
Periodicity, Frequency and Amplitude
A musical note is characterised by three properties; pitch (how high or low the note is), volume (how loud or soft), and its 'colour' or 'timbre'. If the note does not change pitch, we say it has a particular frequency (expressed in cycles per second or hertz, the scientific unit of frequency). The note we hear has a characteristic frequency, most commonly the fundamental frequency of the system from which the note eminates. The 'colour' or 'timbre' of a note is determined by a number of higher frequencies, sounding at the same time as the fundamental. When heard in combination, they give the note its 'colour'. This is why it is possible to distinguish the same note produced on different musical instruments. In some cases the higher frequencies are harmonics of the fundamental that are characteristic of the generating system. The harmonics associated with a vibrating string are one example. In other cases, the higher frequencies may be associated with a secondary process only weakly coupled to the primary resonator. The chiff from certain organ pipes would be a good example. We will consider 'timbre' later. For the moment, we examine vibrations and frequency in a more general way. We will consider the simple pendulum in some detail because of its historical importance. Other systems, that exhibit similar behaviour, we discuss towards the end of this section. Free Oscillations  The Simple Pendulum Galileo's study of the simple pendulum introduces us, as it introduced the modern scientific world, to a simple mechanical oscillator with a characteristic frequency. Because frequency is the reciprocal of the period, the system is said to be periodic or to have a characteristic periodicity. Crudely put, 'frequency' is the number of times a particular event occurs within a given time interval, while 'period' is the time interval between each of those events. It was the periodicity that struck Galileo as interesting and begged the question, why? For our mathematical discussion, we 'idealise' the pendulum (see illustration above). It is a spherical weighted bob attached to one end of an inextensible, massless string, whose other end is fixed to a point on the ceiling. We are viewing the pendulum normal to the plane of motion. By making the bob spherical and the string massless, the centre of gravity of the bob is the centre of gravity of the whole pendulum. By making the string inextensible we know that the energy in the system is described only by the motion and position of the bob. When not in motion, the bob hangs directly below the fixed point of attachment (see lefthand illustration above). We ignore any resistance to the bob's motion from the air through which it moves. In our diagram above, the two light blue bobs and strings (on the right), represent (i) the position from which the bob is released, and (ii) the position to which the pendulum swings after release. It is immaterial whether it is released from the right or from the left. The period is the time the bob takes to travel from the release point and back again. We have also marked two blue lines; the lower dark blue line marks the centre of gravity of the hanging resting bob (the lowest point the centre of gravity of the bob reaches during its motion) while the upper light blue line marks the centre of gravity of the bob at the moment of release and again when it is half way through a period of swing (which is the same height as the centre of gravity of the bob was when it is released). We have indicated the weight of the bob (the product of its mass, m, and the acceleration due to gravity, g), the distance from the upper fixed point to the centre of gravity of the bob, l, and the angle the string makes when the bob is drawn to one side and released. This we denote θ. The total angle of swing is 2θ. When the bob is drawn to one side and released, we notice a number of things about the way the bob swings. At the moment of release the bob is not moving, but it quickly picks up speed until it reaches the lowest point of its swing, after which it slows its speed until, reaching the point that is the reflection of the point of release in the vertical through the point where the string is attached to the ceiling, the bob finally comes, momentarily, to a stop. It then reverses its direction of motion and moves back towards the point of release. From the observation that an object released a distance above the ground falls towards it, we can conclude that the motion of the bob is basically the same effect but the bob is constrained, by the string, to move in an arc rather than in a straight line towards the ground below. The centre of gravity of the bob is similarly constrained to move between the dark and light horizontal blue lines. The dynamic behaviour of objects in freefall gives us a clue as to how we might analyse the dynamics of our pendulum. When objects fall under gravity they accelerate. The longer they fall and the further they fall, the faster they travel. Isaac Newton, who first understood how bodies move under gravity, incorporated the work of Galileo and others into three Laws of Motion:
Let us now look at the forces acting on the bob in greater detail. All forces act through the centre of gravity of the bob. The vertical downward force is that due to gravity acting on the mass of the bob. The bob is released when the string is at an angle θ to the vertical. The tension in the string, T, acts to constrain the centre of gravity of the bob to movement in the arc of a circle centred on the point of attachment on the ceiling. Because it acts perpendicularly to the direction of motion, it cannot accelerate the bob along its path.
We 'resolve' the forces in two perpendicular directions and use Newton's second law of motion in each direction. The tension in the string acts only perpendicularly to the path. The downward force due to gravity must be resolved into two forces, one acting in a direction perpendicular to the path, and the other acting in a direction parallel to the path. Any net force will be associated with acceleration in that direction. perpendicular to the path:
(the term on the right is the product of the mass of the bob m and the centripetal acceleration, the square of the speed v divided by the radius l). parallel to the path:  mg sin (θ) = ma where a is the acceleration along the path. Note the minus sign in the second equation. As the angle between the string and the vertical increases the bob is slowed. The force retards the motion. If we call the lateral displacement x, and assume that x is much smaller than l, then
We can now write the linear differential equation that describes the motion of our pendulum.
