10. Oscillatory Motion

10.1 Oscillatory Motion¶

A motion repeating itself is referred to as periodic or oscillatory motion. An object in such motion oscillates about an equilibrium position due to a restoring force or torque. Such force or torque tends to restore (return) the system toward its equilibrium position no matter in which direction the system is displaced. This motion is important to study many phenomena including electromagnetic waves, alternating current circuits, and molecules. For a vibration to occur, two quantities are necessary to be present—stiffness and inertia.

10.2 Free Vibrations¶

When a system vibrates, a restoring force must be present. In addition to that force, there is always a retarding or damping force such as friction. If the effect of the damping force is small and can be neglected, then the motion is classified as free and undamped motion. Otherwise, the motion is classified as free damped motion. In both cases, the motion is known as free vibration since no forces other than the restoring and damping forces exist during vibration. If a driving force that does positive work on the system exists, the motion is classified as forced vibration.

This force may be applied externally to the system or sometimes is produced within the system. In this chapter, the case in which a restoring force is directly proportional to the displacement is considered. The resulting motion is then known as a harmonic vibration and the system is said to be linear. If the restoring force depends on the displacement in some other way, the resulting motion is known as anharmonic vibration and the system is said to be nonlinear.

10.3 Free Undamped Vibrations¶

This kind of motion is known as the simple harmonic motion. Next, we will examine examples of such motion in physics.

10.3.1 Mass Attached to a Spring¶

Consider a block of mass m attached to a light spring of spring constant k that is fixed at the other end (see Figure 1). Suppose that the system lies on a frictionless horizontal surface. For small displacements, the restoring force acting on the block by the spring is given by Hook’s law

As we’ve mentioned in 4.1 Introduction, if the block is displaced slightly to the right (for example to $x=A$), the restoring spring force will accelerate the block to the left transferring its potential energy into kinetic energy As the block reaches its equilibrium position $x=0$, all of its potential energy will be transformed into kinetic energy and it will overshoot to the other side. Again, as it moves left, the spring force decelerates the block to the right, transferring its kinetic energy into potential energy until all of its energy is potential at $x= -A$ where it comes to rest. At that point, it accelerates back to $x=0$ and regains all of its kinetic energy where it overshoots again to $x=A$. Therefore, stiffness restores the mass where inertia is responsible for the mass to overshoot. From Newton’s second law we, have

or

or

where $\omega _{n}=\sqrt{k/m}$ is called the natural angular frequency of the system. The general solution of this equation is of the form

where $A_{1}$ and $A_{2}$ are arbitrary constants that can be found from the initial conditions. Therefore, there are many possible motions with the same angular frequency $\omega _{n}$. By multiplying and dividing (5) by $\sqrt{A_{1}^{2}+A_{2}^{2}}$, you can show that the solution may be written as

where $A=\sqrt{A_{1}^{2}+A_{2}^{2}}$ is called the amplitude of motion and $\phi =\tan ^{-1}A_{2}/A_{1}$ is called the phase constant. In general, $\phi$ is chosen such that $0\le \phi \le \pi. A$ and $\phi$ can be determined from the initial conditions, i.e., from the values of the displacement and velocity when the motion starts. The mass therefore oscillates between A and $-A$. The quantity $(\omega _{n}t-\phi )$ is called the phase angle. If this angle is increased by $2\pi$, all physical quantities such as the displacement, velocity, and acceleration repeat themselves. The plot of x versus t is shown in Figure 2. If A is fixed and $\phi$ is changed the motion will be the same except that the same physical quantities will appear either earlier or later than the preceding motion.

10.3.1.1 The Period and Frequency of Motion¶

The period of motion is the time required for one complete cycle or oscillation. Since the phase angle is changed by $2\pi$ after one complete cycle, we have for the mass–spring system,

or

The frequency is defined as the number of complete cycles per unit time

This frequency is called the natural frequency of the motion. The unit of the frequency is cycles/s or hertz (Hz).

10.3.1.2 The Phase Difference¶

The phase constant $\phi$ is important when comparing two or more oscillations of the same frequency Suppose a certain vibration has $\phi =0$, this means that at $t=0$ the displacement is maximum $x=A$. If a second vibration has also $\phi =0$, then the two vibrations are said to be in phase (see Figure 3 part a). Otherwise, the two vibrations are out of phase. If the phase constant of the second vibration is $\phi >0$, then the second vibration is leading the first vibration in phase by $\phi$. If $\phi <0$, then the second vibration is lagging the first by $\phi$. If $\phi =\pm \pi$, the two vibrations are said to be in antiphase with each other (see Figure 3 part b).

