A child on a jhoola in a garden is pushed high, and then the pushing stops. The swings grow smaller and smaller, and after a minute or two the jhoola hangs almost still. Yet a friend counting "one, two" finds that every swing, large or small, takes about the same time.
Where the energy goes
An ideal SHM goes on for ever with the same amplitude, because its mechanical energy \(E = \tfrac12 kA^2\) stays constant. A real oscillator is never alone. Air drags on a pendulum bob, the pivot rubs, and a spring warms a little as it flexes. Each of these forces opposes the motion, so each does negative work. From the chapter Potential Energy and Conservation of Energy, the work of the non-conservative forces equals the change in mechanical energy: \(W_{nc} = \Delta E\). So \(E\) falls, and since \(E = \tfrac12 kA^2\), the amplitude falls too.
Damping is the steady loss of the mechanical energy of an oscillator to forces that oppose its motion. Such forces are friction, air drag and forces inside the material. Damping makes the amplitude of the oscillation decrease with time. An oscillation losing energy in this way is a damped oscillation.
The energy does not vanish. It becomes heat in the air, the pivot and the spring, and a little of it leaves as sound. A guitar string that sounds and fades has passed its energy to the air as sound and to its body and pegs as heat.
What causes damping in an oscillating system?
A drag that grows with the speed
A body moving slowly through air or a liquid feels a drag roughly proportional to its speed and opposite to its velocity. Write it as \(F_d = -bv\). The constant \(b\) is the damping constant: the larger it is, the stronger the drag. Its unit is N per (m/s), which is kg/s. This simple law describes a pendulum in air and a block moving in oil quite well.
Put this drag beside the spring force on a block of mass \(m\) on a spring of constant \(k\). Newton's second law gives \(m\,d^2x/dt^2 = -kx - b\,dx/dt\). Move every term to one side.
For small \(b\) the solution of this equation is an SHM whose amplitude shrinks exponentially: \[ x = A_0\,e^{-bt/2m}\sin(\omega' t + \varphi), \qquad \omega' = \sqrt{\dfrac{k}{m} - \dfrac{b^2}{4m^2}}. \] Here \(A_0\) is the amplitude at \(t = 0\), and \(\omega_0 = \sqrt{k/m}\) is the natural frequency of the undamped oscillator, in rad/s. The motion is a sine curve squeezed between the two curves \(\pm A_0e^{-bt/2m}\), called the envelope.
To check the solution, put \(x = A_0e^{-\gamma t}\sin\omega' t\) into the equation, with \(\gamma = b/2m\). The terms in \(e^{-\gamma t}\cos\omega' t\) cancel only if \(\gamma = b/2m\). The terms in \(e^{-\gamma t}\sin\omega' t\) then cancel only if \(\omega'^2 = k/m - \gamma^2\). Both conditions are met, so the trial solution is exact.
Set \(b = 0.2\,\text{kg/s}\) in the figure above. The amplitude halves every \(6.9\,\text{s}\), yet each swing still takes \(1.00\,\text{s}\). Raise \(b\) to \(1\,\text{kg/s}\): the amplitude now halves in only \(1.4\,\text{s}\), while the period grows by just \(0.3\,\%\). Damping attacks the amplitude strongly and the period hardly at all. This is the jhoola of the opening: its swings shrink, but each one keeps its time.
What is the main effect of a light damping force on a simple harmonic oscillator?
A \(0.2\,\text{kg}\) block on a spring of \(20\,\text{N/m}\) oscillates in air with damping constant \(0.04\,\text{kg/s}\), starting with amplitude \(5\,\text{cm}\). Find the natural and damped angular frequencies, the time for the amplitude to halve, and the number of oscillations in that time. Take \(\ln 2 = 0.693\).
\(m = 0.2\,\text{kg}\), \(k = 20\,\text{N/m}\), \(b = 0.04\,\text{kg/s}\). The amplitude is \(A_0e^{-bt/2m}\), and it halves when \(e^{-bt/2m} = 1/2\).
\(\omega_0 = \sqrt{k/m}\), \(\omega' = \sqrt{\omega_0^2 - (b/2m)^2}\), and \(bt_{1/2}/2m = \ln 2\).
\(\omega_0 = \sqrt{100} = 10\,\text{rad/s}\) and \(b/2m = 0.04/0.4 = 0.1\,\text{s}^{-1}\). Then \(\omega' = \sqrt{100 - 0.01} \approx 9.9995\,\text{rad/s}\), and \(t_{1/2} = 0.693/0.1 = 6.93\,\text{s}\). The period is \(2\pi/\omega' \approx 0.628\,\text{s}\), so \(6.93/0.628 \approx 11\) oscillations fit in that time.
The amplitude falls from \(5\,\text{cm}\) to \(2.5\,\text{cm}\) in \(6.93\,\text{s}\), over about \(11\) oscillations, while the frequency stays \(10\,\text{rad/s}\) to four figures.
The amplitude of a damped oscillator falls to half its first value in \(10\,\text{s}\). How long does it take to fall to one quarter of its first value, in seconds?