The linear differential equation has constant coefficients and has a general solution of the form: x = A cos (ω_{o}t) + B sin (ω_{o}t)
Alternatively, the solution may be expressed in a simpler form: x = A cos (ω_{o}t + φ). where A is the amplitude, ω_{o} is the natural angular frequency of the system and φ is the initial phase of motion. Note that we have not changed the number of arbitrary constants. In the first solution the constants are A and B while in the second solution the two constants are A and φ. In both cases, the value of these arbitrary constants is determined by the initial conditions (for example: how far the bob is displaced before release) and not by the characteristics of the system itself (for example: mass of the bob or the length of the pendulum).
The amplitude for the simple pendulum is the maximum lateral displacement of the bob from its position when the pendulum is at rest. Because cos (ω_{o}t + φ) takes values between 1 and 1, we find that x, the position of the pendulum, takes positive values on one side of the equilibrium line and negative values on the other and the equilibrium line intersects the line of lateral displacement where x = 0. The periodicity of this solution arises from the fact that cos (ω_{o}t + φ) = cos (ω_{o}t + φ + 2π), which reduces to ω_{o}t_{1} = 2π + ω_{o}t_{2}. T _{o}, the period, is given by the equation
therefore
Linear Differential Equations with Complex Variables The general solution of our linear differential equation can be solved using exponential functions and complex variables. If the solution contains only zeroth and first powers of x, it can be separated into its real and imaginary parts. The real part of the solution must satisfy the real part of the original linear differential equation and the imaginary part must satisfy the imaginary part of the original differential equation. Indeed, the real parts of the complex variables will then be the values of amplitude, frequency, position, motion, amplitude and so forth we found earlier. In certain circumstances, it is easier to manipulate the solution expressed this way than when it is written in terms of sine and cosine functions. However, the solutions are equivalent because, from Euler's Formula, exp ( j ω_{o}t) = cos (ω_{o}t) + j sin (ω_{o}t) and exp ( j ω_{o}t) = cos (ω_{o}t)  j sin (ω_{o}t) where j is the square root of 1 x = A cos (ω_{o}t + φ) = Re[ A exp ( j ω_{o}t + j φ) ] = Re[ A exp ( j φ) exp ( j ω_{o}t) ] where Re[] means 'take only the real part of the complex variable'. The complex variables displacement, velocity and acceleration may be thought of as vectors rotating in the complex plane (called 'phasors') with real angular velocity ω_{o}. The real time dependence of each may be found by taking the projection of the complex quantity on the real axis (i.e. Re[]). There is a fourth form of the solution where ω_{o} is a complex number. In this case the solution may be written: x = C exp ( j φ) exp ( j ω_{o}t ) + [ C exp ( j φ) exp ( j ω_{o}t ) ]* where the second term, the expression within [ ]*, is the complex conjugate of the first term. The sum of a number and its complex conjugate is twice the real part of the number. Therefore, x is real, but we still have two unknowns, C and φ. 'Live' Demonstration of Simple Harmonic Motion The behaviour of solutions of the differential equation associated with simple harmonic motion (SHM), is easier to understand by relating SHM to circular motion. This relationship is elegantly demonstrated on Professor FuKwun Hwang's web site with a 'live' graphical demonstration. The Energy Equation Let us return to our differential equations in its most general form.
Writing the equation y_{2} = f (y) and multiplying each side by 2y_{1}, where y_{1} = dy / dt, we get 2 y_{1} y_{2} = 2 f (y) y_{1} Integrating, with respect to t,
This is the general form of the energy equation. Taking the differential equation we met earlier
and using the method outlined above,
Integrating with respect to t, (dx / dt) ^{2} = ω_{o}^{2} (a ^{2}  x ^{2}) where a is a constant of integration (dx / dt) ^{2} = v ^{2} from which we find that the velocity is zero when x = a or x =  a. For the simple pendulum these are the points of release and of momentary rest half way through the period of the swing. The maximum velocity of the bob occurs when x = 0 which is the lowest point on the arc of swing. Here the kinetic energy of the bob is at its maximum. The general expression for the kinetic energy of the bob is mv ^{2} / 2. Its maximum value is m(ω_{o}a) ^{2} / 2. From the principle of the conservation of energy, the potential energy is the maximum energy less the kinetic energy. m(ω_{o}a) ^{2} / 2  mv ^{2} / 2 = m(ω_{o}x) ^{2} / 2 The difference in potential energy of the bob when it is released and when it passes through the lowest point of the swing is the energy available for convertion into kinetic energy. Since the total energy (the sum of the potential energy and the kinetic energy) is constant during the period of the swing (because we have assumed that there is no loss of energy to the surrounding air due to air resistance) it is clear that it is also proportional to a ^{2}, the square of the amplitude. The energy equation may be used to find the relationship between position and time. The solution is: x = a cos (ω_{o}t + φ) which is the result we found earlier, and where ω_{o} can be the positive or negative root of ω_{o}^{2}.
