A mass m is suspended at the end of a spring, its weight stretches the spring
by a length L to reach a static state (the equilibrium position of the system).
Let u(t) denote the displacement, as a function of time, of the mass relative
to its equilibrium position. Recall that the textbook’s convention is that
downward is positive Then, u > 0 means the spring is stretched beyond its
equilibrium length, while u < 0 means that the spring is compressed. The
mass is then set in motion (by any one of several means).
© 2008 Zachary S Tseng B▯3 ▯ 1 The equations that govern a mass-spring system
At equilibrium: (by Hooke’s Law)
mg = kL
While in motion:
m u″ + γ u′ + k u = F(t)
This is a second order linear differential equation with constant coefficients.
It usually comes with two initial conditions: u(t0) = u 0, and u′(t0) = u′ 0.
Summary of terms:
u(t) = displacement of the mass relative to its equilibrium position.
m = mass (m > 0)
γ = damping constant (γ ≥ 0)
k = spring (Hooke’s) constant (k > 0)
g = gravitational constant
L = elongation of the spring caused by the weight
F(t) = Externally applied forcing function, if any
u(t0) = initial displacement of the mass
u′(0) = initial velocity of the mass
© 2008 Zachary S Tseng B▯3 ▯ 2 Undamped Free Vibration (γ = 0, F(t) = 0)
The simplest mechanical vibration equation occurs when γ = 0, F(t) = 0.
This is the undamped free vibration. The motion equation is
m u″ + k u = 0.
The characteristic equation is mr + k = 0. Its solutions are= ± i .
The general solution is then
u(t) = C c1s ω t +0C sin ω2t. 0
Where ω 0 is called the natural frequency of the system. It is the
frequency at which the system tends to oscillate in the absence of any
damping. A motion of this type is called simple harmonic motion.
Comment: Just like everywhere else in calculus, the angle is measured in
radians, and the (angular) frequency is given in radians per second. The
frequency is not given in hertz (which measures the number of cycles or
revolutions per second). Instead, their relation is: 2π radians/sec = 1 hertz.
The (natural) period of the oscillation is given by (seconds).
© 2008 Zachary S Tseng B▯3 ▯ 3 To get a clearer picture of how this solution behaves, we can simplify it with
trig identities and rewrite it as
u(t) = Rcos (ω t 0 δ ).
The displacement is oscillating steadily with constant amplitude of
R = C +C 1 2 .
The angle δ is the phase or phase angle of displacement. It measures how
much u(t) lags (when δ > 0), or leads (when δ < 0) relative to cos(ωt),
which has a peak at t = 0. The phase angle satisfies the relation
tanδ = .
More explicitly, it is calculated by:
δ =tan −1 2, if C > 0,
C 1 1
δ =tan +π , if C1< 0,
δ = , if C = 0 and C > 0,
2 1 2
δ =− π
2 , if C1= 0 and C 2 0,
The angle is undefined if C1= C 2 0.
© 2008 Zachary S Tseng B▯3 ▯ 4 An example of simple harmonic motion:
Graph of u(t) = cos(t) − sin(t)
Amplitude: R = 2
Phase angle: δ = −π/4
© 2008 Zachary S Tseng B▯3 ▯ 5 Damped Free Vibration (γ > 0, F(t) = 0)
When damping is present (as it realistically always is) the motion equation
of the unforced mass▯spring system becomes
m u″ + γ u′ + k u = 0.
Where m, γ, k are all positive constants. The characteristic equation is m r +
γr + k = 0. Its solution(s) will be either negative real numbers, or complex
numbers with negative real parts. The displacement u(t) behaves differently
depending on the size of γ relative to m and k. There are three possible
classes of behaviors based on the possible types of root(s) of the
Case I. Two distinct (negative) real roots
When γ > 4mk, there are two distinct real roots, both are negative. The
displacement is in the form
r t r t
u(t) =Ce +C e 1 2
1 2 .
