefficient $\beta$.

The peak described by equation (3.20) is referred to as a Lorentzian profile. This is seen in a large variety of situations where we have a resonance.

We finally consider the power drawn by the oscillator from the external force. The instantaneous power $P(t)=F(t) \dot{x}(t)$ has a value

$$P(t)= [F \cos(\omega t)] [- \mid \tilde{x}\mid \omega \sin(\omega t + \phi)] \,.$$ (3.22)

The average power is the quantity of interest, we study this as a function of the frequency. Calculating this we have

Using equation (3.14) we have

$$\mid \tilde{x}\mid \sin \phi = \frac{- 2 \beta \omega}{(... ...mega^2)^2 + 4 \beta^2 \omega^2} \left(\frac{F}{m} \right)$$ (3.24)

which gives the average power

$$\langle P \rangle(\omega) = \frac{ \beta \omega^2}{(\omeg... ...)^2 + 4 \beta^2 \omega^2} \left(\frac{F^2}{m} \right) \,.$$ (3.25)

The solid curve in Figure 3.4 shows the average power as a function of $\omega$. Here again, a prominent, sharp peak is seen only if $\beta \ll \omega_0$. In the mild damping limit, in the vicinity of the maxima we have

$$\langle P \rangle(\omega) \approx \frac{ \beta }{(\omega_0-\omega)^2 + \beta^2 } \left(\frac{F^2}{4 m} \right) \,.$$ (3.26)

which again is a Lorentzian profile. For comparison we have also plotted the Lorentzian profile as a dashed curve in Figure 3.4.

Figure 3.4: Power resonance

Problem 1: Plot the response, $x(t)$, of a forced oscillator with a forcing $3\cos 2t$ and natural frequency $\omega_0=3$ Hz with initial conditions, $x(0)=3$ and $\dot{x}(0)=0$, for two different resistances, $\beta=1$ and $\beta=0.5$. Plot also for fixed resistance, $\beta=0.5$ and different forcing amplitudes $f_0=1,3,5$ and 9.

Solution 1: The evolution is shown in the Fig. 3.5. Notice that the transients die and the steady state is achieved relatively sooner in the case of larger resistance, $\beta=1$. Furthermore, the steady state is reached quicker in the case of larger forcing amplitude. See the variation of steady state amplitudes for different parameters.

Figure 3.5: Forced oscillations with different resistances and forcing amplitudes

Problem 2: The galvanometer: A galvanometer is connected with a constant-current source through a switch. At time t=0, the switch is closed. After some time the galvanometer deflection reaches its final value $\theta_{max}$. Taking damping torque proportional to the angular velocity draw deflection of the galvanometer from the initial position of rest (i.e. $\theta = 0$, $\dot{\theta} = 0$) to its final position $\theta =\theta_{max}$, for the underdamped, critically damped and overdamped cases.

Solution 2: We solve the forced oscillator equation with constant forcing (i.e. driving frequency =0) and given initial conditions and plot the various evolutions. Figure 3.6 shows the galvanometer deflection as a function of time for some arbitrary values of $\theta_{max}$, damping coefficient and natural frequency.

Figure 3.6: Galvanometer deflection

Problems

1. An oscillator with $\omega_0=2 \pi \, {\rm s}^{-1}$ and negligible damping is driven by an external force $F(t)=a \, \cos \omega t$. By what percent do the amplitude of oscillation and the energy change if $\omega$ is changed from $\pi \, {\rm s}^{-1}$ to $3 \pi/2 \, {\rm s}^{-1}$? ( $71.4 \%, 114 \%$)

2. An oscillator with $\omega_0=10^4 \, {\rm s}^{-1}$ and $\beta=1 \, {\rm s}^{-1}$ is driven by an external force $F(t)=a \, \cos \omega t$. [a.] Determine $\omega_{max}$ where the power drawn by the oscillator is maximum? [b.] By what percent does the power fall if $\omega$ is changed by $\Delta \omega=0.5 \, {\rm s}^{-1}$ from $\omega_{max}$?[c.] Consider $\beta=0.1 \, {\rm s}^{-1}$ instead of . ( ([a.] , $33.3 \%$, $96.2 \%$)

