## Section13.7Dissipation of Energy and Q Factor

### Subsection13.7.1Energy of an Underdamped Oscillator

Just as for any other system, the energy of a damped oscillator is the sum of its kinetic and potential energies. We will consider only the underdamped case, since, it is the only one that actually oscillates - the other two cases, the overdamped and the critically damped, do not oscillate.

$$E = \dfrac{1}{2}m v^2 + \dfrac{1}{2}k x^2,\label{eq-Energy-of-an-Underdamped-Oscillator-main}\tag{13.7.1}$$

where $x$ and $v$ are for underdamped oscillator are

\begin{align*} \amp x = A\,e^{-\gamma t/2}\, \cos( \omega^{\prime} t + \phi ), \\ \amp v = -\dfrac{\gamma}{2}\, x - \omega^{\prime}\,A\,e^{-\gamma t/2}\, \sin( \omega^{\prime} t + \phi ), \end{align*}

When you insert these in Eq. (13.7.1), you get a really ugly expression, which shows that energy of the oscillator oscillates in time with decreasing amplitudes, but at twice the frequency of the oscillation of $x \text{.}$

But if you have a lightly damped oscillator, i.e., if $\gamma \lt\lt \omega \text{,}$ meaning that the tendency to damp is considerably small compared to the tendency to oscillate, then, we can show that, on average, energy dies out exponentially with time.

$$E(t) \approx E(0)\, e^{-\gamma\, t},\label{eq-dissipation-of-energy-energy-t}\tag{13.7.2}$$

where $E(0)$ is the energy at initial time, which for a lightly damped oscillator is

\begin{equation*} E(0) \approx \dfrac{1}{2}k A^2. \end{equation*}

Eq. (13.7.2) shows that energy of a lightly damped oscillator reduces to about one-third in time, $\tau\text{,}$ equal to inverse of $\gamma\text{.}$

\begin{equation*} \tau = \dfrac{1}{\gamma}. \end{equation*}

This $\tau$ is sometimes called time constant for energy decay. It gives us time for energy to decay to $1/e\approx 1/2.7$ of the original energy.

### Subsection13.7.2Quality Factor of Underdamped Oscillator

Every physical system that oscillates, will have damping. That is, real oscillating systems are actually underdamped oscillators. Since, the amplitude of oscillations decreases with time, an important feature of a good oscillator will be to oscillate for many cycles before losing significant amount of energy.

Quality factor provides a standard way to talk about the goodness of an oscillator. We denote Quality factor by $Q\text{.}$ A simple measure of Quality of an oscillator would be to divide the average energy in a particular cycle by the energy lost in the cycle, so that the less the energy is lost in a cycle, the more persistent are the oscillations, and therefore, higher the Quality.

For technical reasons, we define $Q$ factor by dividing the energy at the beginning of a cycle to the energy lost in $\frac{1}{2\pi}$ of a cycle rather than one cycle. The factor is not critical to the definition, since when we use $Q$ factor, we are mostly interested in order of magnitude. But with $\frac{1}{2\pi}$ of a cycle, we do get simple formula for $Q$ when we apply this to a lightly damped oscillator as we will see below. Suppose $E(t)$ be the energy at time $t\text{,}$ then we would compute $Q$ by

$$Q = \dfrac{E(t)}{\left| \Delta E\left(\text{during }t\text{ to }t+ \dfrac{1}{\omega^\prime}\right) \right| }.\tag{13.7.3}$$

You could think of this definition as also telling us about amount of energy lost in $\frac{1}{2\pi}$ of a cycle.

\begin{equation*} \left| \Delta E \text{ lost in next 1/2}\pi\text{ of a cycle} \right| = Q E. \end{equation*}

### Subsection13.7.3Quality Factor of a Lightly Damped Oscillator

The exact equations given above for $Q$ of an underdamped oscillator are too complicated to be of much use, especilly since, most times, we use the concept of $Q$ to characterize relatively good oscillators. These oscillators are called light damped oscillators. We identify an oscillator to be a lightly damped if the following condition holds.

$$\gamma \lt\lt \omega^\prime.\tag{13.7.4}$$

Under these conditions, energy on average decreases exponentially and is given by

$$E(t) = E(0) e^{-\gamma t}.\tag{13.7.5}$$

It is then easy to work out energy $\Delta E$ lost during $t$ to $t + \frac{1}{\omega} \text{,}$ i.e., within $1/2\pi$ of one cycle. Dividing $E(t)$ by $\Delta E$ and simplifying, we get quality factor to be:

$$Q = \frac{ E(0)e^{-\gamma t} }{ E(0)e^{-\gamma (t+1/\omega^\prime)} - E(0)e^{-\gamma t} } = \frac{1}{ e^{-\gamma/\omega^\prime} -1 } \approx \dfrac{\omega^\prime}{\gamma}\tag{13.7.6}$$

Good oscillators such as tuning forks and guitar strings have $Q$ values in the thousands. Laser cavities have much higher $Q$ values, exceeding $10^7\text{.}$ Of course, the undamped oscillator has zero $\gamma\text{,}$ and hence infinite $Q\text{.}$ There is no $Q$ for the critically damped and overdamped cases since they do not oscillate. You might say, their $Q = 0\text{.}$

A copper block of mass $1.5\text{ kg}$ is attached to a spring of stiffness $450\text{ N/m}$ and hung from a platform above a beaker that contains a thick liquid so that the block oscillates entirely in the liquid with a damping constant of $3.0\text{ kg/s}\text{.}$ What is the Quality of this oscillator?

Hint

Check if lightly damped condition applies

8.7

Solution

Let us first find out if the oscillator is a lightly damped oscillator.

\begin{align*} \amp \omega = \sqrt{\dfrac{k}{m}} = \sqrt{\dfrac{450}{1.5}} = 17.32. \\ \amp \gamma = \dfrac{b}{m} = \dfrac{3.0}{1.5} = 2. \end{align*}

Since $2 \lt\lt 17.32 \text{,}$ we can treat this oscillator as a lightly damped oscillator. Therefore, we use simpler formula for $Q\text{.}$

\begin{equation*} Q = \dfrac{\omega}{\gamma} = \dfrac{17.32}{2} = 8.7. \end{equation*}

A piano technician uses a $440\text{-Hz}$ standard tuning fork to tune piano strings. She notices that after $2\text{ sec}$ only $\dfrac{1}{3}$ of sound energy remains. What is the Q of the tuning fork?

Hint

Use energy disspation to find $\gamma\text{.}$

$5000$

Solution

From the rate at which energy dissipates, we can get $\gamma\text{.}$

\begin{equation*} e^{-\gamma \times 4} = \dfrac{1}{3}. \end{equation*}

Taking natural log of both sides gives

\begin{equation*} -2\, \gamma = \ln(1/3)=-\ln(3). \end{equation*}

Therefore,

\begin{equation*} \gamma = \dfrac{\ln(3)}{2} = 0.55\text{ s}^{-1} \end{equation*}

Using this in the formula for $Q$ of underdamped oscillators, we get

\begin{equation*} Q = \dfrac{\omega}{\gamma} = \dfrac{2\pi\times 440\text{ s}^{-1} }{0.55\text{ s}^{-1}} \approx 5000. \end{equation*}

A copper block of mass $1.5\text{ kg}$ is attached to a spring of stiffness $450\text{ N/m}$ and hung from a platform above a beaker that contains a thick liquid so that the block oscillates entirely in the liquid with a damping constant of $45.0\text{ kg/s}\text{.}$ What is the Quality of this oscillator?

Hint

Check if lightly damped condition applies.

0.3

Solution

Let us first find out if the oscillator is a lightly damped oscillator.

\begin{align*} \amp \omega = \sqrt{\dfrac{k}{m}} = \sqrt{\dfrac{450}{1.5}} = 17.32. \\ \amp \gamma = \dfrac{b}{m} = \dfrac{45.0}{1.5} = 30. \end{align*}

Since $\omega \gt \gamma/2\text{,}$ this is an underdamped oscillator, but since $17.32 \sim 30/2 \text{,}$ it is not lightly damped. We will then find $\omega^{\prime}\text{.}$

\begin{equation*} \omega^{\prime} = \sqrt{17.32^2 - 15^2 } = 8.7. \end{equation*}

Therefore, we us a loosely approximate formula for $Q\text{.}$

\begin{equation*} Q \sim \dfrac{\omega^{\prime}}{\gamma} = \dfrac{8.7}{30} = 0.3. \end{equation*}

This is not a very good oscillator and $Q$ of $0.3$ has little practical use.

