Mechanical Vibrations

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MECHANICAL VIBRATIONS

CHAPTER I

KINEMATICS OF VIBRATION

1. Definitions.—A vibration in its general sense is a periodic motion, i.e., a motion which repeats itself in all its particulars after a certain interval of time, called the period of the vibration and usually designated by the symbol T. A plot of the displacement x against the time t may be a curve of considerable complication. As an example, Fig. 1a shows the motion curve observed on the bearing pedestal of a steam turbine.

FIG. 1.—A Periodic and a harmonic function, showing the period T and the amplitude x0.

The simplest kind of periodic motion is a harmonic motion: in it the relation between x and t may be expressed by as shown in Fig. 1b, representing the small oscillations of a simple pendulum. The maximum value of the displacement is x0 the amplitude of the vibration.

The period T usually is measured in seconds; its reciprocal f = 1/T is the frequency of the vibration, measured in cycles per second. In some publications this is abbreviated as cyps and pronounced as it is written. In the German literature cycles per second are generally called Hertz in honor of the first experimenter with radio waves (which are electric vibrations).

In Eq. (1) there appears the symbol ω, which is known as the circular frequency and is measured in radians per second. This rather unfortunate name has become familiar on account of the properties of the vector representation, which will be discussed in the next section. The relations between ω, f, and T are as follows. From Eq. (1) and Fig. 1b it is clear that a full cycle of the vibration takes place when ωt has passed through 360 deg. or 2π radians. Then the sine function resumes its previous values. Thus, when ωt = 2π, the time interval t is equal to the period T or

Since f is the reciprocal of T,

For rotating machinery the frequency is often expressed in vibrations per minute, denoted as v.p.m. = 30ω/π.

In a harmonic motion for which the displacement is given by x = x0 sin ωt, the velocity is found by differentiating the displacement with respect to time,

so that the velocity is also harmonic and has a maximum value ωx0

The acceleration is also harmonic and with the maximum value ω²x0.

Consider two vibrations given by the expressions x1 = α sin ωt and x2 = b sin (ωt + φ) which are shown in Fig. 2, plotted against ωt as abscissa. Owing to the presence of the quantity φ, the two vibrations do not attain their maximum displacements at the same time, but the one is φ/ω sec. behind the other. The quantity is known as the phase angle or phase difference between the two vibrations. It is seen that the two motions have the same ω and consequently the same frequency f. A phase angle has meaning only for two motions of the same frequency; if the frequencies are different, phase angle is meaningless.

FIG. 2.—Two harmonic motions including the phase angle φ

Example: A body, suspended from a spring, vibrates vertically up and down between two positions 1 and 1 ½ above the ground. During each second it reaches the top position (1½ in, above ground) twenty times. What are T, f,, and x0?

Solution: x0 = ¼ in, T = ½0_ sec., f = 20 cycles per second, and ω = 2πf = 126 radians per second.

FIG. 3—A harmonic vibration represented by the horizontal projection of a rotating vector.

2. The Vector Method of Representing Vibrations.(Fig. 3) rotate with uniform angular velocity ω in a counterclockwise direction. When time is reckoned from the horizontal position of the vector as a starting point, the horizontal projection of the vector can be written as

and the vertical projection as

Either projection can be taken to represent a reciprocating motion; in the following discussion, however, we shall consider only the horizontal projection.

This representation has given rise to the name circular frequency for. The quantity ω, being the angular speed of the vector, is measured in radians per second; the frequency f in this case is measured in revolutions per second. Thus it can be seen immediately that ω = 2πf.

The velocity of the motion x = a cos ωt is and can be represented by (the horizontal projection of) a vector of length aω, rotating with the same angular velocity ω as the displacement vector but situated always 90 deg. ahead of that vector. The acceleration is —aω² cos ωt and is represented by (the horizontal projection of) a vector of length aω² rotating with the same angular speed ω and 180 deg. ahead of the position or displacement vector or 90 deg. ahead of the velocity vector (Fig. 4). The truth of these statements can be easily verified by following the various vectors through one complete revolution.

