** Higher order linear ODEs**

**2.4. MECHANICAL VIBRATIONS 81 Note that the errors that we get from the approximation build up So after a very long time, the**

behavior of the real system might be substantially different from our solution. Also we will see that in a mass-spring system, the amplitude is independent of the period. This is not true for a pendulum. Nevertheless, for reasonably short periods of time and small swings (for example if the pendulum is very long), the approximation is reasonably good.

In real world problems it is often necessary to make these types of simplifications. We must understand both the mathematics and the physics of the situation to see if the simplification is valid in the context of the questions we are trying to answer.

### 2.4.2

### Free undamped motion

In this section we will only consider free or unforced motion, as we cannot yet solve nonhomoge- neous equations. Let us start with undamped motion wherec= 0. We have the equation

mx00+kx= 0. If we divide bymand letω0=

√

k/_{m}_{, then we can write the equation as}

x00+ω2_{0}x= 0.
The general solution to this equation is

x(t)= Acos(ω0t)+Bsin(ω0t).

By a trigonometric identity, we have that for two different constantsC andγ, we have Acos(ω0t)+Bsin(ω0t)=Ccos(ω0t−γ).

It is not hard to compute thatC = √

A2+_{B}2 _{and tan}γ = B/_{A}_{. Therefore, we let}_{C} _{and}γ _{be our}

arbitrary constants and writex(t)=Ccos(ω0t−γ).

Exercise2.4.1: Justify the above identity and verify the equations for C andγ. Hint: Start with

cos(α−β)=cos(α) cos(β)+sin(α) sin(β)and multiply by C. Then think what shouldαandβbe. While it is generally easier to use the first form with AandBto solve for the initial conditions, the second form is much more natural. The constantsCand γhave very nice interpretation. We look at the form of the solution

x(t)=Ccos(ω0t−γ).

We can see that theamplitudeisC,ω0 is the (angular) frequency, andγis the so-calledphase shift.

The phase shift just shifts the graph left or right. We callω0thenatural (angular) frequency. This

entire setup is usually calledsimple harmonic motion.

Let us pause to explain the wordangularbefore the wordfrequency. The units ofω0are radians

radians, the usual frequency is given by ω0

2π. It is simply a matter of where we put the constant 2π,

and that is a matter of taste.

Theperiodof the motion is one over the frequency (in cycles per unit time) and hence _{ω}2π

0. That

is the amount of time it takes to complete one full cycle.

Example 2.4.1: Suppose thatm=2 kg andk=8N/_{m}_{. The whole mass and spring setup is sitting}

on a truck that was traveling at 1m/_{s}_{. The truck crashes and hence stops. The mass was held in place}

0.5 meters forward from the rest position. During the crash the mass gets loose. That is, the mass is
now moving forward at 1m/_{s}_{, while the other end of the spring is held in place. The mass therefore}

starts oscillating. What is the frequency of the resulting oscillation and what is the amplitude. The units are the mks units (meters-kilograms-seconds).

The setup means that the mass was at half a meter in the positive direction during the crash and
relative to the wall the spring is mounted to, the mass was moving forward (in the positive direction)
at 1m/_{s}_{. This gives us the initial conditions.}

So the equation with initial conditions is

2x00+8x=0, x(0)= 0.5, x0(0)= 1. We can directly compute ω0 =

√

k/_{m} =

√

4 = 2. Hence the angular frequency is 2. The usual
frequency in Hertz (cycles per second) is2/_{2}_{π}=1/_{π}_{≈} _{0}._{318.}

The general solution is

x(t)= Acos(2t)+Bsin(2t).

Letting x(0)=0.5 meansA= 0.5. Then x0_{(t)}= _{−2(0}._{5) sin(2t)}+_{2B}_{cos(2t). Letting} _{x}0_{(0)}=_{1 we}

get B = 0.5. Therefore, the amplitude isC = √

A2+_{B}2 = √_{0}._{25}+_{0}._{25} = √_{0}._{5} _{≈} _{0}._{707. The}

solution is

x(t)= 0.5 cos(2t)+0.5 sin(2t). A plot ofx(t) is shown in on the facing page.