The energy falls twice as fast
For light damping each oscillation is nearly an SHM, so its energy is still close to \(\tfrac12 kA^2\) with the current amplitude. The energy goes as the square of the amplitude, so its decay is squared too.
\[ E = \tfrac12 kA^2 \]
Why. the energy of an SHM of amplitude \(A\), from the chapter Dynamics of SHM.
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Your turn. square the amplitude \(A = A_0e^{-bt/2m}\). Which line is it?
\[ E = \tfrac12 kA_0^2\,e^{-bt/m} = E_0\,e^{-bt/m} \]
Why. substitute step 2 into step 1, with \(E_0 = \tfrac12 kA_0^2\) the energy at \(t = 0\).
\[ t_{1/2}(E) = \dfrac{m\ln 2}{b} = \tfrac12\,t_{1/2}(A) \]
Read it. \(E\) halves when \(bt/m = \ln 2\), which takes half the time that the amplitude needs.
The rate of loss follows at once: \(P = -dE/dt = (b/m)E\). The oscillator loses energy fastest when it has the most, at the start. In the example above, the block starts with \(E_0 = \tfrac12 \times 20 \times 0.05^2 = 0.025\,\text{J}\). After \(6.93\,\text{s}\) the amplitude has halved, so the energy is a quarter, \(0.00625\,\text{J}\).
The amplitude of a lightly damped pendulum falls to half in \(4\,\text{s}\). How long does its energy take to fall to half, in seconds?
Counting the loss swing by swing
Look at the maxima of a damped oscillation, one period \(T' = 2\pi/\omega'\) apart. Each maximum is \(e^{-bT'/2m}\) times the one before, whatever its size. So the ratio of any maximum to the next is a constant, \(A_n/A_{n+1} = e^{bT'/2m}\). Its natural logarithm is called the logarithmic decrement, and it is written \(\delta\) here.
The swings of a damped pendulum to the right reach \(5.0\), \(4.0\) and \(3.2\,\text{cm}\) in turn. Its period is \(0.5\,\text{s}\). Find the logarithmic decrement, the value of \(b/2m\), and the swing five periods after the \(3.2\,\text{cm}\) one.
The ratios are \(5.0/4.0 = 1.25\) and \(4.0/3.2 = 1.25\): the same, as expected.
\(\delta = \ln 1.25 \approx 0.223\), and \(b/2m = \delta/T' = 0.223/0.5 \approx 0.446\,\text{s}^{-1}\).
Five more periods divide the swing by \(1.25^5 \approx 3.05\), so it is \(3.2/3.05 \approx 1.05\,\text{cm}\).
Too much damping stops the oscillation
The formula \(\omega' = \sqrt{\omega_0^2 - (b/2m)^2}\) carries a warning. As \(b\) grows, \(\omega'\) shrinks. When \(b/2m\) reaches \(\omega_0\), that is at \(b = 2\sqrt{mk}\), \(\omega'\) becomes zero. There is then no oscillation at all. The body released from an extreme only slides back towards the mean position. Three cases follow.
- For \(b < 2\sqrt{mk}\) the motion is underdamped: the body oscillates with a shrinking amplitude, crossing the mean position again and again.
- For \(b = 2\sqrt{mk}\) the motion is critically damped: the body returns to the mean position in the least possible time without crossing it.
- For \(b > 2\sqrt{mk}\) the motion is overdamped: the body returns without crossing, but slowly, because the heavy drag holds it back.
Three identical blocks, each on a spring of natural period \(1\,\text{s}\), are pulled \(4\,\text{cm}\) aside and released in three liquids. The damping constants are \(1.3\), \(12.6\) and \(37.7\,\text{kg/s}\). Which block is the first to settle within \(0.2\,\text{cm}\) of the mean position and stay there?
Critical damping is the designer's target whenever a body must come to rest quickly and without swinging. The needle of a meter should settle on its reading at once. A car should settle after a pothole without bouncing, and a door closer should shut a door firmly but without a bang.
One wheel of a car carries \(250\,\text{kg}\) on a suspension spring of \(40\,000\,\text{N/m}\). What damping constant makes this motion critically damped, in kg/s?
The needle of an electrical meter must come to rest at its reading as quickly as possible, without swinging past it. What damping should the designer choose?
Each wheel of an auto-rickshaw sits on a spring, and beside the spring is a shock absorber: a piston that pushes oil through small holes. The oil gives a drag proportional to the speed of the piston, exactly a \(-bv\) force. Without it, a pothole would set the rickshaw bouncing for many seconds. With \(b\) near \(2\sqrt{mk}\), the body rises once and settles, and the passengers stay in their seats.
- Thinking that damping changes the period a lot.For light damping \(\omega' = \sqrt{\omega_0^2 - (b/2m)^2}\) is almost \(\omega_0\). The amplitude changes; the period barely does.
- Using \(e^{-bt/2m}\) for the energy.The energy goes as \(A^2\), so it falls as \(e^{-bt/m}\), twice as fast as the amplitude.
- Thinking that the heaviest damping stops a body fastest.Critical damping is fastest. Heavier damping removes the overshoot but makes the return slow.