FIG. 4.—Displacement, velocity, and acceleration are perpendicular vectors.

This vector method of visualizing reciprocating motions is very convenient. For example, if a point is simultaneously subjected to two motions of the same frequency which differ by the phase angle φ, namely, a cos ωt and b cos ( ωt — φ), the addition of these two expressions by the methods of trigonometry is wearisome. However, the two vectors are easily drawn up, and the total motion is represented by the geometric sum of the two vectors as shown in the upper part of is considered to rotate in a counterclockwise direction with the uniform angular velocity co, and the horizontal projections of the various vectors represent the displacements as a function of time. This is shown in the lower part of Fig. 5. The line α-α .

That this vector addition gives correct, results is evident, because a cos ωt -vector and b cos (ωt — φ-vector. The horizontal projection of the geometric sum of these two vectors is evidently equal to the sum of the horizontal projections of the two component vectors, which is exactly what is wanted.

FIG. 5.—TWO vibrations are added by adding their vectors geometrically.

Addition of two vectors is permissible only if the vibrations are of the same frequency. The motions α sin ωt and a sin 2ωt can be represented by two vectors, the first of which rotates with an angular speed ω and the second with twice this speed, i.e., with 2ω. The relative position of these two vectors in the diagram is changing continuously, and consequently a geometric addition of them has no meaning.

A special case of the vector addition of Fig. 5, which occurs rather often in the subsequent chapters, is the addition of a sine and a cosine wave of different amplitudes: a sin ωt and b cos ωt. For this case the two vectors are perpendicular, so that from the diagram of Fig. 6 it is seen at once that

where tan φ = b/a.

FIG. 6.—Addition of a sine and cosine wave of different amplitudes.

Example: What is the maximum amplitude of the sum of the two motions

Solution: The first motion is represented by a vector 5 in. long which maybe drawn vertically and pointing downward. Since in this position the vector has no horizontal projection, it represents the first motion at the instant t = 0. At that instant the second motion is x2= 10 sin 1, which is represented by a vector of 10 in. length turned 1 radian (57 deg.) in a counterclockwise direction with respect to the first vector. The graphical vector addition shows the sum vector to be 13.4 in. long.

FIG. 7.—Vector diagrams illustrating the mechanism of beats.

3. Beats.—If the displacement of a point moving back and forth along a straight line can be expressed as the sum of two terms, a sin ω1t + b sin ω2t, where ω1 ≠ ω2, the motion is said to be the " -vector rotates with ω2. If ω2 is nearly equal to ω2, the two vectors will retain sensibly the same relative position during one revolution, i.e., changes. Therefore the resulting motion can be described approximately as a sine wave with a frequency ω1 and an amplitude varying slowly between (b + a) and (b — a), or, if b = a, between 2α and 0 (Figs. 7 and 8).

This phenomenon is known as beats. The beat frequency is the number of times per second the amplitude passes from a minimum through a maximum to the next minimum (A to B in -vector. Thus the beat frequency is seen to be ω1 — ω2.

FIG. 8.—Beats.

Example: A body describes simultaneously two vibrations, x1 = 3 sin 40t and x2 = 4 sin 41t, the units being inches and seconds. What is the maximum and minimum amplitude of the combined motion and what is the beat frequency?

Solution: The maximum amplitude is 3 + 4 = 7 in.; the minimum is 4 — 3 = 1 in. The circular frequency of the beats ωb = 41 — 40 = 1 radian per second. Thus fb = ωb/2π = 1/2π cycles per second. The period Tb or duration of one full beat is Tb = 1/fb = 6.28 sec.