3. A mildly damped oscillator driven by an external force is known to have a resonance at an angular frequency somewhere near $\omega=1 {\rm MHz}$ with a quality factor of $1100$. Further, for the force (in Newtons)
   
   $$F(t)=10 \, \cos( \omega t)$$
   
   
   the amplitude of oscillations is $8.26 \,mm$ at $\omega=1.0 \, {\rm KHz}$ and $1 .0\, \mu m$ at $100 \, {\rm MHz}$.

   a.

   What is the spring constant of the oscillator?

   b.

   What is the natural frequency $\omega_0$ of the oscillator?

   c.

   What is the FWHM?

   d.

   What is the phase difference between the force and the oscillations at $\omega=\omega_0+{\rm FWHM}/2$?

4. Show that, $x(t)={f\over{\omega_0^2-\omega^2}}(\cos\omega t-\cos\omega_0 % t)$, is a solution of the undamped forced system, , with initial conditions, . Show that near resonance, $\omega \rightarrow \omega_0$, $x(t)\approx{f\over{2\omega_0}}t\sin\omega_0t$, that is the amplitude of the oscillations grow linearly with time. Plot the solution near resonance. (Hint: Take $\omega=\omega_0-\Delta\omega$ and expand the solution taking $\Delta\omega\rightarrow 0$.)

5. Find the driving frequencies corresponding to the half-maximum power points and hence find the FWHM for the power curve of Fig. 3.4.

6. Show that the average power loss due to the resistance dissipation is equal to the average input power calculated in the expression (3.25).

7. (a) Evaluate average energies at frequencies, $\omega_{AmRes}=\sqrt{\omega_0^2-2\beta^2}$ (at the amplitude resonance) and $\omega_{PoRes}=\omega_0$ (at the power resonance). Show that they are equal and independent of $\omega_0$.
   (b) Find the value of the forcing frequency, , for which the energy of the oscillator is maximum.
   (c) What is the value of the maximum energy?
   ((a) $mf^2/8\beta^2$,
   (b) $\omega_{EnRes}^2=2\omega_0\sqrt{\omega_0^2-\beta^2}-\omega_0^2$, $\omega_{AmRes}<\omega_{EnRes}<\omega_{PoRes}$,
   (c) $mf^2/16(\omega_0\sqrt{\omega_0^2-\beta^2}-\omega_0^2+\beta^2)$.)

8. A massless rigid rod of length $l$ is hinged at one end on the wall. (see figure). A vertical spring of stiffness $k$ is attached at a distance $a$ from the hinge. A damper is fixed at a further distance of $b$ from the spring providing a resistance proportional to the velocity of the attached point of the rod. Now a mass $m(<0.1 ka^2/gl)$ is plugged at the other end of the rod. Write down the condition for critical damping (treat all angular displacements small). If mass is displaced $\theta_0$ from the horizontal, write down the subsequent motion of the mass for the above condition.
   

9. A critically damped oscillator has mass 1 kg and the spring constant equal to 4 N/m. It is forced with a periodic forcing $F(t)=2\cos t\cdot \cos 2t$ N. Write the steady state solution for the oscillator. Find the average power per cycle drawn from the forcing agent.

10. A horizontal spring with a stiffness constant $9$ N/m is fixed on one end to a rigid wall. The other end of the spring is attached with a mass of $1$ kg resting on a frictionless horizontal table. At $t=0$, when the spring -mass system is in equilibrium and is perpendicular to the wall, a force $F(t)=8\cos 5t$ N starts acting on the mass in a direction perpendicular to the wall. Plot the displacement of the mass from the equilibrium position between $t=0$ and $t=2\pi$ neatly.