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# Differential Equations

2nd Order Equations

# Mechanical Oscillations

Page 1
Theory
Page 2
Problems 1-3

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.

Figure 1.

### 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 (Figure $$2$$):

${x\left( t \right) = A\cos \left( {\omega t + {\varphi _0}} \right)\;\;\text{or}\;\;}\kern-0.3pt{x\left( t \right) = A\sin \left( {\omega t + {\varphi _0}} \right).}$

In these formulas, $$A$$ means the amplitude of oscillation, $${\omega t + {\varphi _0}}$$ is the phase of oscillation, $${{\varphi _0}}$$ is the initial phase at time $$t = 0.$$

Figure 2.

The variable $$\omega$$ is called the circular or cyclic frequency of oscillation. It is related to the period of oscillation $$T$$ by the formula

$\omega = \frac{{2\pi }}{T}.$

If the displacement $$x\left( t \right)$$ is known, then sequentially differentiating, we can find the velocity and acceleration of the body:

${v\left( t \right) = x’\left( t \right) }={ – A\omega \sin \left( {\omega t + {\varphi _0}} \right),}$
${a\left( t \right) = x^{\prime\prime}\left( t \right) = v’\left( t \right) }={ – A{\omega ^2}\cos\left( {\omega t + {\varphi _0}} \right).}$

This shows that the displacement $$x\left( t \right)$$ and acceleration $$x^{\prime\prime}\left( t \right)$$ satisfy the differential equation

$x^{\prime\prime} + {\omega ^2}x = 0,$

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:

$F = – kx,$

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

${mx^{\prime\prime} = – kx,\;\; }\Rightarrow {x^{\prime\prime} + \frac{k}{m}x = 0.}$

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

$\omega = \sqrt {\frac{k}{m}} .$

The period of oscillation, respectively, will be equal to

$T = \frac{{2\pi }}{\omega } = 2\pi \sqrt {\frac{m}{k}} .$

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

$T = 2\pi \sqrt {\frac{L}{g}} ,$

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

$T = 2\pi \sqrt {\frac{I}{{mga}}} ,$

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 by $${F_\text{C}}$$) is proportional to the velocity of the body, i.e.

${F_\text{C}} = – cx’.$

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

${mx^{\prime\prime} + cx’ + kx = 0,\;\;}\Rightarrow {x^{\prime\prime} + \frac{c}{m}x’ + \frac{k}{m}x = 0.}$

We introduce the following notations: $${\large\frac{c}{m}\normalsize} = 2\beta,$$ $${\large\frac{k}{m}\normalsize} = \omega_0^2.$$ Here $${\omega_0}$$ is the natural frequency of the undamped oscillator (previously, we denoted it as $$\omega\text{),}$$ $$\beta$$ is the damping coefficient. In the new notations, the differential equation looks like

$x^{\prime\prime} + 2\beta x’ + \omega _0^2x = 0.$

We will seek the solution of this equation as a function

$x\left( t \right) = A{e^{\lambda t}}.$

The derivatives are given by

${x’\left( t \right) = A\lambda {e^{\lambda t}},\;\;\;}\kern-0.3pt {x^{\prime\prime}\left( t \right) = A{\lambda ^2}{e^{\lambda t}}.}$

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

${{A{\lambda ^2}{e^{\lambda t}} + 2\beta A\lambda {e^{\lambda t}} }+{ \omega _0^2A{e^{\lambda t}} }={ 0,\;\;}}\Rightarrow {{\lambda ^2} + 2\beta \lambda + \omega _0^2 = 0.}$

The roots of this equation are

${D = 4{\beta ^2} – 4\omega _0^2,\;\;}\Rightarrow {{\lambda _{1,2}} = \frac{{ – 2\beta \pm \sqrt {4{\beta ^2} – 4\omega _0^2} }}{2} }={ – \beta \pm \sqrt {{\beta ^2} – \omega _0^2} .}$

It can be seen that depending on the sign of the radicand $${{\beta ^2} – \omega _0^2}$$ there may be three different types of solutions.

### Case 1. Overdamping: $$\beta \gt {\omega _0}$$

In this case (the case of strong damping), the radicand is positive: $${{\beta ^2} \gt \omega _0^2}.$$ The roots of the characteristic equation are real and negative. The general solution of the differential equation has the form

$x\left( t \right) = {C_1}{e^{{\lambda _1}t}} + {C_2}{e^{{\lambda _2}t}},$

where the coefficients $${C_1},{C_2},$$ as usual, depend on the initial conditions.

