15.7.1 Lightly Damped Oscillators

A lightly damped oscillator, which oscillates many times before stopping, is one for which b/2m≪ω0. In that case, ω≈ω0 is a good approximation. That is, light damping does not affect the oscillation frequency.

FIGURE 15.22 is a graph of the position x(t) for a lightly damped oscillator, as given by Equation 15.55. To keep things simple, we’ve assumed that the phase constant is zero. Notice that the term Ae−bt/2m, which is shown by the dashed line, acts as a slowly varying amplitude:

xmax(t)=Ae−bt/2m (15.57)

where A is the initial amplitude, at t=0. The oscillation keeps bumping up against this line, slowly dying out with time.

FIGURE 15.22 Position-versus-time graph for a lightly damped oscillator.

The figure illustrates the amplitude versus time graph for damped oscillations.
FIGURE 15.22 Full Alternative Text

A slowly changing line that provides a border to a rapid oscillation is called the envelope of the oscillations. In this case, the oscillations have an exponentially decaying envelope. Make sure you study FIGURE 15.22 long enough to see how both the oscillations and the decaying amplitude are related to Equation 15.55.

Changing the amount of damping, by changing the value of b, affects how quickly the oscillations decay. FIGURE 15.23 shows just the envelope xmax(t) for several oscillators that are identical except for the value of the damping constant b. (You need to imagine a rapid oscillation within each envelope, as in Figure 15.22.) Increasing b causes the oscillations to damp more quickly, while decreasing b makes them last longer.

FIGURE 15.23 Oscillation envelopes for a mass of 1.0 kg with several values of b.

The figure illustrates a position versus time graph with a line and three curves for different values of b.
FIGURE 15.23 Full Alternative Text

MATHEMATICAL ASIDE

Exponential decay

Exponential decay occurs in a vast number of physical systems of importance in science and engineering. Mechanical vibrations, electric circuits, and nuclear radioactivity all exhibit exponential decay.

The mathematical analysis of physical systems frequently leads to solutions of the form

u=Ae−v/v0=A exp(−v/v0)

where exp is the exponential function. The number e=2.71828… is the base of natural logarithms in the same way that 10 is the base of ordinary logarithms.

The figure illustrates an exponential decay graph.

A graph of u illustrates what we mean by exponential decay. It starts with u=A at v=0 (because e0=1) and then steadily decays, asymptotically approaching zero. The quantity v0 is called the decay constant. When v=v0, u=e−1A=0.37A. When v=2v0, u=e−2A=0.13A.

The decay constant v0 must have the same units as v. If v represents position, then v0 is a length; if v represents time, then v0 is a time interval. In a specific situation, v0 is often called the decay length or the decay time. It is the length or time in which the quantity decays to 37% of its initial value.

No matter what the process is or what u represents, a quantity that decays exponentially decays to 37% of its initial value when one decay constant has passed. Thus exponential decay is a universal behavior. The decay curve always looks exactly like the figure shown here. Once you’ve learned the properties of exponential decay, you’ll immediately know how to apply this knowledge to a new situation.

The mechanical energy of a damped oscillator is not conserved because of the drag force. We previously found the energy of an undamped oscillator to be E=12k A2. This is still valid for a lightly damped oscillator if we replace A with the slowly decaying amplitude xmax. Thus

E(t)=12k(xmax)2=12k(Ae−bt/2m)2=12k A2e−bt/m (15.58)

Here A is the initial amplitude, so 12k A2 is the initial energy, which we call E0. Let’s define the time constant τ (also called the decay constant or the decay time) to be

τ=mb (15.59)

Because b has units of kg/s, τ has units of seconds. With this, we can write the energy decay as

E(t)=E0e−t/τ (15.60)

In other words, a lightly damped oscillator’s mechanical energy decays exponentially with time constant τ.

As FIGURE 15.24 shows, the time constant is the amount of time needed for the energy to decay to e−1, or 37%, of its initial value. We say that the time constant τ measures the “characteristic time” during which the energy of the oscillation is dissipated. Roughly two-thirds of the initial energy is gone after one time constant has elapsed, and nearly 90% has dissipated after two time constants have gone by.

FIGURE 15.24 Energy decay of a lightly damped oscillator.

The figure illustrates the energy decay graph of a lightly damped oscillator.
FIGURE 15.24 Full Alternative Text

For practical purposes, we can speak of the time constant as the lifetime of the oscillation—about how long it lasts. Mathematically, there is never a time when the oscillation is “over.” The decay approaches zero asymptotically, but it never gets there in any finite time. The best we can do is define a characteristic time when the motion is “almost over,” and that is what the time constant τ does.

EXAMPLE 15.10 A damped pendulum

A 500 g mass swings on a 60-cm-long string as a pendulum. The amplitude is observed to decay to half its initial value after 35 oscillations.

  1. What is the time constant for this oscillator?

  2. At what time will the energy have decayed to half its initial value?

MODEL The motion is a lightly damped oscillation.

SOLVE

  1. The initial amplitude at t=0 is xmax=A. After 35 oscillations the amplitude is xmax=12A. The period of the pendulum is

    T=2πLg=2π0.60 m9.8 m/s2=1.55 s

    so 35 oscillations have occurred at t=54.2 s.

    The amplitude of oscillation at time t is given by Equation 15.57: xmax(t)=Ae−bt/2m=Ae−t/2τ. In this case,

    12A=Ae−(54.2 s)/2τ

    Notice that we do not need to know A itself because it cancels out. To solve for τ, we take the natural logarithm of both sides of the equation:

    ln(12)=−ln 2=ln e−(54.2 s)/2τ=−54.2 s2τ

    This is easily rearranged to give

    τ=54.2 s2 ln 2=39 s

    If desired, we could now determine the damping constant to be b=m/τ=0.013 kg/s.

  2. The energy at time t is given by

    E(t)=E0e−t/τ

    The time at which an exponential decay is reduced to 12E0, half its initial value, has a special name. It is called the half-life and given the symbol t1/2. The concept of the half-life is widely used in applications such as radioactive decay. To relate t1/2 to τ, we first write

    E(at t=t1/2)=12E0=E0e−t1/2/τ

    The E0 cancels, giving

    12=e−t1/2/τ

    Again, we take the natural logarithm of both sides:

    ln(12)=−ln 2=ln e−t1/2/τ=−t1/2/τ

    Finally, we solve for t1/2:

    t1/2=τ ln 2=0.693τ

    This result that t1/2 is 69% of τ is valid for any exponential decay. In this particular problem, half the energy has been dissipated at

    t1/2=(0.693)(39 s)=27 s

REVIEW The oscillator loses energy faster than it loses amplitude. This is what we should expect because the energy depends on the square of the amplitude.