The phenomenon can be observed in a great many cases (pages 109, 402). For audio or sound vibrations it is especially notable. Two tones of slightly different pitch and of approximately the same intensity cause fluctuations in the total intensity with a frequency equal to the difference of the frequencies of the two tones. For example, beats can be heard in electric power houses when a generator is started. An electric machine has a magnetic hum, of which the main pitch is equal to twice the frequency of the current or voltage, usually 120 cycles per second. Just before a generator is connected to the line the electric frequency of the generator is slightly different from the line frequency. Thus the hum of the generator and the hum of the line (other generators or transformers) are of different pitch, and beats can be heard.

The existence of beats can he shown also by trigonometry. Let the two vibrations be α sin ω1t and b sin ω2t, where ω1 and ω2 are nearly equal and ω2 — ω1 = Δ ω.

Then

Applying formula (6) the resultant vibration is

where the phase angle φ can be calculated but is of no interest in this case. The amplitude, given by the radical, can be written which expression is seen to vary between (α + b) and (α — b) with a frequency Δω.

4. A Case of Hydraulic-turbine Penstock Vibration.—Adirect application of the vector concept of vibration to the solution of an actual problem is the following.

In a water-power generating station the penstocks, i.e., cyps. beat between the penstock and transformer hums could be plainly heard. The essential parts of the turbine are shown schematically in Fig. 9, which is drawn in a horizontal plane, the turbine shaft being vertical. The water enters from the penstock I into the spiral case II; there the main stream splits into 18 partial streams on account of the 18 stationary, non-rotating guide vanes. The water then enters the 17 buckets of the runner and finally turns through an angle of 90 deg. to disappear into the vertical draft tube III.

cycles per second. If all these impulses had the same phase, they would all add up arithmetically and give a very strong disturbance in the penstock.

FIG. 9.—Explains the vibration in the penstock of a Francis hydraulic turbine.

Assume that stream a experiences the maximum value of its impulse when the two vanes 1 and 1 line up. Then the maximum value of the impulse in stream b takes place somewhat earlier (to be exact, 1/(17 X 18)th revolution earlier, at the instant that the two vanes 2 and 2 are lined up).

The impulse of stream a travels back into the penstock with the velocity of sound in water (about 4,000 ft./sec.)* and the same is true for the impulse of stream b. However, the path traveled by the impulse of b is somewhat longer than the path for α, the difference being approximately one-eighteenth part of the circumference of the spiral case. Because of this fact, the impulse b will arrive in the penstock later than the impulse a.

In the machine in question it happened that these two effects just canceled each other so that the two impulses a and b arrived at the cross section AA of the penstock simultaneously, i.e., in the same phase. This of course is true not only for a and b but for all the 18 partial streams. In the vector representation the impulses behave as shown in Fig. 10a, the total impulse at AA being very large.

FIG. 10.—The 18 partial impulses at the section AA of Fig. 9 for a 17-bucket runner (a) and for a 16-bucket runner (b).

In order to eliminate the trouble, the existing 17-bucket runner was removed from the turbine and replaced by a 16-bucket runner. This does not affect the time difference caused by the different lengths of the paths a, b, etc., but it does change the interval of time between the impulses of two adjacent guide vanes. In particular, the circumferential distance between the bucket 2 and guide vane 2 becomes twice as large after the change. In fact, at the instant that rotating bucket 1 gives its impulse, bucket 9 also gives its impulse, whereas in the old construction bucket 9 was midway between two stationary vanes (Fig. 9).

It was a fortunate coincidence that half the circumference of the spiral case was traversed by a sound wave in about ½ X 1/113 sec, so that the two impulses due to buckets 1 and 9 arrived in the cross section AA in phase opposition (Fig. 106). The phase angle between the impulses at section AA of two adjacent partial streams is thus one-ninth of 180 degrees, and the 18 partial impulses arrange themselves in a circular diagram with a zero resultant.