A mass▯spring system with such type displacement function is called
overdamped. Note that the system does not oscillate; it has no periodic
components in the solution. In fact, depending on the initial conditions the
mass of an overdamped mass▯spring system might or might not cross over
its equilibrium position. But it could cross the equilibrium position at most
© 2008 Zachary S Tseng B▯3 ▯ 6 Figures: Displacement of an Overdamped system
Graph of u(t) = e − e −2t
Graph of u(t) = − e + 2e
© 2008 Zachary S Tseng B▯3 ▯ 7 Case II. One repeated (negative) real root
When γ = 4mk, there is one (repeated) real root. It is negative: r = .
The displacement is in the form
u(t) = C e + C te . rt
A system exhibits this behavior is called critically damped. That is, the
damping coefficient γ is just large enough to prevent oscillation. As can be
seen, this system does not oscillate, either. Just like the overdamped case,
the mass could cross its equilibrium position at most one time.
Comment: The value γ 2= 4mk → γ = 2 mk is called critical damping. It
is the threshold level below which damping would be too small to prevent
the system from oscillating.
© 2008 Zachary S Tseng B▯3 ▯ 8 Figures: Displacement of a Critically Damped system
−t / 2 − t / 2
Graph of u(t) = e + t e
Graph of u(t) = e −t / − t e − t / 2
© 2008 Zachary S Tseng B▯3 ▯ 9 Case III. Two complex conjugate roots
When γ < 4mk, there are two complex conjugate roots, where their common
real part, λ, is always negative. The displacement is in the form
u(t) = C e 1os ▯t + C e sin ▯t. 2 λt
A system exhibits this behavior is called underdamped. The name means
that the damping is small compares to m and k, and as a result vibrations will
occur. The system oscillates (note the sinusoidal components in the
solution). The displacement function can be rewritten as
u(t) = Re cos (▯t − δ ).
The formulas for R and δ are the same as in the previous (undamped free
vibration) section. The displaceλ tt function is oscillating, but the
amplitude of oscillation, Re , is decaying exponentially. For all particular
solutions (except the zero solution that corresponds to the initial conditions
u(t0) = 0, u′( 0 ) = 0), the mass crosses its equilibrium position infinitely
Damped oscillation: u(t) = e cos(2t)
© 2008 Zachary S Tseng B▯3 ▯ 10 The displacement of an underdamped mass▯spring system is a quasi▯periodic
function (that is, it shows periodic▯like motion, but it is not truly periodic
because its amplitude is ever decreasing so it does not exactly repeat itself).
It is oscillating at quasi▯frequency, which is ▯ radians per second. (It’s just
the frequency of the sinusoidal components of the displacement.) The peak▯
to▯peak time of the oscillation is the quasi▯period:q (seconds).
In addition to cause the amplitude to gradually decay to zero, damping has
another, more subtle, effect on the oscillating motion: It immediately
decreases the quasi▯frequency and, therefore, lengthens the quasi▯period
(compare to the natural frequency and natural period of an undamped
system). The larger the damping constant γ, the smaller quasi▯frequency and
the longer the quasi▯period become. Eventually, at the critical damping
threshold, when γ = 4mk , the quasi▯frequency vanishes and the
displacement becomes aperiodic (becoming instead a critically damped
Note that in all 3 cases of damped free vibration, the displacement function
tends to zero as t → ∞. This behavior makes perfect sense from a
conservation of energy point▯of▯view: while the system is in motion, the
damping wastes away whatever energy the system has started out with, but
there is no forcing function to supply the system with additional energy.
Consequently, eventually the motion comes to a halt.
© 2008 Zachary S Tseng B▯3 ▯ 11 Example: A mass of 1 kg stretches a spring 0.1 m. The system has a
damping constant of γ = 14. At t = 0, the mass is pulled down 2 m and
released with an upward velocity of 3.5 m/s. Find the displacement function.
What are the system’s quasi▯frequency and quasi▯period?
m = 1, γ = 14, L = 0.1;
mg = 9.8 = kL = 0.1 k → 98 = k.
The motion equation is u″ + 14u′ + 98u = 0 , and
the initial conditions aru(0) = 2, u′(0) = −3.5 .
The roots of characteristic polynomial are r = −7 ± 7i: