What is Damped Harmonic Motion

Have you ever tried swinging? The ones that move back and forth as you exert force on them with your limbs. But what happens when you stop pushing forward? The swing swings back and forth for a while before coming to a halt. Why is that? When we swing a pendulum, we know that it will ultimately come to rest due to air pressure and friction at the support. This motion is described as damped harmonic motion. 

Another case is we know that the motion of a simple pendulum, swinging in the air, dies out eventually. Why does it happen? This is because the air drag and the friction at the support oppose the motion of the pendulum and dissipate its energy gradually. The pendulum is said to execute damped oscillations. In damped oscillations, the energy of the system is dissipated continuously; but, for small damping, the oscillations remain approximately periodic. The dissipating forces are generally frictional forces.

Equation Of Damped Oscillation

let us consider a system as shown in Fig. 14.19. Here a block of mass \(m\) connected to an elastic spring of spring constant \(k\) oscillates vertically. If the block is pushed down a little and released, its angular frequency of oscillation is
\(\omega_0=\sqrt{\frac{k}{m}}\), as seen in Eq. \((14.20)\).

The damping force depends on the nature of the surrounding medium. If the block is immersed in a liquid, the magnitude of damping will be much greater and the dissipation of energy much faster. The damping force is generally proportional to the velocity of the bob. [Remember Stokes’ Law, Eq. (10.19)] and acts opposite to the direction of velocity. If the damping force is denoted by \(F_d\), we have
\(
\mathbf{F}_d=-b \mathbf{v} \dots(14.30)
\)

where the positive constant \(b\) depends on the characteristics of the medium (viscosity, for example) and the size and shape of the block, etc. Eq. \((14.30)\) is usually valid only for small velocity.

When the mass \(m\) is attached to the spring (hung vertically as shown in Fig. 14.19) and released, the spring will elongate a little and the mass will settle at some height. This position, shown by \(\mathrm{O}\) in Fig 14.19, is the equilibrium position of the mass. If the mass is pulled down or pushed up a little, the restoring force on the block due to the spring is \(\mathbf{F}_{\mathrm{S}}=-k \mathbf{x}\), where \(\mathbf{x}\) is the displacement” of the mass from its equilibrium position. Thus, the total force acting on the mass at any time \(t\), is \(\mathbf{F}=-k \mathbf{x}-b \boldsymbol{v}\).

If \(a(t)\) is the acceleration of mass at time \(t\), then by Newton’s Law of Motion applied along the direction of motion, we have
\(m a(t)=-k x(t)-b v(t) \dots(14.31)\)
Here we have dropped the vector notation because we are discussing the one-dimensional motion.

Using the first and second derivatives of \(x(\mathrm{t})\) for \(v(t)\) and \(a(t)\), respectively, we have
\(
m \frac{\mathrm{d}^2 x}{\mathrm{~d} t^2}+b \frac{\mathrm{d} x}{\mathrm{~d} t}+k x=0 \dots(14.32)
\)

The solution of Eq. (14.32) describes the motion of the block under the influence of a damping force which is proportional to velocity. The solution is found to be of the form
\(
x(t)=A_0 e^{-b t / 2 m} \cos \left(\omega^{\prime} t+\phi\right) \dots(14.33)
\)
where \(A_0\) is the amplitude and \(\omega^{\prime}\) is the angular frequency of the damped oscillator given by,
\(
\omega^{\prime}=\sqrt{\frac{k}{m}-\frac{b^2}{4 m^2}}=\sqrt{\omega_0{ }^2-(b / 2 m)^2} \dots(14.34)
\)
In this function, the cosine function has a period \(2 \pi / \omega^{\prime}\) but the function \(x(t)\) is not strictly periodic because of the factor \(e^{-b t / 2 m}\) which decreases continuously with time.