Figure 3.

It follows from this expression that there are no oscillations and the system returns to equilibrium exponentially, i.e. aperiodically (Figure $$3$$).

### Case 2. Critical Damping: $$\beta = {\omega _0}$$

In the limiting case when $$\beta = {\omega _0},$$ the roots of the characteristic equation are real and coincide:

${{\lambda _1} = {\lambda _2} = – \beta }={ – {\omega _0}.}$

Here the solution is given by the formula

$x\left( t \right) = \left( {{C_1}t + {C_2}} \right){e^{ – {\omega _0}t}}.$

In this mode, the value of $$x\left( t \right)$$ may even increase at the beginning of the process because of the linear factor $${{C_1}t + {C_2}}.$$ But in the end the deflection $$x\left( t \right)$$ decreases rapidly due to the exponential decay with a characteristic time $$\tau = {\large\frac{{2\pi }}{{{\omega _0}}}\normalsize}.$$ 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 $${\lambda_1},$$
and will be given by the formula

$\require{cancel} {\tau = \frac{{2\pi }}{{{\lambda _1}}} } = {\frac{{2\pi }}{{\beta – \sqrt {{\beta ^2} – \omega _0^2} }} } = {\frac{{2\pi \left( {\beta + \sqrt {{\beta ^2} – \omega _0^2} } \right)}}{{\cancel{\beta ^2} – \cancel{\beta ^2} + \omega _0^2}} } = {\frac{{2\pi }}{{{\omega _0}}}\left[ {\frac{\beta }{{{\omega _0}}} + \sqrt {{{\left( {\frac{\beta }{{{\omega _0}}}} \right)}^2} – 1} } \right] } = {\frac{{2\pi }}{{{\omega _0}}}\Phi \left( {\frac{\beta }{{{\omega _0}}}} \right).}$

The function $$\Phi \left( {\large\frac{\beta }{{{\omega _0}}}\normalsize} \right)$$ 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 $${\large\frac{\beta }{{{\omega _0}}}\normalsize}$$ is $$1,$$ and $${\large\frac{\beta }{{{\omega _0}}}\normalsize} \gt 1$$ in the case of the aperiodic damping (Case $$1\text{).}$$ Therefore for the aperiodic damping mode, we can write

${\tau = \frac{{2\pi }}{{{\omega _0}}}\Phi \left( {\frac{\beta }{{{\omega _0}}}} \right) }\gt{ \frac{{2\pi }}{{{\omega _0}}}.}$

Figure 4.

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: $$\beta \lt {\omega _0}$$

Here the roots of the characteristic equation are complex conjugate:

${\lambda _{1,2}} = – \beta \pm i\sqrt {\omega _0^2 – {\beta ^2}} .$

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

${x\left( t \right) }={ {e^{ – \beta t}}\left[ {{C_1}\cos \left( {{\omega _1}t} \right) }\right.}+{\left.{ {C_2}\sin\left( {{\omega _1}t} \right)} \right],}$

where the oscillation frequency $${\omega_1}$$ is equal to

${\omega _1} = \sqrt {\omega _0^2 – {\beta ^2}} .$

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

${x\left( t \right) }={ A{e^{ – \beta t}}\cos \left( {{\omega _1}t + {\varphi _0}} \right),}$

where $${\varphi_0}$$ is the initial phase of the oscillations and $$A\cos {\varphi _0}$$ is the initial amplitude of the oscillations. We see that classical damped oscillations occur in this mode. Here the oscillation frequency $${\omega_1}$$ is less than the harmonic frequency $${\omega_0},$$ and the oscillation amplitude decreases exponentially with $${e^{ – \beta t}}.$$

### Forced Oscillations. Resonance

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

$F\left( t \right) = {F_0}\cos \left( {\omega t} \right).$

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

${x^{\prime\prime} + \omega _0^2x }={ \frac{{{F_0}}}{m}\cos \left( {\omega t} \right).}$

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

${{x_0}\left( t \right) }={ A\sin \left( {{\omega _0}t + {\varphi _0}} \right),}$

where the amplitude $$A$$ and phase $${\varphi _0}$$ are determined by initial conditions.

