Mechanical Oscillations

Oscillatory processes are widespread in nature and technology. In astronomy, planets revolve around the sun, variable stars, such as Cepheids, periodically change their brightness, motion of the moon causes the tides. In geophysics, periodic processes occur in climate change, in the behavior of ocean currents, and in the dynamics of cyclones and anticyclones. Within living organisms, there are dozens of different periodic processes with periods from fractions of a second up to a year, etc.

We begin by considering the simplest oscillating system - a harmonic oscillator. Free Harmonic Oscillations

An example of such a simple system is the mass m, attached to a spring of stiffness k (Figure 1). In the ideal case (neglecting air resistance and friction), such a system will perform undamped harmonic oscillations, in which the displacement x is described by the cosine or sine function:

In these formulas, A means the amplitude of oscillation, ωt + φ0 is the phase of oscillation, φ0 is the initial phase at time t = 0. The variable ω is called the circular or cyclic frequency of oscillation. It is related to the period of oscillation T by the formula

Fig.1 Fig.2If the displacement x(t) is known, then sequentially differentiating, we can find the velocity and acceleration of the body:

This shows that the displacement x(t) and acceleration x''(t) satisfy the differential equation

which is called the equation of harmonic oscillations. The solution of this equation are mentioned above cosine or sine functions.

In the case of a mass on a spring, the restoring force for small oscillations obeys Hooke's law:

where k is the stiffness of the spring. Here the coordinate x = 0 corresponds to the point of equilibrium, in which the force of gravity is balanced by the initial tension of the spring. Then, according to Newton's second law, the movement of the mass will be described by the differential equation

Thus, the mass on the spring will perform undamped harmonic oscillations with the circular frequency

The period of oscillation, respectively, will be equal to

A similar analysis of other oscillatory system - a simple (mathematical) pendulum - leads to the following formula for the oscillation period:

where L is the length of the pendulum, g is the acceleration of gravity.

In the case of a compound or physical pendulum, the period of oscillation is given by

where I is the moment of inertia of the pendulum about the pivot point, m is the mass of the pendulum, a is the distance between the pivot point and the center of mass of the pendulum. Damped Oscillations

In real systems, there is always a resistance or friction, which leads to a gradual damping of the oscillations. In many cases, the resistance force (denoted Fc) is proportional to the velocity of the body, i.e.

Then, taking into account the force of resistance, the differential equation for the "mass-spring" system is written as

We denote: c/m = 2β, k/m = ω02. Here ω0 is the natural frequency of the undamped oscillator

(previously, we denoted it as ω), β is the damping coefficient. In the new notation, the differential equation looks like this:

We seek the solution of this equation as a function

Derivatives respectively are equal to

Substituting this into the differential equation, we obtain an algebraic characteristic equation:

This equation has the following roots:

It is seen that, depending on the sign of the radicand  β2 − ω02 there may be three different types

of solutions.       Case 1. Overdamping:  β > ω0

In this case (the case of strong damping), the radicand is positive:  β2 > ω02. The roots of the

characteristic equation are real and negative. The general solution of the differential equation has the form

where the coefficients C1, C2, as usual, depend on the initial conditions. It follows from this expression that there are no oscillations and the system returns to equilibrium exponentially, i.e. aperiodically (Figure 3).

Fig.3 Fig.4      Case 2. Critical Damping:  β = ω0

In the limiting case when  β = ω0, the roots of the characteristic equation are real and coincide:

Here the solution is given by the formula

In this mode, the value of x(t) may even increase at the beginning of the process because of the linear factor C1t + C2. But in the end the deflection x(t) decreases rapidly due to the exponential decay with a characteristic time τ = 2π/ω0. Note that in this critical mode the relaxation occurs faster than in the case of the aperiodic damping (Case 1). Indeed, in this mode, the relaxation time will be determined by the smaller (in absolute value) root λ1, and will be given by the formula

The function Ô(β/ω0) included in this expression is monotonically increasing. It is always greater than or equal to 1, as shown in Figure 4. In the critical case (Case 2) the ratio β/ω0 is 1, and β/ω0 > 1 in the case of the aperiodic damping (Case 1). Therefore, for the aperiodic damping mode, we can write

Thus, the critical damping mode provides the fastest possible return of the system to equilibrium. This is often used, for example, in door closing mechanisms.       Case 3. Underdamping:  β < ω0

Here the roots of the characteristic equation are complex conjugate:

The general solution of the differential equation is oscillatory in nature and can be written as

where the oscillation frequency ω1 is equal to

The resulting formula can be written in a somewhat different form:

where φ0 is the initial phase of the oscillations and Acosφ0 is the initial amplitude of the oscillations. We see that classical damped oscillations occur in this mode. Here the oscillation

frequency ω1 is less than the harmonic frequency ω0, and the oscillation amplitude decreases exponentially with  exp(− βt). Forced Oscillations. Resonance

Suppose that an external force, which varies with time according to a harmonic law with frequency ω, acts on the oscillatory system:

In the case of an undamped oscillator, the following differential equation can be written based on Newton's second law:

According to the general theory, the solution of this equation is the sum of the general solution of the homogeneous equation and a particular solution of the nonhomogeneous equation.