In general, for free undamped motion, a solution of the form x(t)= Acos(ω0t)+Bsin(ω0t),

corresponds to the initial conditions x(0)= Aandx0_{(0)}= ω

0B. Therefore, it is easy to figure outA

andBfrom the initial conditions. The amplitude and the phase shift can then be computed fromA
andB. In the example, we have already found the amplitudeC. Let us compute the phase shift. We
know that tanγ =B/_{A}=_{1. We take the arctangent of 1 and get}π/_{4}_{or approximately 0.785. We still}

need to check if thisγis in the correct quadrant (and addπtoγif it is not). Since bothAandBare positive, thenγshould be in the first quadrant,π/4radians is in the first quadrant, soγ= π/4.

Note: Many calculators and computer software do not only have theatanfunction for arctangent, but also what is sometimes calledatan2. This function takes two arguments,BandA, and returns aγin the correct quadrant for you.

2.4. MECHANICAL VIBRATIONS 83 0.0 2.5 5.0 7.5 10.0 0.0 2.5 5.0 7.5 10.0 -1.0 -0.5 0.0 0.5 1.0 -1.0 -0.5 0.0 0.5 1.0

Figure 2.2: Simple undamped oscillation.

### 2.4.3

### Free damped motion

Let us now focus on damped motion. Let us rewrite the equation
mx00+cx0+kx=0,
as
x00+2px0+ω2_{0}x= 0,
where
ω0 =
r
k
m, p=
c
2m.
The characteristic equation is

r2+2pr+ω2_{0} =0.
Using the quadratic formula we get that the roots are

r =−p± q

p2_{−}ω2
0.

The form of the solution depends on whether we get complex or real roots. We get real roots if and only if the following number is nonnegative:

p2−ω2_{0} =
_{c}
2m
2
− k
m =
c2−4km
4m2 .
The sign of p2−ω2

0is the same as the sign ofc

2_{−}_{4km. Thus we get real roots if and only if}_{c}2_{−}_{4km}

Overdamping

Whenc2 −4km > 0, we say the system isoverdamped. In this case, there are two distinct real rootsr1 andr2. Both roots are negative: As

q

p2_{−}ω2

0is always less thanp, then−p±

q

p2_{−}ω2
0is

negative in either case.

The solution is 0 25 50 75 100 0 25 50 75 100 0.0 0.5 1.0 1.5 0.0 0.5 1.0 1.5

Figure 2.3: Overdamped motion for several different

initial conditions.

x(t)=C1er1t+C2er2t.

Since r1,r2 are negative, x(t) → 0 as t → ∞.

Thus the mass will tend towards the rest position as time goes to infinity. For a few sample plots for different initial conditions, see .

Do note that no oscillation happens. In fact, the graph will cross thexaxis at most once. To see why, we try to solve 0=C1er1t+C2er2t. Therefore,

C1er1t =−C2er2t and using laws of exponents we

obtain

−C1

C2

= e(r2−r1)t.

This equation has at most one solutiont≥ 0. For some initial conditions the graph will never cross thexaxis, as is evident from the sample graphs. Example 2.4.2: Suppose the mass is released from rest. That is x(0)= x0andx0(0)= 0. Then

x(t)= x0 r1−r2

r1er2t−r2er1t.

It is not hard to see that this satisfies the initial conditions. Critical damping

When c2−4km = 0, we say the system iscritically damped. In this case, there is one root of multiplicity 2 and this root is−p. Our solution is

x(t)=C1e

−pt_{+}

C2te

−pt_{.}

The behavior of a critically damped system is very similar to an overdamped system. After all a critically damped system is in some sense a limit of overdamped systems. Since these equations are really only an approximation to the real world, in reality we are never critically damped, it is a place we can only reach in theory. We are always a little bit underdamped or a little bit overdamped. It is better not to dwell on critical damping.

2.4. MECHANICAL VIBRATIONS 85