10.3.1.3 The Velocity and Acceleration¶

The velocity of the mass is

This can also be written as

The acceleration of the mass is

or

Hence, the velocity and acceleration also vary harmonically with time with amplitudes $\omega _{n}A$ and $\omega _{n}^{2}A$, respectively, but they all have the same angular frequency From Eqs. 10.5 and 10.7 you can see that the velocity leads the displacement by $\pi /2$ or 90. The acceleration on the other hand leads the velocity by $\pi /2$ and the displacement by $\pi$ or 180. Figure 10.4 shows the displacement, velocity, and acceleration versus time.

10.3.1.4 Boundary Conditions¶

Boundary conditions are used to find A and $\phi$ for a specific vibration. Suppose that the vibration is measured when the stopwatch is set to zero, i.e., at $t=0$ and that at that instant the mass is released from rest at a distance of $x=A_{1}$ from its equilibrium position. Substituting these conditions into Eqs. 10.3 and 10.4, we have

Dividing (15) by (14) gives

Squaring and adding Eqs. 10.9 and 10.8 gives

or

10.3.2 Simple Harmonic Motion and Uniform Circular Motion¶

Consider a circle of radius A centered at the $\mathrm {x}$ and $\mathrm {y}$ axes as shown in Figure 8. Let A be the position vector of a particle $\mathrm {P}$ rotating with a constant angular speed $\omega _{n}$ in the anticlockwise direction. The particle is thus in uniform circular motion. Suppose $\mathrm {P}$ starts the rotation at $t=0$ at an angle of $\phi$ measured from the positive $\mathrm {x}$-axis. At any time, the angular position of the particle is given by $(\omega _{n}t+\phi )$, therefore the vector position of the particle at any time is

Hence,

and

That is, as $\mathrm {P}$ moves in uniform circular motion, its projection $\mathrm {P}'$ on the x-axis moves in simple harmonic motion where the radius of the circle is equal to the amplitude of motion. The projection of $\mathrm {P}$ along the $\mathrm {y}$-axis also undergoes simple harmonic motion. Thus, uniform circular motion may be considered as a combination of the simple harmonic motions of the projections of $\mathrm {P}$ on each axis. These two simple harmonic motions have equal amplitudes and angular frequencies but are in quadrature with each other (they differ in phase by $\pi /2$). The linear tangential velocity of the particle in this uniform circular motion is given by

The $\mathrm {x}$ component of the velocity is from Figure 9 given by

The acceleration of the particle in uniform circular motion is just the radial (centripetal) acceleration that is given by

The $\mathrm {x}$ components of the acceleration (see Figure 10) is

Hence as you can see, the displacement, velocity, and acceleration of the projection of $\mathrm {P}$ onto the $\mathrm {x}$ (or $\mathrm {y}$ axis) are the same as that of a simple harmonic motion. From this, we conclude that the simple harmonic motion can be represented as the projection of uniform circular motion along a diameter of the circle.

10.3.3 Energy of a Simple Harmonic Oscillator¶

Since in a simple harmonic oscillator, there aren’t any dissipative forces, the total mechanical energy of the system is conserved and is equal to the sum of its kinetic and potential energies, that is

Thus,

or

The equation of motion of a simple harmonic oscillator can be obtained from the total mechanical energy of the system as follows:

or

Hence

where $\omega _{n}=\sqrt{k/m}$. As the mass moves, its kinetic energy is transformed into potential energy and vice versa. Figure 10.11 shows the kinetic energy and potential energy of the system as a function of time and as a function of the displacement respectively Note that the variation of U and K with time is at twice the angular frequency of the variation of x, v, and a with time. This is because the potential energy is converted to kinetic energy twice in each cycle. The velocity of the simple harmonic oscillator can be obtained from the total energy of the system. From (60), we have

Hence, the maximum speed is at $x=0$ and is zero at $x=\pm A$ which are called the turning points as discussed in Chap. chap444.

10.3.4 The Simple Pendulum¶

The simple pendulum is an example of an angular vibration in which the restoring effect is due to a restoring torque. A simple pendulum consists of a mass (called the bob) suspended by a light string of length L that is fixed at the other end (see Figure 13). If the mass is pulled to the right or left from its equilibrium position and released, then the pendulum will swing in a vertical plane about an axis passing through O. The resulting motion is then a periodic or oscillatory motion. The restoring torque is due to gravity and is given by

The minus sign indicates that the torque is a restoring torque, since it always tends to decrease $\theta$. The moment of inertia of the bob about an axis passing through $\mathrm {O}$ is

