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3.6 Vibrating Mechanical Systems
Undamped Vibrations
A spring of length l 0 units is suspended from a support. When an object of mass m is attached to the spring, the spring stretches to a length l 1 units. If the object is then pulled down (or pushed up) an additional y 0 units at time t = 0 and then released, what is the resulting motion of the object? That is, what is the position y(t) of the object at time t > 0? Assume that time is measured in seconds
We begin by analyzing the forces acting on the object at time t > 0. First, there is the weight of the object (gravity): F 1 = mg.
This is a downward force. We choose our coordinate system so that the positive direction is down. Next, there is the restoring force of the spring. By Hooke’s Law, this force is proportional to the total displacement l 1 + y(t) and acts in the direction opposite to the displacement: F 2 = −k[l 1 + y(t)] with k > 0.
The constant of proportionality k is called the spring constant. If we assume that the spring is frictionless and that there is no resistance due to the surrounding medium (for example, air resistance), then these are the only forces acting on the object. Under these conditions, the total force is
F = F 1 + F 2 = mg − k[l 1 + y(t)] = (mg − kl 1 ) − ky(t).
Before the object was displaced, the system was in equilibrium, so the force of gravity, mg plus the force of the spring, −kl 1 , must have been 0:
mg − kl 1 = 0.
Therefore, the total force F reduces to
F = −ky(t).
By Newton’s Second Law of Motion, F = ma (force = mass × acceleration), we have
ma = −ky(t) and a = − k m y(t).
Therefore, at any time t we have
a = y′′(t) = − k m y(t) or y′′(t) + k m y(t) = 0.
When the acceleration is a constant negative multiple of the displacement, the object is said to be in simple harmonic motion.
Since k/m > 0, we can set ω =
k/m and write this equation as
y′′(t) + ω^2 y(t) = 0, (1)
a second order linear homogeneous equation with constant coefficients. The characteristic equation is r^2 + ω^2 = 0
and the characteristic roots are ±ωi. The general solution of (1) is
y = C 1 cos ωt + C 2 sin ωt.
In Exercises 3.6 (Problem 5) you are asked to show that the general solution can be written as
y = A sin (ωt + φ 0 ), (2)
where A and φ 0 are constants with A > 0 and φ 0 ∈ [0, 2 π). For our purposes here, this is the preferred form. The motion is periodic with period T given by
T = 2 π ω ,
a complete oscillation takes 2 π/ω seconds. The reciprocal of the period gives the number of oscillations per second. This is called the frequency , denoted by f :
f = ω 2 π.
Since sin (ωt + φ 0 ) oscillates between − 1 and 1,
y(t) = A sin (ωt + φ 0 )
oscillates between −A and A. The number A is called the amplitude of the motion. The number φ 0 is called the phase constant or the phase shift. The figure gives a typical graph of (2).
t
A
A
y
Figure 1
Newton’s Second Law F = ma = my′′^ then gives
my′′(t) = −ky(t) − cy′(t)
which can be written as
y′′^ + c m y′^ + k m y = 0. (c, k, m all constant) (3)
This is the equation of motion in the presence of a damping factor.
The characteristic equation r^2 + (^) m cr + (^) mk = 0
has roots
r = −c^ ±
c^2 − 4 km 2 m. There are three cases to consider:
c^2 − 4 km < 0 , c^2 − 4 km > 0 , c^2 − 4 km = 0.
Case 1: c^2 − 4 km < 0. In this case the characteristic equation has complex roots:
r 1 = − c 2 m
- iω, r 2 = − c 2 m − iω where ω =
4 km − c^2 2 m
The general solution is
y = e(−c/^2 m)t^ (C 1 cos ωt + C 2 sin ωt)
which can also be written as
y(t) = A e(−c/^2 m)t^ sin (ωt + φ 0 ) (4)
where, as before, A and φ 0 are constants, A > 0 , φ 0 ∈ [0, 2 π). This is called the underdamped case. The motion is similar to simple harmonic motion except that the damping factor e(−c/^2 m)t^ causes y(t) → 0 as t → ∞. The oscillations continue indefinitely with constant frequency f = ω/ 2 π but diminishing amplitude Ae(−c/^2 m)t. This motion is illustrated in Figure 2.
t
y
Figure 2
Case 2: c^2 − 4 km > 0. In this case the characteristic equation has two distinct real roots: r 1 = −c +
c^2 − 4 km 2 m ,^ r^2 =^
−c −
c^2 − 4 km 2 m. The general solution is
y(t) = y = C 1 er^1 t^ + C 2 er^2 t. (5) This is called the overdamped case. The motion is nonoscillatory. Since √ c^2 − 4 km <
c^2 = c,
r 1 and r 2 are both negative and y(t) → 0 as t → ∞.
Case 3: c^2 − 4 km = 0. In this case the characteristic equation has only one real root: r 1 = −c 2 m, and the general solution is y(t) = y = C 1 e−(c/^2 m)^ t^ + C 2 t e−(c/^2 m)^ t. (6)
This is called the critically damped case. Once again, the motion is nonoscillatory and y(t) → 0 as t → ∞.
In both the overdamped and critically damped cases, the object moves back to the equilibrium position (y(t) → 0 as t → ∞). The object may move through the equilibrium position once, but only once. Two typical examples of the motion are shown in Figure 3.
t
y
t
y
Figure 3
Forced Vibrations
The vibrations that we have considered thus far result from the interplay of three forces: gravity, the restoring force of the spring, and the retarding force of friction or the surround- ing medium. Such vibrations are called free vibrations.
A typical illustration of the motion is given in Figure 4.
t
y
Figure 4
Exercises 3.
- An object is in simple harmonic motion. Find an equation for the motion given that the period is 14 π and, at time t = 0, y = 1, y′^ = 0. What is the amplitude? What is the frequency?
- An object is in simple harmonic motion. Find an equation for the motion given that the frequency is 1 /π and, at time t = 0, y = 0, y′^ = −2. What is the amplitude? What is the period?
- An object is in simple harmonic motion with period T and amplitude A. What is the velocity at the equilibrium point y = 0?
- An object in simple harmonic motion passes through the equilibrium point y = 0 at time t = 0 and every three seconds thereafter. Find the equation of motion given that y(0) = 5.
- Show that simple harmonic motion y(t) = C 1 cos ωt + C 2 sin ωt can be written as: (a) y(t) = A sin(ωt + φ 0 ); (b) y(t) = A cos(ωt + ψ 0 ).
- What is the effect of an increase in the resistance constant c on the amplitude and frequency of the vibrations given by (4)?
- Show that the motion given by (5) can pass through the equilibrium point at most once. How many times can the motion change directions?
- Show that the motion given by (6) can pass through the equilibrium point at most once. How many times can the motion change directions?
- Show that if γ 6 = ω, then the method of undetermined coefficients applied to (7) gives z = F^0 /m ω^2 − γ^2 cos γt.
- Show that if ω/γ is rational, then the vibrations given by (8) are periodic.
- Show that if γ = ω, then the method of undetermined coefficients applied to (7) gives z = 2 Fωm^0 t sin ωt.