An oscillator has $Q=100\text{.}$ What fraction of energy does the oscillator lose every cycle?

Hint

Use the fact that $1/Q$ lost in $1/2\pi$ of a cycle.

0.063, i.e., 6.3%.

Solution

Recall that $E_0/Q$ is the amount of energy the oscillator loses in $1/2\pi$ of a cycle. Therefore, in one cylce fractional loss is

\begin{equation*} \dfrac{2\pi\times E_0/Q}{E_0} = \dfrac{2\pi}{Q} = \dfrac{2\pi}{100}=0.063. \end{equation*}

That is, the loss of energy in each cycle is 6.3%.

Find the $Q$ factor for the following underdamped oscillators, where $x$ is in $\text{cm}$ and $t$ in $\text{sec}\text{.}$

1. $\displaystyle x(t) = 2 \exp{(-0.1\, t)}\cos(2\pi t)$
2. $\displaystyle x(t) = 2 \exp{(-0.02\, t)}\cos(t)$
3. $\displaystyle x(t) = 2 \exp{(-0.25\, t)}\cos(200\pi t)$
4. $\displaystyle x(t) = 2 \exp{(-0.015\, t)}\cos\left(2000\pi t+\frac{\pi}{2}\right)$
Hint

Read off $\gamma$ and $\omega^\prime\text{.}$

(a) 31.4; (b) 25; (c) 1,257; (d) 209,440.

Solution

(a) We read off $\gamma$ and $\omega^\prime$ from the given solution.

\begin{equation*} \dfrac{\gamma}{2} = 0.1\text{ sec}^{-1},\ \ \omega^\prime=2\pi\text{ sec}^{-1}. \end{equation*}

Therefore,

\begin{equation*} Q = \dfrac{\omega^\prime}{\gamma} = 10\pi = 31.4 \end{equation*}

(b) Here,

\begin{equation*} \dfrac{\gamma}{2} = 0.02\text{ sec}^{-1},\ \ \omega^\prime=1\text{ sec}^{-1}. \end{equation*}

Therefore,

\begin{equation*} Q = \dfrac{\omega^\prime}{\gamma} = 25. \end{equation*}

(c)

\begin{equation*} Q = \dfrac{200\pi}{0.5} = 1,257. \end{equation*}

(d)

\begin{equation*} Q = \dfrac{2000\pi}{0.03} = 209,440. \end{equation*}

The position of a 250-gram lightly damped oscillator is given by the following function of time, $x(t) = 2\exp{(-0.01\,t)}\cos(2\pi t)\text{,}$ where $t$ is in seconds and $x$ in meters.

(a) Plot $x$ vs $t$ for at least 4 cycles.

(b) How long does it take for the envelop of oscillations to drop by $\frac{1}{e}\text{?}$

(c) How long does it take the envelop to drop by a factor $\frac{1}{e^2}\text{?}$

(d) What is the $Q$ factor of the oscillator?

(e) What is the rate at which the energy of the oscillator is dissipated at $t =0\text{?}$

(f) What is the frequency of oscillations in $\text{Hz}\text{?}$

(g) How many oscillations will the oscillator make before 90% of the energy is dissipated?

Hint
Just the definitions.

(b) $100\text{ sec}\text{;}$ (c) $200\text{ sec}\text{;}$ (d) $314\text{;}$ (e) $0.375\text{ J/s}\text{;}$ (f) $1\text{ Hz}\text{;}$ (g) $115\text{.}$

Solution 1 (a)

(a) Skipped, for now.

