Mechanical Vibrations
In this chapter we study some applications of the IVP
ay′′ + by′ + cy = g(t), y(0) = y0, y′(0) =
The springmass system: See ﬁgure below.
Figure (A): a spring in rest, with length l.
Figure (B): we put a mass m on the spring, and the spring is stretched. We call length L the elongation.
Figure (C): The springmass system is set in motion by stretch/squueze it extra, with initial velocity, or with external force.
Force diagram at equilibrium position: mg = F s.
Hooke’s law: Spring force F_{s} = −kL, where L =elongation and k =spring constant.
So: we have mg = kL which give
which gives a way to obtain k by experiment: hang a mass m and measure the elongation L.
Model the motion: Let u(t) be the displacement/position of the mass at time t, assuming the origin u = 0 is at the equilibrium position, and downward is the positive direction.
Total elongation: L + u Total spring force: F_{s} = −k(L + u) Other forces:
Total force on the mass:
Newton’s law of motion ma = gives
Since mg = kL, by rearranging the terms, we get
mu′′ + γu′ + ku = F
where m ia the mass, γ is the damping constant, k is the spring constant, and F is the external force.
Next we study several cases.
Case 1: Undamped free vibration (simple harmonic motion). We assume no damping (γ = 0) and no external force (F = 0). So the equation becomes
mu′′ + ku = 0.
Solve it
General solution
u(t) = c_{1 }cos ω_{0}t + c_{2} sin ω_{0}t.
Four terminologies of this motion: frequency, period, amplitude and phase, deﬁned below.
Frequency:
Period:
Amplitude and phase: We need to work on this a bit. We can write
Now, deﬁne
so we have
A trick to memorize the last term formula: Consider a right triangle, with c_{1} and c_{2} as the sides that form the right angle. Then, the amplitude equals to the length of the hypotenuse, and the phase δ is the angle between side c_{1} and the hypotenuse. Draw a graph and you will see it better.
Problems in this part often come in the form of word problems. We need to learn the skill of extracting information from the text and put them into mathematical terms.
Example 1. A mass weighing 10 lb stretches a spring 2 in. We neglect damping. If the mass is displaced an additional 2 in, and is then set in motion with initial upward velocity of 1 ft/sec, determine the position, frequency, period, amplitude and phase of the motion.
Answer.We see this is free harmonic oscillation. The equation is
mu′′ + ku = 0
with initial conditions (Pay attention to units!)
We need ﬁnd the values for m and k. We have
To ﬁnd k, we see that the elongation is By Hooke’s law, we have
kL = mg, k = mg/L = 60.
Put in these values, we get the equation
So the frequency is and the general solution is
u(t) = c_{1} cos ω_{0}t + c_{2} sin ω_{0}t
We can ﬁnd c_{1} , c_{2} by the ICs:
(Note that c_{1} = u(0) and c_{2} = u′ (0)/ω_{0} .) Now we have the position at any time t
The four terms of the motion are
and
Case II: Damped free vibration. We assume that γ ≠ 0(> 0) and F = 0.
mu′′ + γ u′ + ku = 0
then
We see the type of root depends on the sign of the discriminant − 4km.
So the position function is
This motion is called damped oscillation. We can rewrite it as
Here the term e^{−λt }R is the amplitude, and µ is called the quasi frequency, and the quasi period is The graph of the solution looks like the one for complex roots with negative real part.
Summary: For all cases, since the real part of the roots are always negative, u will go to zero as t grow. This means, if there is damping, no matter how big or small, the motion will eventually come to a rest.
Example 2. A mass of 1 kg is hanging on a spring with k = 1. The mass is in a medium that exerts a viscous resistance of 6 newton when the mass has a velocity of 48 m/sec. The mass is then further stretched for another 2m, then released from rest. Find the position u(t) of the mass.
Answer. We have So the equation for u is
Solve it
By ICs, we have u(0) = c_{1} = 2, and
So the position at any time t is
Forced Vibrations
In this chapter we assume the external force is F (t) = F_{0} cos ωt. (The case where F (t) = F_{0 }sin ωt is totally similar.) The reason for this particular choice of force will be clear later when we learn Fourier series, i.e., we represent periodic functions with the sum of a family of sin and cos functions.