From Newton’s second law in angular form, we have

Hence,

or

This equation does not represent a harmonic motion. That is because the torque is not directly proportional to the angular displacement. Thus, the system is nonlinear. However for small angular displacements, we have $\sin \theta \approx \theta ($since $\sin \theta =\theta -\theta ^{3}/3!+\theta ^{5}/5!\ldots )$ and (93) becomes

or

where $\omega _{n}=\sqrt{g/L}$. Hence for small angular displacements, the motion is a simple harmonic motion. The solution of (95) is of the form

where $\theta _{m}$ is the maximum angular displacement and $\phi$ is the phase constant. The plot of this equation is shown in Figure 14. The period of the simple pendulum is therefore given by

10.3.4.1 Energy¶

The kinetic energy of the simple pendulum is

Taking the reference point of potential energy of the system to be zero when the bob is at the bottom, we have

The total energy is therefore given by

For small $\theta$, we have $\cos \theta \approx 1-\frac{\theta ^{2}}{2}$since $\cos \theta =1-\theta ^{2}/2!+\theta ^{4}/4!\ldots )$ thus

Since

we have

or

Therefore, the total energy of the system is constant. Figure 10.15 shows the variation of the kinetic and potential energies with the displacement.

The equation of motion may also be obtained from energy as follows:

or

10.3.5 The Physical Pendulum¶

The physical pendulum is a rigid body that oscillates about an axis passing through a point in the body other than its center of mass (the center of mass is assumed to be located at the center of gravity). Figure 10.16 shows a rigid body pivoted at point $\mathrm {O}$ that is at a distance d from the center of mass. The equilibrium position of the body is when its center of mass is directly below the pivot O. If the body is displaced either to the right or left from the equilibrium position, a restoring torque due to gravity will act on it. As a result, the body will oscillate in a vertical plane where the axis of rotation is perpendicular to the page. The restoring torque is given by

where M is the mass of the body and d is the moment arm of the tangential component of the weight $(Mg\ \sin \theta )$. From Newton’s second law, we have

For small angular displacements $\sin \theta \approx \theta$ and hence

or

This equation is of a simple harmonic motion with an angular frequency of

and a period of motion of

Thus,

Therefore, the moment of inertia of a body can be found by measuring its period when it is in simple harmonic motion as a physical pendulum. Note that, the simple pendulum is a special case of the physical pendulum since for a simple pendulum of mass m, the moment of inertia is

and thus, the angular frequency is

This angular frequency is of a simple pendulum where d represents the length of the string.

10.3.6 The Torsional Pendulum¶

The torsional pendulum consists of a rigid body suspended by a wire from its center of mass where the other end of the wire is fixed as shown in Figure 19. The body is in equilibrium if the wire is untwisted. If the body is rotated through an angle $\theta$ it will oscillate about its equilibrium position (the line OP) due to a restoring torque exerted by the twisted wire on the body. This torque is found to be directly proportional to the angular displacement of the body. That is

where k is called the torsional constant. Its value depends on the property of the wire. Note that this equation is the rotational analogue of Hook’s law in linear form $(F=-kx)$. From Newton’s second law, we have

or

That gives

or

where $\omega _{n}=\sqrt{k/I}$ and the period is $T=2\pi \sqrt{I/k}.$

10.4 Damped Free Vibrations¶

In this section, we will discuss the case in which the effect of damping that is due to a nonconservative force cannot be neglected. An example of such a force in mechanical systems is the force of friction. In this case, the mechanical energy of the system will be lost, the amplitude of motion will decrease to zero, and the oscillation dies out eventually. Here, we will discuss damping due to friction in the simplest case, where the frictional force is proportional to the first power of the velocity of the oscillating body. An example of such a frictional force is the force that an object experience when moving in a fluid with a low speed and is given by

where b is a positive constant called the damping coefficient. Its SI units is $\mathrm {N}(\mathrm {m}\,\mathrm {s}^{-1})=\mathrm {k}\mathrm {g}\,\mathrm {s}^{-1}$. The negative sign shows that the direction of the force is always opposite to the velocity. Now consider the spring–mass system as shown in Figure 21, the cylinder shown in the figure contains a viscous fluid and a piston moving in it. Such device is known as the viscous damper. The net force on the oscillating body is

hence

or

where $\gamma =b/m$ and $\omega _{n}=\sqrt{k/m}$. The units of $\gamma$ is $\mathrm {s}^{-1}$. This equation is a second order linear differential equation of constant coefficients. We may assume a solution of the form

Substituting this solution into (144) gives the characteristic (auxiliary) equation given by

The roots of this equation are given by

and

From superposition, the general solution is given by

Three possible solutions arise depending on whether the sign of the bracket $(\gamma ^{2}/4-\omega _{n}^{2})$ is positive, negative or zero, i.e., depending on the size of the damping force. The roots $\lambda _{1}$ and $\lambda _{2}$ are either distinct real roots, equal real roots or a conjugate complex roots. Therefore, there are three possible motions of the system.