Solution 2 (b)

(b) The envelop goes as $\sim \exp(-0.01\ t) \text{.}$ Therefore, drops by $1/e$ in time:

\begin{equation*} t = \dfrac{1}{0.01} = 100\text{ sec}. \end{equation*}
Solution 3 (c)

(c) The envelop goes as $\sim \exp(-0.01\ t) \text{.}$ Therefore, drops by $1/e^2$ in time:

\begin{equation*} t = \dfrac{2}{0.01} = 200\text{ sec}. \end{equation*}
Solution 4 (d)

(d) The $Q$ from $\omega^\prime/\gamma$ is

\begin{equation*} Q = \dfrac{2\pi}{0.02} = 31.4. \end{equation*}
Solution 5 (e)

(e) The $\Delta E /\Delta t = -\gamma E(t)$ gives

\begin{equation*} -\gamma E(0) = -0.02\times \dfrac{1}{2}kA^2, \end{equation*}

where $k=\omega^2 m$ with $\omega^2 = {\omega'}^2 + \gamma^2$ and $m=0.25\text{ kg}\text{,}$ and $A=2\text{ m}\text{.}$

\begin{equation*} \omega^2 = 4\pi^2 + 0.02^2 = 39.5\text{ sec}^{-2}. \end{equation*}

Therefore,

\begin{equation*} k = 39.5\times 0.25 = 9.875\text{ N/m}. \end{equation*}

Therefore, rate of energy loss

\begin{equation*} -0.02\times \dfrac{1}{2}\times 9.875 \times 2^2 = -0.395\text{ J/s}. \end{equation*}
Solution 6 (f)

(f) The frequency is obtained from the argument of cosine here.

\begin{equation*} f = \dfrac{\omega^\prime}{2\pi} = 1\text{ Hz}. \end{equation*}
Solution 7 (g)

(g) Since only 10% energy is left, we need to solve the following condition for $t\text{.}$

\begin{equation*} E(t) = 0.1\,E(0), \text{ where } E(t) = E(0)\exp(-0.02\, t), \end{equation*}

since $\gamma = 0.02\text{ sec}^{-1}\text{.}$ Therefore

\begin{equation*} \exp(-0.02\, t) = 0.1. \end{equation*}

Taking natural log both sides we get

\begin{equation*} t = -\dfrac{\ln(0.1)}{0.02} = 115\text{ sec}. \end{equation*}

Now, we divide this time by the period of oscillations, which is $1 \text{ sec}\text{.}$ Therefore, we have $115$ oscillations.

A car of mass $3,500\text{ kg}$ has a suspension system with an effective spring constant of $4.0\times 10^3\text{ N/m} \text{.}$ When new, the damping constant corresponded to critical damping. But, with use, the damping had deteriorated and now the damping is underdamped. Car now takes $1.5\text{ s}$ to damp out 90% of initial energy from a bump in the road. Suppose the shock absorber is designed to remove 90% of initial energy in $1.5\text{ sec}\text{.}$

(a) How much energy does the shock absorber remove in $1.5\text{ sec}$ if the initial displacement is $10\text{ cm}\text{?}$

(b) What is the $Q$ of the suspension system now?

Hint

(a) Initially energy is all potential energy. (b) Use the fact that 90% energy dissipates in 1.5 sec to find $\gamma\text{.}$

(a) $200\text{ J} \text{,}$ (b) $15\text{.}$

Solution

(a) The energy initially is all potential energy.

\begin{equation*} E = \dfrac{1}{2}kA^2 = \dfrac{1}{2}\times 4.0\times 10^4\text{ N/m} \times (0.10\text{ m})^2 = 200\text{ J}. \end{equation*}

(b) From the fact the 90% of energy is $1.5\text{ sec}\text{,}$ we have the following equation.

\begin{equation*} e^{-1.5\,\gamma} = 0.9. \end{equation*}

Taking natural log of both sides we get

\begin{equation*} \gamma = -\dfrac{\ln(0.9)}{1.5} = 0.07\text{ s}^{-1}. \end{equation*}

The angular frequency,

\begin{equation*} \omega^\prime = \sqrt{ k^2/m^2 - (\gamma/2)^2 } = 1.07\text{ s}^{-1}. \end{equation*}

Therefore, $Q$ is

\begin{equation*} Q = \dfrac{1.07}{0.07} =15. \end{equation*}