Free Oscillations  Mass on a Spring The motion of a mass m attached to a massless spring whose spring constant (sometimes called 'stiffness') is K, which moves vertically about its equilibrium position, exhibits simple harmonic motion, where ω_{o}^{2} = (K / m). This analysis combines Newton's second law of motion and Hooke's law of elasticity, the latter stating that the force in a system which tries to restore the system to its original condition when it is distorted is proportional to the distortion. For a mass attached to a vertical spring, the equilibrium position is that where the force due to extension of the spring exactly matches the force due to gravity on the mass attached to it. The frequency varies as the square root of the ratio of two quantities (i) the elasticity or degree of stability (in this case the 'stiffness' of the spring K), and (ii) a coefficient of inertia (in this case the 'inertial mass' m). To produce mechanical waves, a system must have the means of storing potential energy (for example, in a spring) and kinetic energy (for which there must 'inertial mass' which can be 'localised' or 'distributed'). As we saw with the simple pendulum, undamped vibratory motion is associated with the transformation of potential to kinetic energy, or the reverse, while the sum of these two forms of energy remains constant. If the system is damped, and energy is steadily lost to some damping mechanism (generally through frictional forces, for example, air resistance), the system may continue its vibratory motion but with a steadily reducing amplitude. A mass m, attached to a horizontal massless spring with spring constant K, free to move on a horizontal frictionless surface, also exhibits SHM. In this case, because the force due to gravity acting on the mass is perpendicular to the direction of the spring, the equilibrium position of the system as a whole is that where the spring is stretched to its natural length, i.e. the greatest length where the spring generates no force. Some authors use 'compliance' C in place of 'stiffness' K where C = 1 / K. In this caseω_{o}^{2} = (1 / Cm). Free Oscillations  Spring of Air A piston is free to move in a vertical cylinder above a volume of air trapped below it. As the piston moves to compress the air below it, a force acting against the motion acts like a spring. In its equilibrium position, the piston slightly compresses the gas in the cylinder to balance the gravitational force due to the mass of the piston. The problem turns out to be no more than finding the correct expression for the 'spring constant' of the air. Let m be the mass of the piston, L and S the length and surface crosssectional area of of the cylinder, p_{a} the atmospheric pressure and γ the ratio of the specific heats of air. The relationship for adiabatic compression between the pressure in the cylinder p_{i} and the volume of air in the cylinder V_{i}, which from simple geometry equals SL, is p_{i} V_{i}^{γ} = a constant. The differential equation for the motion of the piston is
where dp_{i} is the increase in the air pressure in the cylinder, above that when the piston is in its equilibrium position, acting against the piston when, moving a distance x downwards from its equilibrium position, it compresses the air in the cylinder. Downward motion is associated with x increasingly negative and p_{i} increasingly positive. dp_{i} = p_{a} γ x / L from which
The Action of Frictional Forces Friction between solid bodies (called 'sliding friction') is usually independent of their relative velocity. However, a particle moving through a fluid (a gas or a liquid) experiences frictional forces, (e.g. internal damping saps energy from the oscillator and converts it into heat), that are generally proportional to the speed and act to oppose the motion. Let us reexamine the dynamics of a mass attached to a massless spring moving horiontally in the presence of a frictional force of this kind.