The analysis as given would indicate that after the change in the runner had been made the vibration would be totally absent. However, this is not to be expected, since the reasoning given is only approximate, and many effects have not been considered (the spiral case has been replaced by a narrow channel, thus neglecting curvature of the wave front, reflection of the waves against the various obstacles, and effect of damping). In the actual case the amplitude of the vibration on the penstock was reduced to one-third of its previous value, which constituted a satisfactory solution of the problem.

5. Representation by Complex Numbers.—It was shown in the previous sections that rotating vectors can represent harmonic motions, that the geometric addition of two vectors corresponds to the addition of two harmonic motions of the same frequency, and that a differentiation of such a motion with respect to time can be understood as a multiplication by and a forward turning through 90 dog. of the representative vector. These vectors, after a little practice, afford a method of visualizing harmonic motions which is much simpler than the consideration of the sine waves themselves.

For numerical calculations, however, the vector method is not well adapted, since it becomes necessary to resolve the vectors into their horizontal and vertical components. For instance, if two motions have to be added as in Fig. 5, we write

meaning geometric , i.e., the amplitude of the sum motion, we write

is the geometric sum of ax in the x-direction and ay in the y-direction. Then

and the length is consequently

This method is rather lengthy and loses most of the advantage due to the introduction of vectors.

There exists, however, a simpler method of handling the vectors numerically, employing imaginary numbers. A complex number can be represented graphically by a point in a plane where the real numbers 1, 2, 3, etc., are plotted horizontally and the imaginary numbers are plotted vertically. With the notation these imaginary numbers are j, 2j, 3j, etc. In Fig. 11, for example, the point 3 + 2j is shown. In joining that point with the origin, the complex number can be made to represent a vector. If the angle of the vector with the horizontal axis is α and the length of the vector is a, it can be written as

FIG. 11. —A vector represented by a point in the complex plane.

Harmonic motions are represented by rotating vectors. A substitution of the variable angle ωt for the fixed angle α in the last equation leads to representing a rotating vector, the horizontal projection of which is a harmonic motion. But this horizontal projection is also the real part of (7). Thus if we say that a vector represents a harmonic motion, we mean that the horizontal projection of the rotating vector represents that motion. Similarly if we state that a complex number represents a harmonic motion we imply that the real part of such a number, written in the form (7) represents that motion.

Example: Solve the example of page 6 by means of the complex method.

Solution: The first vector is represented by —5j and the second one by —l0j cos 57° + 10 sin 57° = —5.4j + 8.4. The sum of the two is 8.4 — 10.4j= 13.4 in.

Differentiate (7) which gives the result since by definition of j we have j² = —1. It is thus seen that differentiation of the complex number (7) is equivalent to multiplication by jω.

In vector representation, differentiation multiplies the length of the vector by ω and turns it ahead by 90 deg. Thus we are led to the conclusion that multiplying a complex number by j is equivalent to moving it a quarter turn ahead without changing its absolute value. That this is so can be easily verified:

which Fig. 12 actually shows in the required position.

FIG. 12.—Multiplying a complex number by j amounts to turning its vector ahead through 90 deg.

In making extended calculations with these complex numbers the ordinary rules of algebra are followed. With every step we may remember that the motion is represented by only the real part of what we are writing down. Usually, however, this is not done: the algebraic manipulations are performed without much recourse to their physical meaning and only the final answer is interpreted by considering its real part.

For simple problems it is hardly worth while to study the complex method, since the solution can be obtained just as easily without it. However, for more complicated problems, such as are treated in Sec. 24, for example, the labor-saving brought about by the use of the complex notation is substantial.

The expression (7) is sometimes written in a different form:

or, if for simplicity a = 1 and ωt = α,

The right-hand side of this equation is an ordinary complex number, but the left-hand side needs to be interpreted, as follows. The Maclaurin series development of ex is

Substituting x = jα this becomes

The right-hand side is a complex number, which by definition is the meaning of eiα. But we recognize the brackets to be the Maclaurin developments of cos α. and sin α, so that formula (8a) follows.