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

${x_1}\left( t \right) = B\cos \left( {\omega t} \right).$

The derivatives of this function are

${{x’_1}\left( t \right) = – B\omega \sin \left( {\omega t} \right),\;\;\;}\kern-0.3pt {{x^{\prime\prime}_1}\left( t \right) = – B{\omega ^2}\cos\left( {\omega t} \right).}$

Substituting into the differential equation, we get

${{- B{\omega ^2}\cos\left( {\omega t} \right) }+{ \omega _0^2B\cos\left( {\omega t} \right) }={ \frac{{{F_0}}}{m}\cos\left( {\omega t} \right),\;\;}}\Rightarrow { – B{\omega ^2} + \omega _0^2B = \frac{{{F_0}}}{m},\;\;}\Rightarrow {B = \frac{{{F_0}}}{{m\left( {\omega _0^2 – {\omega ^2}} \right)}}.}$

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

${x\left( t \right) = {x_0}\left( t \right) + {x_1}\left( t \right) } = {A\sin \left( {{\omega _0}t + {\varphi _0}} \right) }+{ \frac{{{F_0}}}{{m\left( {\omega _0^2 – {\omega ^2}} \right)}}\cos \left( {\omega t} \right).}$

We see from this expression that the second term showing the effect of the external force increases dramatically as $$\omega \to {\omega _0}.$$ This phenomenon is called resonance. In this simple model, the amplitude $$x\left( t \right)$$ 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:
${x^{\prime\prime} + 2\beta x’ + \omega _0^2x }={ \frac{{{F_0}}}{m}\cos \left( {\omega t} \right).}$
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

${x^{\prime\prime} + 2\beta x’ + \omega _0^2x }={ \frac{{{F_0}}}{m}{e^{i\omega t}}.}$

We will seek a particular solution in the form

${x_1}\left( t \right) = B{e^{i\left( {\omega t + \varphi } \right)}},$

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

${{x’_1}\left( t \right) = i\omega B{e^{i\left( {\omega t + \varphi } \right)}},\;\;\;}\kern-0.3pt {{x^{\prime\prime}_1}\left( t \right) = – {\omega ^2}B{e^{i\left( {\omega t + \varphi } \right)}}.}$

Substituting this into the differential equation, we obtain

${- {\omega ^2}B{e^{i\left( {\omega t + \varphi } \right)}} }+{ 2\beta i\omega B{e^{i\left( {\omega t + \varphi } \right)}} }+{ \omega _0^2B{e^{i\left( {\omega t + \varphi } \right)}} } = {\frac{{{F_0}}}{m}{e^{i\omega t}},}$
$\Rightarrow {\left( { – {\omega ^2} + 2i\beta \omega + \omega _0^2} \right) \cdot}\kern0pt{ B\cancel{e^{i\omega t}}{e^{i\varphi }} }={ \frac{{{F_0}}}{m}\cancel{e^{i\omega t }}, }$
${\Rightarrow \left( {\omega _0^2 – {\omega ^2}} \right)B + 2i\beta \omega B }={ \frac{{{F_0}}}{m}{e^{ – i\varphi }}.}$

By de Moivre’s formula

${e^{ – i\varphi }} = \cos \varphi – i\sin\varphi .$

Therefore, we can write:

${\left( {\omega _0^2 – {\omega ^2}} \right)B + 2i\beta \omega B }={ \frac{{{F_0}}}{m}\left( {\cos \varphi – i\sin\varphi } \right).}$

Equating separately the real and imaginary parts, we obtain

$\left\{ \begin{array}{l} \left( {\omega _0^2 – {\omega ^2}} \right)B = \frac{{{F_0}}}{m}\cos \varphi \\ 2\beta \omega B = \frac{{{F_0}}}{m}\sin\varphi \end{array} \right..$

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

${{{B^2}\left[ {{{\left( {\omega _0^2 – {\omega ^2}} \right)}^2} + 4{\beta ^2}{\omega ^2}} \right] }={ {\left( {\frac{{{F_0}}}{m}} \right)^2},\;\;}}\Rightarrow {{B \;\text{=}\;}\kern-0.3pt{ \frac{{{F_0}}}{{m\sqrt {{{\left( {\omega _0^2 – {\omega ^2}} \right)}^2} + 4{\beta ^2}{\omega ^2}} }}.}}$