The general solution of the homogeneous equation has been obtained above. It is written as

where the amplitude A and phase φ0 are determined by initial conditions.

Let us find a particular solution of the nonhomogeneous differential equation. We seek it in the form

The derivatives of this function are

Substituting into the differential equation, we get

Hence the general solution of the nonhomogeneous equation can be written as

We see from this expression that the second term showing the effect of the external force increases dramatically when ω → ω0. This phenomenon is called resonance. In this simple model, the amplitude x(t) becomes equal to infinity, if the frequency of the external force is equal

to the frequency of free oscillations of the system.

The physical model of the forced oscillations will be more realistic if we consider the damping of oscillations. Then, Newton's second law yields the following differential equation:

The solution of this equation is also represented as the sum of the general solution of the homogeneous equation and a particular solution of the nonhomogeneous equation.

The solution of the homogeneous equation, as shown above, includes three possible scenarios (aperiodic damping mode, critical damping and the oscillatory solution in the case of underdamping).

Find a particular solution of the nonhomogeneous equation. It is more convenient to use the complex form of the differential equation, which can be written as

We seek a particular solution in the form

that is, suppose that the oscillations in the system will occur with the frequency ω of the external force, and perhaps with some phase shift φ. As a result, we have

Substituting this into the differential equation, we obtain

by de Moivre's formula

Therefore, we can write:

Equating separately the real and imaginary parts, we obtain

From this system we find the coefficient B and the angle φ. Squaring both sides and adding, we get:

We find the angle φ by dividing the second equation by the first one:

Thus, a particular solution of the nonhomogeneous equation in the complex form is given by

where the shift angle shift φ is calculated by the formula obtained above. Accordingly, the real part of the solution can be written as

The final answer is the sum of two terms:

where xh(t) is the general solution of the homogeneous equation, which describes the damped oscillator without external force.

Note that due to the decay, the solution of the homogeneous equation xh(t) will tend to zero. Therefore, in steady state the oscillations will depend only on the external force, that is to be determined by the second component of the general solution:

where , β is the damping coefficient.

This formula also describes the phenomenon of resonance, and the maximum amplitude of the steady-state oscillations at resonance will be finite and equal to

The dependence of the amplitude xmax of steady oscillations on the frequency ω of the external force near resonance for different damping coefficients β is shown below in Figure 5. These curves are called resonance curves.

Resonance properties of an oscillatory system can be evaluated using the quality factor (Q factor). The Q factor indicates how many times the amplitude of the forced oscillations at resonance exceeds the amplitude far from resonance.

As the frequency ω of the external force approaches zero, the amplitude of oscillations

approaches :

Therefore, the Q factor of a mechanical oscillatory system is equal to

where β is the damping coefficient.

The Q factor is a very useful feature. From the energy point of view, it shows the ratio of energy stored in an oscillatory system to the energy that the system loses for a single oscillation period.

The energy losses are also characterized by the logarithmic decrement δ. The relation between the quality factor Q and the logarithmic decrement δ (for small δ) is expressed by a simple formula:

Fig.5 Fig.6   Example 1 A ring of radius R performs small oscillations around the pivot point O (Figure 6). Determine the period of oscillation.

Solution. The ring, suspended at the point O, is a physical pendulum. The period of oscillation is determined by the formula

     

where I is the moment of inertia of the ring about its center, m is the mass of the ring, a is the distance from the pivot point to the centre of the ring.

The moment of inertia of a ring of mass m is equal to I0 = mR2. Since the distance from the centre of the ring to the pivot point is equal to R, then using the parallel axis theorem (aka Huygens-Steiner theorem), we find the total moment of inertia of the pendulum:

     

Given that a = R, we obtain the following expression for the oscillation period:

     

   Example 2 A mass is suspended on two springs connected in series. The stiffness of one spring is twice more than of the other: k2 = 2k1. How does the period of oscillation change, if the springs are connected in parallel (Figure 7)?

Fig.7Solution. Calculate the equivalent stiffness in the case of serial and parallel connection of springs.

In the case of series connection, the elastic force in each spring is equal to the force of gravity (without taking into account the weight of the springs). The total extension is the sum of the extensions of each spring:

     

Then the equivalent stiffness is given by

     

When connected in parallel, the extension of both springs is the same, and the total elastic force will be equal to the sum of the forces in each spring:

     

Hence, the equivalent stiffness of the springs connected in parallel is given by

     

The period of oscillation of the springs connected in series is

     

and in the case of parallel connection:

     

Now we can find how the period of oscillation changes when transitioning from series to parallel connection of the springs:

     

Given that the stiffness of one spring is twice more than the other, we obtain:

     

   Example 3 Find the Q factor of an oscillator, if after 50 oscillations the amplitude of the displacement has decreased by 2 times.

Solution. We first calculate the logarithmic decrement δ. By definition, the logarithmic decrement is proportional to the natural logarithm of the ratio of the amplitudes x0 and xN of two oscillations, separated by N periods:

     

In our case it is equal to

     

Then the Q-factor of the system is