A simple graphical representation of the result can be made in the complex plane of Fig. 11 or 12. Consider the circle with unit radius in this plane. Each point on the circle has a horizontal projection cos α and a vertical projection sin α and thus represents the number, cos α. + j sin α. = eiα. Consequently the number eiα is represented by a point on the unit circle, α radians away from the point +1. If α is now made equal to ωt, it is seen that eiωt represents the rotating unit vector of which the horizontal projection is a harmonic vibration of unit amplitude and frequency ω.

On page 52 we shall have occasion to make use of Eq. (8a).

6. Work Done on Harmonic Motions.—A concept of importance for many applications is that of the work done by a harmonically varying force upon a harmonic motion of the same frequency.

Let the force P = P0 sin (ωt + φ) be acting upon a body for which the motion is given by x = x0 sin ωt. The work done by the force during a small displacement dx is Pdx,

During one cycle of the vibration, ωt varies from 0 to 2π and consequently t varies from 0 to 2π/ω. The work done during one cycle is:

A table of integrals will show that the first integral is zero and that the second one is π, so that the work per cycle is

This result can also be obtained by a graphical method, which interprets the above calculations step by step, as follows.

0 (Fig. 13). Now resolve the force into its components P0 cos φ in phase with the motion, and P0 sin φ, 0. This is permissible for the same reason that geometric addition of vectors is allowed, as explained in Sec. 2. Thus the work done splits up into two parts, one part due to a force in phase with the motion and another part due to a force 90 deg. ahead of the motion.

FIG. 13.—A force and a motion of the same frequency.

Consider the first part as shown in Fig. 14a, in which the ordinates are the displacement x and the in phase component of the force. Between A and B the force is positive, say upward, and the body is moving in an upward direction also; positive work is done. Between B and C, on the other hand, the body moves downward toward the equilibrium point while the force is still positive (upward, though of gradually diminishing intensity), so that negative work is done. The work between A and B cancels that between B and C, and over a whole cycle the work done is zero. If a liar manic force acts on a body subjected to a harmonic motion of the same frequency, the component of the force in phase with the displacement does no work.

FIG. 14.—A force in phase with a displacement does no work over a full cycle; a force 90 deg. out of phase with a displacement does a maximum amount of work.

It was shown in Sec. 2 that the velocity is represented by a vector 90 deg. ahead of the displacement, so that the statement can also be worded as follows:

A force does work only with that component which is in phase with the velocity.

Next we consider the other component of the force, which is shown in Fig. 14b. During the interval AB the displacement increases so that the motion is directed upward, the force is positive, and consequently upward also, so that positive work is done. In the interval BC the motion is directed downward, but the force points downward also, so that the work done is again positive. Since the whole diagram is symmetrical about a vertical line through B, it is clear that the work done during AB equals that done during BC. The work done during the whole cycle AD is four times that done during AB.

To calculate that amount it is necessary to turn to the definition of work:

This shows that the work done during a cycle is the time integral of the product of force and velocity. The force is (Fig. 14b)

cost² αdα =π.

P = (P0 sin φ); cos ωt and the velocity is v = x0 cos ωt, so that the work per cycle is

The value of the definite integral on the right-hand side can be deduced from Fig. 15, in which curve I represents cos ωt and curve II represents cos² ωt. The curve cos² ωt is sinusoidal about the dotted line AA as center line and has twice the frequency of cos ωt, which can be easily verified by trigonometry:

Consider the quadrangle 1-2-3-4 as cut in two pieces by the curve II and note that these two pieces have the same shape and the same area. The distance 1-4 is unity, while the distance 3-4 is π/2 radians or 90 deg. Thus the area of the entire quadrangle is π/2 and the area of the part under curve II is half of that. Consequently the value of our definite integral taken between the limits 0 and T/4 is π/4, and taken between the limits 0 and T it is π. Thus the work during a cycle is

It will be seen in the next section that a periodic force as well as a periodic motion may be impure, i.e., it may contain higher harmonics in addition to the fundamental harmonic. In this connection it is of importance to determine the work done by a harmonic force on a harmonic motion of a frequency different from that of the force. Let the force vary with a frequency which is an integer multiple of ω, say nω, and let the frequency of the motion be another integer multiple of ω, say mω. It will now be proved that the work done by such a force on such a motion during a full cycle of ω is zero.

