Chapter 1
Harmonic Motion
Modern acoustics encompasses a large number of special areas such as ultrasonics, underwater sound, speech and hearing, noise, vibration, sound recording and acoustics, surface wave phenomena and so on. Underlying these special areas and common to all of them are the basic concepts of the production and propagation of waves in fluids and solids. It is these fundamental concepts which are treated in the first part of this book.
Intimately related to wave production and propagation is the special type of motion referred to as simple harmonic motion. Partly because it is readily visualized and partly because of its mathematical simplicity, simple harmonic motion proves to be useful in the description of a great many diverse physical phenomena. It plays a particularly important role in the study of vibrations and waves. The vibrations of any material object or any small portion of a medium through which a wave is traveling is almost invariably assumed to be simple harmonic or made up of a combination of simple harmonic motions. Thus it is appropriate to start the subject of acoustics by reviewing simple harmonic motion.
1-1. The Simple Harmonic Oscillator
All physical theories are based on models and a useful model that serves
for some real systems, consists of a particle of mass, m, fastened
to a massless spring and positioned on a horizontal frictionless surface
as suggested in Figure 1.1.
Figure 1.1. Harmonic Oscillator
If the mass is set into oscillation along a straight line, which is taken as the x axis, then applying Newton's second law at a general time t, yields the equation
(1.1) m d2x/dt2 = - Kx
where K is the force constant of the spring. This differential equation is called the equation of motion. Any function x(t) which describes how the particle moves must be a solution of this differential equation. Finding a solution to a differential equation is fundamentally a process of trial and error. However, a study of the motion of the real system that is being modeled, usually provides some clues as to the nature of the solution. Also, there are available general methods of finding solutions to differential equations which are successful in many instances. Even though Equation (1.1) is a comparatively simple equation and the reader is probably aware of its solution, it may prove instructive to use one of these general methods to find a solution, since later on the same method will be applied in a less familiar situation. For convenience, let wo = Ö K/m .The equation of motion becomes
(1.2) m d2x/dt2 + wo2 x =0
The general method consists of assuming there is a solution of the form
(1.3) x(t) = å0infinity an tn = a0 + a1 t + a2 t2 + a3 t3 +...
where a0, a1, a2 ...are all constants. If such a solution exists then (differentiating (1.3))
d2x(t)/dt2 = 2 a2 + 6 a3 t + 12 a4 t2 + 20 a5 t3 + ...
Substituting this expression along with (1.3) into equation (1.2) one obtains
[2 a2 + wo2 a0] + [6 a3 + wo2 a1] t + [12 a4 + wo2 a2] t2 + ... =0 .
For (1.3) to be a solution of the equation of motion the above expression must be identically zero, i.e., zero for all possible values of time. This condition would obviously be satisfied if each of the bracketed quantities were equal to zero. For arbitrarily chosen values of a0 and a1, the first bracket can be made zero by choosing a2 = -1/2 wo2 a0 , the second bracket can be made zero by choosing a3 = -(wo2)/6 a1, and so on for the other brackets.
Thus, (1.3) will be a solution of the equation of motion for arbitrary values of a0 and a1 provided the other coefficients have the values determined as indicated above. Substituting these values in (1.3) one obtains after rearranging, the following solution of the equation of motion
x(t) = a0 [ 1 - ( wo t )2/2! + +(wo t)4/4! + ... ] + a1/(wo) [ (wot)/1! + (wo t)3/3! + ... ]
The infinite series contained in the first bracket is a Taylor's expansion
for cos(wo t) while
that in the second bracket is an expansion for sin(wot).
The solution can therefore be written in the more familiar form
(1.4) x(t) = C cos(wot) + D sin (wo t)
where C and D have been used to replace a0 and a1 respectively.
In Equation (1.4), C and D are arbitrary in the sense that Equation (1.4) is a solution of the equation of motion no matter what values are assigned to them. Since the equation of motion is a second order differential equation and since Equation (1.4) has two arbitrary constants, it may be considered the general solution of the differential equation. If the position and velocity of the particle are specified at an instant of time, then these so-called initial conditions determine particular values of C and D and the resulting solution is said to be a particular solution of the differential equation. For example, x = 3 cos (wot) is a particular solution of (1.2) corresponding to releasing the mass from rest at a distance 3 units from its equilibrium position.
For any values of C and D, it is always possible to find a number A
and an angle f such that
C = A cos f and D=-A sin f
. Thus, one may write (1.4) in the alternate form
(1.5) x(t) = A cos (wot + f )
which brings out certain features of the motion. First, it may be seen
that the largest positive value of x is |A|. This quantity which represents
the maximum displacement of the particle from its equilibrium position
is called the amplitude of the motion. It is also referred to as the peak
value of x. Second, if as in Figure 1.2, one plots x as a
Figure 1.2. Motion of a simple harmonic oscillator.
function of time using arbitrarily chosen values of A, woand f , one notes that the motion is oscillatory, and that the time interval between successive passages (in the same direction) of the object through its equilibrium position is
To = 2 p / wo= 2 pÖ (m/K)
This interval is called the period of the motion and its reciprocal
fo = 1/T = 1/(2p ) Ö (K/m)
is called the frequency of the motion. (Often it is more convenient to work with wo and to refer to it as the "frequency''. It is always understood that the actual frequency is wo /(2p )) .
Example 1.1.
A machine of mass 500 kg is to be mounted on a platform supported by four identical springs, as indicated in the sketch at the right. The force constant of each spring is 105 N/m and the platform has a mass of 25 kg.
(a) How far willthe platform move down when the machine
is loaded onto the platform?
(b) When mounting the machine, it is allowed to drop
onto the platform from a small height. What is the frequency of the resulting
oscillation?
Solution
(a) Since the resultant force exerted on the platform by the four springs is 4 times the force of one spring, one can consider the four springs to be equivalent to a single spring of force constant K = 4 x 105 N/m. The force exerted by a linear spring is proportional to its deflection D. The equilibrium condition mg = Kx, first applied to the platform and then to the machine plus platform, yields
(25)(9.8)
= ( 4 x 105 ) D1
(525)(9.8)
= (4 x 105 )D2
from which
D2 - D1 = (500)(9.8)/(4 x 105) = 1.23 x 10-2 m = 1.23 cm .
(b) Assume the machine and platform can be modeled as a particle of mass m=525 kg attached to a spring of force constant K = 4 x 105 N/m . Then
fo = 1/(2p ) Ö (K/m) = 1/(2p ) Ö ( 4 x 105 /525) = 4.4 Hz .