Case 1: With damping.
mu′′ + γ u′ + ku = F_{0} cos ωt.
Solution consists of two parts:
u(t) = uH (t) + U (t), uH (t): the solution of the homogeneous equation, U (t): a particular solution for the nonhomogeneous equation.
From discussion is the previous chapter, we know that uH → 0 as t → +∞ for systems with damping. Therefore, this part of the solution is called the transient solution.
The appearance of U is due to the force term F . Therefore it is called the forced response.
The form of this particular solution is U (t) = A_{1} cos ωt + A_{2} sin ωt. As we have seen, we can rewrite it as U (t) = R cos(ωt − δ) where R is the amplitude and δ is the phase.
We see it is a periodic oscillation for all time t.
As time t → ∞, we have u(t) → U (t). So U (t) is called the steady state.
Case 2: Without damping. The equation now is
mu′′ + ku = F_{0} cos ωt
Let w_{0} = denote the system frequency (i.e., the frequency for the free oscillation). The homogeneous solution is
uH (t) = c_{1} cos ω_{0}t + c_{2} sin ω_{0}t.
The form of the particular solution depends on the value of w. We have two cases.
Case 2A: if w ≠ w_{0} . The particular solution is of the form
U = A cos wt.
(Note that we did not take the sin wt term, because there is no u′ term in the equation.) And U ′′ = −w^{2}A cos wt. Plug these in the equation
m(−w^{2}A cos wt) + kA cos wt = F_{0} cos wt,
Note that if w is close to w_{0 }, then A takes a large value.
General solution u(t) = c_{1} cos w_{0} t + c_{2} sin w_{0}t + A cos wt where c_{1} , c_{2} will be determined by ICs.
Now, assume ICs:
u(0) = 0, u′(0) = 0.
Let’s ﬁnd c_{1} , c_{2} and the solution:
u(0) = 0 : c_{1 }+ A = 0, c_{1} = −A u′ (0) = 0 : 0 + w_{0}c_{2 }+ 0 = 0, c_{2} = 0
Solution
u(t) = −A cos w_{0}t + A cos wt = A(cos wt − cos w_{0}t).
We see that the solution consists of the sum of two cosine functions, with diﬀerent frequencies. In order to have a better idea of how the solution looks like, we apply some manipulation.
Recall the trig identity:
We now have
Since both w0 , w are positive, then w0 + w has larger value than w0 − w. Then, the ﬁrst term can be viewed as the varying amplitude, and the second term
is the vibration/oscillation.
One particular situation of interests: if w_{0} ≠ w but they are very close wo ≈ w, then we have w_{0 }− w << w_{0 }+ w, meaning that w_{0 }− w is much smaller than w_{0} + w. The plot of u(t) looks like (we choose w_{0} = 9, w = 10)
This is called a beat. (One observes it by hitting a key on a piano that’s not tuned, for example.)
Case 2B: If w = w_{0} . The particular solution is U = At cos w_{0}t + B t sin w_{0}t
A typical plot looks like:
This is called resonance. If the frequency of the source term ω equals to the frequency of the system ω_{0}, then, small source term could make the solution grow very large!
One can bring down a building or bridge by small periodic perturbations.
Historical disasters such as the French troop marching over a bridge and the bridge collapsed. Why? Unfortunate for the French, the system frequency of the bridge matches the frequency of their footsteps.
Hermite and Laguerre Special Functions  1 Doc  12 pages 
Hermite and Laguerre Special Functions  2 Doc  6 pages 
Hermite and Laguerre Special Functions  3 Doc  3 pages 
1. What is a linear ordinary differential equation (ODE) of first order? 
2. How do you solve a firstorder linear ODE? 
3. What is a linear ordinary differential equation (ODE) of second order? 
4. How do you solve a secondorder linear ODE? 
5. How are linear ODEs used in physics? 
Hermite and Laguerre Special Functions  1 Doc  12 pages 
Hermite and Laguerre Special Functions  2 Doc  6 pages 
Hermite and Laguerre Special Functions  3 Doc  3 pages 

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