For small \(b\), the angular frequency \(\omega^{\prime} \approx \sqrt{k / m}=\omega_0\). Thus, the system oscillates with almost the natural angular frequency \(\sqrt{k / m}\) (with which the system will oscillate if there is no damping) and with amplitude decreasing with time according to the equation
\(
A=A_0 e^{-\frac{b t}{2 m}}
\)
The amplitude decreases with time and finally becomes zero. Figure (12.20) shows qualitatively the displacement of the particle as a function of time.

If the damping is large the system may not oscillate at all. If displaced, it will go towards the mean position and stay there without overshooting on the other side. The damping for which the oscillation just ceases is called critical damping.

Mechanical energy of the damped oscillator

For a damped oscillator, the amplitude is not constant but depends on time. For small damping, we may use the same expression but regard the amplitude as \(A_0 e^{-t / 2 m}\)
\(
E(t)=\frac{1}{2} k A_0^2 e^{-b t / m} \dots(14.35)
\)
Equation (14.35) shows that the total energy of the system decreases exponentially with time. Note that small damping means that the dimensionless ratio \(\left(\frac{b}{\sqrt{k m}}\right)\) is much less than 1 Of course, as expected if we put \(b=0\), all equations of a damped oscillator in this section reduce to the corresponding equations of an undamped oscillator. Now the mechanical energy of the undamped oscillator is \(1 / 2 k A_0^2\). 

Example 1: For the damped oscillator shown in Fig. 14.19, the mass \(m\) of the block is \(200 \mathrm{~g}, k=90 \mathrm{~N} \mathrm{~m}^{-1}\) and the damping constant \(b\) is \(40 \mathrm{~g} \mathrm{~s}^{-1}\).
Calculate
(a) the period of oscillation.
(b) time taken for its amplitude of vibrations to drop to half of its initial value.
(c) the time taken for its mechanical energy to drop to half its initial value.

Solution:

(a) We see that \(\mathrm{km}=90 \times 0.2=18 \mathrm{~kg} \mathrm{~N}\) \(\mathrm{m}^{-1}=\mathrm{kg}^2 \mathrm{~s}^{-2}\); therefore \(\sqrt{\mathrm{km}}=4.243 \mathrm{~kg} \mathrm{~s}^{-1}\), and \(b=0.04 \mathrm{~kg} \mathrm{~s}^{-1}\). Therefore, \(b\) is much less than \(\sqrt{\mathrm{km}}\). Hence, the time period Tfrom Eq. (14.34) is given by
\(
\begin{aligned}
T & =2 \pi \sqrt{\frac{m}{k}} \\
& =2 \pi \sqrt{\frac{0.2 \mathrm{~kg}}{90 \mathrm{~N} \mathrm{~m}^{-1}}} \\
& =0.3 \mathrm{~s}
\end{aligned}
\)
(b) Now, from Eq. (14.33), the time, \(T_{1 / 2}\), for the amplitude to drop to half of its initial value is given by,
\(
\begin{aligned}
T_{1 / 2} & =\frac{\ln (1 / 2)}{b / 2 \mathrm{~m}} \\
& =\frac{0.693}{40} \times 2 \times 200 \mathrm{~s} \\
& =6.93 \mathrm{~s}
\end{aligned}
\)
(c) For calculating the time, \(t_{1 / 2}\), for its mechanical energy to drop to half its initial value we make use of Eq. (14.35). From this equation we have,
\(
E\left(t_{1 / 2}\right) / E(0)=\exp \left(-b t_{1 / 2} / m\right)
\)
Or
\(
\begin{aligned}
1 / 2 & =\exp \left(-b t_{1 / 2} / m\right) \\
\ln (1 / 2) & =-\left(b t_{1 / 2} / m\right)
\end{aligned}
\)
Or \(\quad t_{1 / 2}=\frac{0.693}{40 \mathrm{~g} \mathrm{~s}^{-1}} \times 200 \mathrm{~g}\)
\(
=3.46 \mathrm{~s}
\)
This is just half of the decay period for amplitude. This is not surprising, because, according to Eqs. (14.33) and (14.35), energy depends on the square of the amplitude. Notice that there is a factor of 2 in the exponents of the two exponentials.