# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

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The spring constant is k = 10 lb/2 in. = 60 lb/ft and the mass is m = w/g = 10/32 lb-sec2/ft. Hence the equation of motion reduces to

u" + 192u = 0, (19)

and the general solution is

u = A cos(8V31) + B sin(^V31).

The solution satisfying the initial conditions u(0) = 1/6 ft and u'(0) = 1 ft/sec is

u = - cos(8V3t)--------- sin(8V3t). (20)

6 8λ/ 3

The natural frequency is ώ0 = V192 = 13.856 rad/sec, so the period is T = 2n/w0 = 0.45345 sec. The amplitude R and phase S are found from Eqs. (17). We have

1 1 19

R2 = +----------=-------, so R = 0.18162 ft.

36 192 576

The second of Eqs. (17) yields tan S = Σ3/4. There are two solutions of this equation, one in the second quadrant and one in the fourth. In the present problem cos S > 0 and sin S < 0, so S is in the fourth quadrant, namely,

S = arctan(V3/4) = 0.40864 rad.

The graph of the solution (20) is shown in Figure 3.8.4.

U R = 0.182 u = 0.182 cos(8V3 t + 0.409)

0 2 ~ N 1 1 ^

1

- 0 . 2 0.5 1 1.5 t

- T = 0.453 ^

FIGURE 3.8.4 An undamped free vibration; u" + 192u = 0, u(0) = 1 /6, u'(0) = 1.

Damped Free Vibrations. If we include the effect of damping, the differential equation governing the motion of the mass is

mu" + y u'+ ku = 0. (21)

192

Chapter 3. Second Order Linear Equations

We are especially interested in examining the effect of variations in the damping coefficient y for given values of the mass m and spring constant k. The roots of the corresponding characteristic equation are

Y ± VY2 - 4km Y / λ , Λ Akm\ ^

ri> r2 - 2m - 2m \1 ± σ1 - ) ¦ (22)

Depending on the sign of y 2 4km, the solution u has one of the following forms:

Y2 4km > 0, u = Aerit + Ber2(; (23)

Y2 4km = 0, u = (A + Bt)eY t/2m; (24)

(4km Y 2)1/2

Y2 4km < 0, u = e~Yt/2m(A cos it + B sin it), i =---------------------------- > 0.

2m (25)

Since m, y, and k are positive, y2 4km is always less than y2. Hence, if y2 4km >

0, then the values of r1 and r2 given by Eq. (22) are negative. If y2 4km < 0, then

the values of r1 and r2 are complex, but with negative real part. Thus, in all cases,

the solution u tends to zero as t ^to; this occurs regardless of the values of the arbitrary constants A and B, that is, regardless of the initial conditions. This confirms our intuitive expectation, namely, that damping gradually dissipates the energy initially imparted to the system, and consequently the motion dies out with increasing time.

The most important case is the third one, which occurs when the damping is small. If we let A = R cos S and B = R sin S in Eq. (25), then we obtain

u = Re~Yt/2m cos(it S). (26)

The displacement u lies between the curves u = ±Re~Yt/2m; hence it resembles a cosine wave whose amplitude decreases as t increases. A typical example is sketched in Figure 3.8.5. The motion is called a damped oscillation, or a damped vibration. The amplitude factor R depends on m, y, k, and the initial conditions.

Although the motion is not periodic, the parameter i determines the frequency with which the mass oscillates back and forth; consequently, i is called the quasi frequency. By comparing i with the frequency «0 of undamped motion, we find that

i _ (4km y2)1/2/2m / _ \ 1/2 _ i _ y^_

^k/m I 4km / 8km

FIGURE 3.8.5 Damped vibration; u = Re Yt/2m cos(it S).

3.8 Mechanical and Electrical Vibrations

193

EXAMPLE

3

The last approximation is valid when y2/4km is small; we refer to this situation as small damping. Thus, the effect of small damping is to reduce slightly the frequency of the oscillation. By analogy with Eq. (18) the quantity Td = 2ο/ii is called the quasi period. It is the time between successive maxima or successive minima of the position of the mass, or between successive passages of the mass through its equilibrium position while going in the same direction. The relation between Td and T is given by

Φ= * = (1 - (l + -?). (28)

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