Let the force be P = Po sin nωt and let the corresponding motion be x = xo sin (mωt + φ). Then the work per cycle is

Since

and since φ is independent of the lime and can be brought in front of the integral sign, the integral splits up into two parts of the form

Both these integrals are zero if n is different from m, which can be easily verified by transforming the integrands as follows:

Since the interval of integration is T = 2π/ω, the sine and cosine functions are integrated over multiples of 2ω, giving a zero result. In order to gain a physical understanding of this fact let us consider the first of the above two integrals with n = 4 and m = 5. This case is represented in Fig. 16, where the amplitudes of the two waves are drawn to different vertical scales in order to distinguish them more easily. The time interval over which the integration extends is the interval AB. The ordinates of the two curves have to be multiplied together and then integrated. Consider two points, one somewhat to the right of A and another at the same distance to the left of C. Near A both waves are positive; near C one is positive and the other is negative, but the absolute values of the ordinates are the same as near A. Therefore the contribution to the integral of an element near A cancels the contribution of the corresponding element near C. This canceling holds true not only for elements very near to A and C but generally for two elements at equal distances to the left from C and to the right from A. Thus the integral over the region AD cancels that over CD. In the same way it can be shown that the integral over CB is zero.

sin nα sin mα dα = 0.

It should be understood that the work is zero only over a whole cycle. Starting at A, both waves (the force and the velocity) are positive, so that positive work is done. This work, however, is returned later on (so that in the meantime it must have been stored in the form of potential or kinetic energy).

This graphical process can be repeated for any combination of integral values of m and n and also for integrals containing a cosine in the integrand. When m becomes equal to n, we have the case of equal frequencies as already considered. Even then there is no work done when the force and displacement are in phase. In case m = n and the force and displacement are 90 deg. out of phase, the work per cycle of the nth harmonic is πP0X0 as before, and since there are n of these cycles in one cycle of the fundamental frequency ω, the work per fundamental cycle is nπP0x0.

The results thus obtained can be briefly summarized as follows:

1. The work done by a harmonic force acting upon a harmonic displacement or velocity of a different frequency from that of the force is zero during a time interval comprising both an integer number of force cycles and a (different) integer number of velocity cycles.

2. The work done by a harmonic force 90 deg. out of phase with a harmonic velocity of the same frequency is zero during a whole cycle.

3. The work done by a harmonic force of amplitude P0 and frequency ω, in phase with a harmonic velocity v0 = X0ω of the same frequency, is πP0V0/ω = πP0V0 over a whole cycle.

Example: A force 10 sin 2π60t (units are pounds and seconds) is acting on a displacement of 1/10 sin [2π60t — 45°] (units are inches and seconds). What is the work done during the first second, and also during the first one-thousandth of a second?

Solution: of work. During the first second there are 60 cycles so that the work performed is 60 X 2 22 = 133 in. lb.

During the first one-thousandth of a second there are 60/1,000 = 0.06 cycle, so that the vectors in the diagram turn through only 0.06 X 360 deg. = 21.6 deg. Formula (9) holds only for a full cycle. For part of a cycle the. integration has to be performed in full:

This is considerably less than one-thousandth part of the work performed in a whole second, because during this particular 1/1,000 sec. the force is very small, varying between 0 and 0.368P0.