Using methods developed by Euler and d'Alembert, we try a complex solution of the form x = A exp ( j φ ) exp ( λ t ), where x is the complex displacement and A exp ( j φ ) is the complex amplitude. We find that ( λ ^{2} + 2 ( R / 2m) λ + ( K / m ) ^{2} ) A exp ( j φ ) exp ( λ t ) = 0. K / m = ω _{o} , the natural angular frequency in the absence of any damping force. The lefthand term, on the lefthand side of the equation, is quadratic and there are two possible values for λ. If ω _{d} is the natural angular frequency of the damped oscillator, where ω _{d} ^{2} = ω _{o} ^{2}  (R / 2m ) ^{2} , the general solution is: x = exp (  Rt / 2m ) { A_{1} exp ( j φ ) exp ( j ω _{d} t ) + A_{2} exp (  j φ ) exp (  j ω _{d} t ) } The real displacement Re [ x ], found by taking the real part of the general solution above, is x = A exp (  Rt / 2m ) cos ( ω _{d} t + φ ) The presence of the exponential term in the solution means that the behaviour is not strictly periodic although the system passes periodically through its equilibrium position x = 0 because where the amplitude is zero, that value remains the same. That period T _{d} = 2 π / ω _{d}. In addition, one can characterise the decay in the amplitude, say the time to reduce the amplitude to 1 / e of its initial value, with a term called decay time, relaxation time or modulus of decay τ = 2m / R It is worth mentioning that frictional decay in acoustics is generally a slow process. In such cases, ω _{d} ^{2} > 0. If however, ω _{d} ^{2} = 0 the system is said to be 'critically' damped and the real displacement is given by: x _{c} = x _{o} ( 1 + ( Rt / 2m ) ) exp (  Rt / 2m ). If ω _{d} ^{2} < 0 the system is said to be overdamped and the system slowly decays passing at most once through its zero position. This behaviour is also called 'deadbeat'. Forced Oscillations If our simple oscillator is driven by an external force f (t), the equation of motion is
The external force f (t) may be impulsive, periodic or nonperiodic. Of particular interest is a driving force that is periodic and starts at a particular time. We wish to investigate the behaviour of the system from when the force is applied. Although the solution consists of two parts, we will consider the transient behaviour later. Let us consider first the steadystate behaviour. Let f (t) = F exp ( j ω t ) where the force function is in its complex form. The linearity of the equation leads us to expect, that in the steady state, both sides of the equation should be harmonic with the same frequency. Replacing the complex variable x by A exp ( j φ ) exp ( j ω t ),
Substituting ω _{o} ^{2} for K / m and α for R / 2m the equation can be rewritten
The real part of this equation, the steadystate displacement, is given by
where Z = R + j ( ω m  K / ω ) = R + j X _{m} . Z is the mechanical impedance and X _{m} is the mechanical reactance. If ω is very small, the real value of x reduces to F / K . In effect we are looking at a situation where the oscillator has been displaced by a constant force, a region where the oscillator is said to be stiffness dominated. This value of x , written x _{s} , is defined as the static displacement. If ω = ω _{o} , the maximum value for the displacement x _{max} = F / ω _{o} R . One particular feature of the system when the value of ω is close to ω _{o} is the shape of the curve that describes the value of x / x _{s} versus ω / ω _{o}. The peak of this curve, 'the resonance peak', becomes wider and higher as the frictional force R increases. This is why the region is described as 'resistance' dominated. Some characterise the sharpness of the curve by the ratio of ω to the full width of the peak at one half the maximum height Δ ω . When ω = ω _{o} , the system is said to be at resonance. The amplification factor Q is defined as the ratio of maximum amplitude at resonance x _{max} to the static displacement x _{s} . Q = x _{max} / x _{s} = K / ω _{o} R
When ω >> ω _{o} , i.e. the frequency of the external force is much greater than the natural frequency of the system undergoing free oscillation, the displacement falls to zero and the system is described as being mass dominated. The ratio x _{o} / x is less than unity where ω ^{2} > ( 2 ω _{o}^{2}  ( R / m ) ^{2} ) One can also examine the phase angle between the displacement and the driving force for different frequencies.
Because tan (  θ ) =  tan ( θ ) we see that the displacement lags behind the driving force. When ω << ω _{o} the phase shift tends to 0° ; when ω = ω _{o} the phase shift is 90° ( π / 2 radians ); and when ω >> ω _{o} the phase shift tends towards 180° ( π radians). The frequency response of a simple oscillator can be 'described' in many ways, each providing its own 'physical' insight. For example, one might consider how the mechanical impedance or its reciprocal, the mechanical admittance (or mobility), varies with frequency. These are complex quantities and we need to consider the behaviour both of their real and imaginary parts.
Let us now briefly mention the other part of our solution, which describes the transient behaviour at and shortly after the application of the driving force. As the system moves from its steady state free oscillation to its new steady state, analysis of its behaviour can be involved. The time scale of its intermediate behaviour depends on a number of parameters, particularly, the degree of damping in the system. A heavily damped system reaches its new steady state quickly because any transient behaviour decays rapidly. When damping is light, transient behaviour is prolonged. If the driving and natural frequencies are close, the system can exhibit 'beating' with a frequency equal to the difference between the two frequencies. In every case, the system will, after some time, find its new steadystate behaviour. The linearity of the equation
leads us to expect a general solution that combines the solution for a free damped oscillator (the homogeneous equation) and the steady state solution for an oscillator undergoing forced oscillation (the particular solution). Our main task would be to choose suitable values for the initial conditions which determine the values of the arbitrary constants A and φ. 