We find the angle $$\varphi$$ by dividing the second equation by the first one:

${\frac{{2\beta \omega }}{{\omega _0^2 – {\omega ^2}}} }={ \frac{{\sin \varphi }}{{\cos\varphi }} }={ \tan \varphi ,\;\;}\Rightarrow {\varphi = \arctan \frac{{2\beta \omega }}{{\omega _0^2 – {\omega ^2}}}.}$

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

${{x_1}\left( t \right) }={ \frac{{{F_0}}}{{m\sqrt {{{\left( {\omega _0^2 – {\omega ^2}} \right)}^2} + 4{\beta ^2}{\omega ^2}} }}\cdot}\kern0pt{ {e^{i\left( {\omega t + \varphi } \right)}},}$

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

${\text{Re}\left[ {{x_1}\left( t \right)} \right] } = {\frac{{{F_0}}}{{m\sqrt {{{\left( {\omega _0^2 – {\omega ^2}} \right)}^2} + 4{\beta ^2}{\omega ^2}} }} \cdot}\kern0pt{ \cos\left[ {i\left( {\omega t + \varphi } \right)} \right].}$

The final answer is the sum of two terms:

${x\left( t \right) }={ {x_0}\left( t \right) + \text{Re}\left[ {{x_1}\left( t \right)} \right] } = {{x_0}\left( t \right) }+{ \frac{{{F_0}}}{{m\sqrt {{{\left( {\omega _0^2 – {\omega ^2}} \right)}^2} + 4{\beta ^2}{\omega ^2}} }} \cdot}\kern0pt{ \cos\left[ {i\left( {\omega t + \varphi } \right)} \right],}$

where $${x_0}\left( t \right)$$ 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 $${x_0}\left( t \right)$$ 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:

${x\left( t \right) }={ \frac{{{F_0}}}{{m\sqrt {{{\left( {\omega _0^2 – {\omega ^2}} \right)}^2} + 4{\beta ^2}{\omega ^2}} }} \cdot}\kern0pt{ \cos\left[ {i\left( {\omega t + \varphi } \right)} \right],}$

where $$\varphi =$$ $$\arctan {\large\frac{{2\beta \omega }}{{\omega _0^2 – {\omega ^2}}}\normalsize},$$ $$\beta$$ 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

${x_{\max }}\left( {\omega = {\omega _0}} \right) = \frac{{{F_0}}}{{2m\beta {\omega_0} }}.$

Figure 5.

The dependence of the amplitude $${x_{\max }}$$ of steady oscillations on the frequency $$\omega$$ of the external force near resonance for different damping coefficients $$\beta$$ 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 $$\omega$$ of the external force approaches zero, the amplitude of oscillations approaches $${\large\frac{{{F_0}}}{{m\omega _0^2}}\normalsize} :$$

${\lim\limits_{\omega \to 0} {x_{\max }} } = {\lim\limits_{\omega \to 0} \frac{{{F_0}}}{{m\sqrt {{{\left( {\omega _0^2 – {\omega ^2}} \right)}^2} + 4{\beta ^2}{\omega ^2}} }} }={ \frac{{{F_0}}}{{m\omega _0^2}}}$

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

${Q = \frac{{{x_{\max }}\left( {\omega = {\omega _0}} \right)}}{{{x_{\max }}\left( {\omega \to 0} \right)}} } = {\frac{{\frac{{{F_0}}}{{2m\beta {\omega _0}}}}}{{\frac{{{F_0}}}{{m\omega _0^2}}}} } = {\frac{{{\omega _0}}}{{2\beta }},}$

where $$\beta$$ 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 $$\delta.$$ The relationship between the quality factor $$Q$$ and the logarithmic decrement $$\delta$$ (for small $$\delta$$) is expressed by the simple formula:

$Q = \frac{\pi }{\delta }.$

## Solved Problems

Click on problem description to see solution.

### ✓Example 1

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

### ✓Example 2

A mass is suspended on two springs connected in series. The stiffness of one spring is twice more than of the other: $${k_2} = 2{k_1}.$$ How does the period of oscillation change if the springs are connected in parallel (Figure $$7$$)?

### ✓Example 3

Find the $$Q$$ factor of an oscillator, if after $$50$$ oscillations the amplitude of the displacement has decreased twice.

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Theory
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Problems 1-3