7. Non-harmonic, Periodic Motions.—A periodic motion has the property of repeating itself in all details after a certain time interval, the period of the motion. All harmonic motions are periodic, but not every periodic motion is harmonic. For example, Fig. 17 represents the motion the superposition of two sine waves of different frequency. It is periodic but not harmonic. The mathematical theory shows

FIG. 17.—The sum of two harmonic motions of different frequencies is not a harmonic motion.

that any periodic curve f(t) of frequency co can be split up into a series of sine curves of frequencies ω, 2ω, 3ω, 4ω, etc. Or provided that f(t) repeats itself after each interval T = 2π/ω. The amplitudes of the various waves A1, A2,..., and their phase angles φ1, φ2,..., can be determined analytically when f(t) is given. The series (10) is known as a Fourier series.

The second term is called the fundamental or first harmonic of f(t) and in general the (n + 1)st term of frequency nω is known as the nth harmonic of f(t). Since

the series can also be written as

The constant term b0 represents the average height of the curve f(t) during a cycle. For a curve which is as much above the zero line during a cycle as it is below, the term b0 is zero. The amplitudes a1 an, b1 bn can be determined by applying the three energy theorems of pages 18 and 19.

Consider for that purpose f(t) to be a force, and let this non-harmonic force act on a point having the harmonic velocity sin nωt. Now consider the force f(t) as the sum of all the terms of its Fourier series and determine the work done by each harmonic term separately. All terms of the force except an sin nωt and bn cos nωt are of a frequency different from that of the velocity sin nωt so that no work per cycle is done by them. Moreover, bn cos nωt is 90 deg. out of phase with the velocity so that this term does not do any work either. Thus the total work done is that of the force an sin nωt on the velocity sin nωt cycle of the nω-frequency. Per cycle of the fundamental frequency (which is n times as long), the work is πan/ω.

Thus the amplitude an is found to be ω/π times as large as the work done by the complete non-harmonic force f(t) on a velocity sin nωt during one cycle of the force. Or, mathematically

By assuming a velocity cos nωt instead of sin nωt and repeating the argument, the meaning of bn is disclosed as

The relations between an, bn and the quantities An, φn of Eq. (10) are as shown in Eq. (6), page 6, so that

Thus the work done by a non-harmonic force of frequency ω upon a harmonic velocity of frequency nω is merely the work of the component of the nth harmonic of that force in phase with the velocity; the work of all other harmonics of the force is zero when integrated over a complete force cycle.

With the aid of the formulas (11) it is possible to find the cn and bn for any periodic curve which may be given. The branch of mathematics which is concerned with this problem is Known as harmonic analysis.

Example: The curve c of Fig. 254 (page 426)shows approximately the damping force caused by turbulent air on a body in harmonic motion. If the origin of coordinates of Fig. 254 is displaced one-quarter cycle to the left, the mathematical expression for the curve is

Find the amplitudes of the various harmonics of this curve.

Solution: The curve to be analyzed is an antisymmetric one, i.e., the values of f(ωt) are equal and opposite at two points at ωt at equal distances on both sides of the origin. Sine waves are antisymmetric and cosine waves arc symmetric. An antisymmetric curve cannot have cosine components. Hence, all bn are zero. This can be further verified by sketching the integrand of Eq. (11b) in the manner of Fig. 16 and showing that the various contributions to the integral cancel each other. The constant term b0 = 0, because the curve has no average height. For the sine components we find

The integrands can be transformed by means of the last formula on page 17,

The indefinite integral of this is

The harmonic an is 1/π times the definite integrals.

Since F(2π) = F(0), we have

For even values for n the an thus is zero, while only for odd values of n the harmonic exists. In particular for n = 1, we have for the fundamental harmonic

Thus the amplitude of the fundamental harmonic is 85 percent of the maximum amplitude of the curve itself.

The evaluation of the integrals (11) by calculation can be done only for a few simple shapes of f(t). If f(t) is a curve taken from an actual vibration record or from an indicator diagram, we do not even possess a mathematical expression for it. However, with the aid of the curve so obtained the integrals can be determined either graphically or numerically or by means of a machine, known as a harmonic analyzer.