10.4.1 Light Damping (Under-Damped) $(\gamma <2\omega _{n})$¶

If $\gamma <2\omega _{n}$ the resulting roots are complex roots given by

and

where

Hence, (149) may be written as

Since $e^{\pm ix}=\cos x\pm i\sin x$ we have

where $A_{1}=C_{1}+C_{2}$ and $A_{2}=i(C_{1}-C_{2})$. As mentioned earlier (156) can be written as

where A is the initial amplitude of motion. $Ae^{\frac{-\gamma }{2}t}$ is called the amplitude of motion and $\phi$ is the phase constant and $\omega _{D}$ is the angular frequency of the damped motion. This equation shows that the system oscillates in a decreasing harmonic motion where the amplitude of motion decreases exponentially with time until eventually the oscillation dies out (see Figure 22). The dashed lines in Figure 22 are called the envelope of the oscillation curve. The period of motion in light damping is therefore given by

If $b=0$ and thus $\gamma =0$ the period of motion is reduced to that of a simple harmonic oscillator. If $\gamma \ll \omega _{D}$, the situation is referred to as very light damping and $\omega _{D}\approx \omega _{n}$. Furthermore if there are two amplitudes $A_{a}$ and $A_{b}$ separated by the period of motion, then their ratio is given by

A quantity known as the logarithmic decrement is defined as

10.4.2 Critically Damped Motion $(\gamma =2\omega _{n})$¶

If $\gamma =2\omega _{n}$, then the roots are equal real roots

In that case, the motion decays without oscillation (see Figure 23) and the general solution of (144) is

$C_{1}$ and $C_{2}$ are found from boundary conditions. If at $t=0, x=A$, and $v=0,$ then

and

or

That gives

10.4.3 Over Damped Motion (Heavy Damping) $(\gamma >2\omega _{n})$¶

If $\gamma >2\omega _{n}$, the roots are distinct real roots given by

and

The general solution is given by

or

where

$C_{1}$ and $C_{2}$ are found from boundary conditions. As critical damping, the resulting motion here is nonperiodic but the system returns to its equilibrium position at large values of t unlike critical damping (see Figure 24).

10.4.4 Energy Decay¶

In damped free vibrations, the total mechanical energy is not constant since the damping force opposes the motion and dissipates the energy of the system. Now, consider the mass–spring system, the total mechanical energy of the system is

The rate of change of energy is

For damped vibrations in which the damping force is directly proportional to the velocity, we have

Hence,

Thus, the energy decreases with time in any damped motion and the rate in which it decreases is not uniform.

10.5 Forced Vibrations¶

In the previous sections, only free vibrations have been considered (i.e., vibrations in which only a restoring and damping force act within the system during motion). This section considers the case in which an external driving force is applied to the vibrator. This force is given as a function of time and we have

Here, we will consider the case in which the force is a simple periodic force given by

where $F_{0}$ is the amplitude and $\omega$ is the driving frequency. This force does positive work on the system to balance the energy loss due to damping. Substituting (188) into (187) gives

or

Let us assume that the solution of (187) is given by

then, we have

and

Substituting into (187) gives

That gives

and

Solving for $C_{1}$ and $C_{2}$ gives

and

Hence,

The term in brackets is of the form $A_{1}\cos \omega t+A_{2}\sin \omega t$ and thus it can be written as $A'\cos (\omega t-\phi )$ where

i.e.,

and

where $0\le \phi \le \pi$. Hence,

If the driving force is applied for a long time compared with the time that the damped vibration dies out, then the system will eventually vibrate at the same frequency of the deriving force. Therefore, the general solution of (144) is called the transient solution since it approaches zero in a relativity short time whereas (189) is called the steady-state solution where the system oscillates with the same frequency as the deriving force. Therefore, the amplitude of a steady-state vibration is

When the deriving frequency $\omega$ approaches the natural frequency of the system $\omega _{D}$, the amplitude of the resulting forced oscillation will increase. This is known as resonance. If the damping is very light, the amplitude reaches its peak when the deriving frequency is nearly equal to the natural frequency $\omega _{n}$. As the damping becomes heavier, the maximum amplitude shifts to lower frequencies (see Figure 25). In the case where there is no damping at all $(b=0)$, the amplitude of resonance is infinite at $\omega =\omega _{n}.$

Problems