Such a harmonic analyzer operates on the same principle as Watt’s steam-engine indicator. The indicator traces a closed curve of which the ordinate is the steam pressure (or piston force) and the abscissa is the piston displacement. The area of this closed curve is the work done by the piston force per cycle. The formulas (11) state that the coefficients an or bn are ω/π times the work done per cycle by the force f(t) on a certain displacement of which the velocity is expressed by sin nωt. To obtain complete correspondence between the two cases, we note that sin nωt , so that (11a) can be written in the modified form

FIG. 18.—The harmonic analyzer, an instrument operating on the same principle as Watt’s steam-engine indicator.

indicates that the integration extends over the closed curve described during one cycle of the force f(t).

The machine is shown schematically in Fig. 18. A is a cardboard template representing one cycle of the curve f(t) which is to be analyzed. The template A is fastened to a rack and a pinion B, which is rotated by an electric motor. The arm C is guided so that it can move in its longitudinal direction only and is pressed lightly against the template by a spring. Thus the vertical motion of the pen D on the arm C is expressed by f(t). The table, or platen, E moves horizontally and is driven by a scotch crank and gear which is connected by suitable intermediate gears to B so that E oscillates n times, while A moves through the length of the diagram. The machine has with it a box of spare gears so that any gear ratio n from 1 to 30 can be obtained by replacing one gear in the train by another.

The horizontal motion of the platen E is expressed by sin nωt or by cos nωt, depending on the manner in which the gears are interlocked. The point D will thus trace a closed curve on the platen, for which the area equals an or bn (multiplied by a constant 1/nπ ). Instead of actually tracing this curve, the instrument usually carries a planimeter of which one point is attached to E and the other end to D, so that the area is given directly by the planimeter reading.

Harmonic analyzers have been built on other principles as well. An interesting optical method using the sound tracks of motion picture films was invented by Wente and constructed by Montgomery, both of the Bell Telephone Laboratories. The reference to this paper is given in the Bibliography.

[Eqs. (10) and (10a)], without giving information on the phase angles φn [or the ratios an/bn Eq. (10)], are available on the market. They have been developed by the Western Electric Company (model RA-277 to be used in conjunction with model RA-246) for sound or noise analysis and require the original curve to be available in the form of an electric voltage, varying with the time, such as results from an electric vibration pickup (page 81) or a microphone. This voltage, after proper amplification, is fed into an electric network known as a band-pass filter, which suppresses all frequencies except those in a narrow band of a width equal to 5 cycles per second. This passing band of frequencies can be laid anywhere in the range from 10 to 10,000 cycles per second. When a periodic (steady-state) vibration or noise is to be Fourier-analyzed, a small motor automatically moves the pass band across the entire spectrum and the resulting analysis is drawn graphically by a stylus on a strip of waxed paper, giving the harmonic amplitudes vs. the frequency from 10 to 10,000 cycles per second, all in a few minutes. The record is immediately readable.

Another electrical analyzer, operating on about the same principle but without graphic recording, is marketed by the General Radio Company, Cambridge, Mass.

There are several methods for calculating the Fourier coefficients numerically, i.e., methods to evaluate the integrals (11) in eases where the function f(t) is given only in the form of a curve. For convenience; we rewrite Eq. (11), by taking as the abscissa not the time t, but rather the combination φ = ωt, which is an angle. With the latter, Eqs. (11a and b) become

In order to calculate these integrals numerically, we divide the base length 2π of the curve in a number N of equal parts, each of which is 2π/N = Δ. (In the particular example that follows, N = 48 and Δ = 7.5 deg.) The ordinates of the curve f(φ) at these, N points arc designated as y0, y1, y2..., etc., so that y2 = f(kΔ). With this notation we can replace the above continuous integrals by finite sums, which are approximately equal to these integrals:

In case the subdivision of the base of the curve becomes finer and finer, i.e., N becomes greater and Δ smaller,