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CHAPTER 14 – OSCILLATIONSPHYS 212 S’14

CHAPTER 14OSCILLATIONS

In this chapter, we will briefly study the dynamics of systems that undergo periodic motion. This material can all be developed using methods of PHYS 211. We require

it as an introduction to the study of waves.

Slide 1


14.1 Simple Harmonic Motion

Slide 2

Oscillation refers to repetitive motion about an equilibrium position. A pendulum, or a mass on a spring, are the most obvious examples.

The period 𝑇 of oscillation is the time required to complete exactly one cycle. The number of oscillations in one second is the frequency 𝑓. By these definitions, we can see that

𝑓 =1

𝑇and 𝑇 =

1

𝑓

The SI unit for frequency is the Hertz, Hz:1 Hz = 1 cycle per second = 1 s-1



Slide 3

The most important type of oscillation is simple harmonic motion (SHM), illustrated here. In SHM, the position is a sinusoidal function of time. We can describe this by the function

𝑥 𝑡 = 𝐴cos2𝜋𝑡

𝑇= 𝐴 cos(2𝜋𝑓𝑡)

The parameter 𝐴 is the amplitude of the SHM – the maximum displacement from equilibrium (𝐴 has represented an area up until now, but there’s no area to be concerned with in this chapter)

Note that the argument of the cosine function must be in radians. Set your calculators accordingly!



Slide 4

Recalling that one full cycle consists of 2𝜋 radians, we can introduce the angular frequency, 𝜔 = 2𝜋𝑓 = 2𝜋/𝑇 (units of radians per second). This provides a third expression for SHM:

𝑥 𝑡 = 𝐴 cos𝜔𝑡

In PHYS 211, we learned how to determine an object’s velocity based on its position –through differentiation:

𝑣 𝑡 =𝑑𝑥

𝑑𝑡= −𝜔𝐴 sin𝜔𝑡

Since the sine function varies between ±1, the

maximum speed is 𝜔𝐴 =2𝜋𝐴

𝑇= 2𝜋𝑓𝐴. This

occurs when the position is at its equilibrium point.


14.2 SHM and Circular Motion

Slide 5

The Phase ConstantOur 𝑥(𝑡) equation for SHM is not valid in all cases – it implies that the object is at its point of maximum amplitude at time 𝑡 = 0:

𝑥 𝑡 = 0 = 𝐴 cos 0 = 𝐴

To be completely general, we should include a phase constant in order to satisfy the initial conditions of the oscillation:

𝑥 𝑡 = 𝐴 cos(𝜔𝑡 + 𝜑0)𝑣 𝑡 = −𝜔𝐴 sin(𝜔𝑡 + 𝜑0)

The quantity 𝜑 = 𝜔𝑡 + 𝜑0 is called the phase of the oscillation


Problem #1: SHM

Slide 6

An object in SHM has amplitude 8.0 cm and frequency 0.50 Hz. At 𝑡 = 0 s it has its most negative position. Write the function 𝒙(𝒕) that describes the object’s position in meters.

Solution: in class

RDK EX 14.9


14.3 Energy in Simple Harmonic Motion

Slide 7

When an object undergoes SHM, there must be some energy associated with the motion. Let’s use the example of a mass 𝑚 that is attached to a spring of force constant 𝑘.

The kinetic energy, as usual, is 𝐾 =1

2𝑚𝑣2.

The potential energy is associated with the stretching or compressing of the spring. In PHYS 211, we learned that this is

𝑈 =1

2𝑘 ∆𝑥 2, where ∆𝑥 is the

displacement from equilibrium. We can arbitrarily set 𝑥 = 0 at equilibrium, in which

case 𝑈 =1

2𝑘𝑥2.



Slide 8

The total mechanical energy is therefore

𝐸 =1

2𝑚𝑣2 +

1

2𝑘𝑥2

Inserting our equations for 𝑥 𝑡 and 𝑣(𝑡), this becomes

𝐸 =1

2𝑚 −𝜔𝐴 2 sin2 𝜔𝑡 +

1

2𝑘𝐴2 cos2 𝜔𝑡

However, we know that 𝐸 must be constant in both time and position. This can only be the case if

𝜔 =𝑘

𝑚

and thus 𝑓 =1

2𝜋

𝑘

𝑚and 𝑇 = 2𝜋

𝑚

𝑘

Note that the frequency does not depend on the amplitude.



Slide 9

We can find a simple expression for 𝐸 by noting that the kinetic energy must be zero at the turning points of the mass (since the speed, and thus the kinetic energy, is zero here by definition). These turning points occur at 𝑥 = ±𝐴. Then,

𝐸(𝑥 = ±𝐴) = 𝑈(𝑥 = ±𝐴) =1

2𝑘𝐴2

Since 𝐸 must be constant in space, this expression is valid at any point in the oscillation. That is,

𝐸 =1

2𝑘𝑥2 +

1

2𝑚𝑣2 =

1

2𝑘𝐴2

The speed at any point 𝑥 is therefore

𝑣 𝑥 =𝑘

𝑚(𝐴2 − 𝑥2) = 𝜔 (𝐴2 − 𝑥2)


14.4 The Dynamics of SHM

Slide 10

We have avoided a key point so far in this chapter. We’ve claimed that a mass on a spring executes SHM based entirely on that fact that the motion looks sinusoidal. It’s time to prove this!Hooke’s law tells us that when a spring of force constant 𝑘 is stretched or compressed by a distance ∆𝑥, there is a restoring force

𝐹sp = −𝑘∆𝑥

As before, we’ll assume that the equilibrium position is 𝑥 = 0. Thus, ∆𝑥 = 𝑥, and

𝐹sp = −𝑘𝑥

Since there are no other forces acting on the mass, Newton’s 2nd law tells us that 𝐹sp = 𝑚𝑎. Therefore, the acceleration of the mass as a

function of its position is

𝑎 = −𝑘

𝑚𝑥 = −𝜔2𝑥



Slide 11

Since acceleration is the 2nd derivative of position, we can write this equation as

𝑑2𝑥

𝑑𝑡2 = −𝜔2𝑥

This is a second-order differential equation. There’s a formal method to solve it, but we can just use a qualitative argument: we seek a function 𝑥 𝑡 for which the second derivative is proportional to −𝑥 𝑡 (the proportionality constant being 𝜔2). Since 𝑑2

𝑑𝑡2 [cos𝜔𝑡] = −𝜔2 cos𝜔𝑡, our sinusoidal function is the solution

that we seek. You can verify for yourself that the amplitude 𝐴 and phase constant 𝜑0 don’t affect the validity of this argument.



Slide 12

Although we only considered a mass-spring system in the last couple of slides, the end result is valid in many other physical situations.

What we actually proved is this:

If an object has an acceleration that is proportional to the negative of its position, then it must undergo simple harmonic motion, with an angular frequency equal to the square root of the proportionality constant.


Problem #2: Oscillations

Slide 13

A block attached to a spring with unknown spring constant oscillates with a period of 2.0 seconds. What is the period if…a) …the mass is doubled?b) …the mass is halved?c) …the amplitude is doubled?d) …the spring constant is doubled?

Solution: in class

RDK EX 14.11


14.6 The Pendulum

Slide 14

An oscillating pendulum looks like it undergoes SHM. Let’s see if this is actually the case.The pendulum consists of a mass 𝑚 (called the “bob”) at the end of a string of length 𝐿. The bob’s position is described by the arc length 𝑠, which is zero at equilibrium (when the pendulum hangs straight down).The bottom figure provides a free-body diagram at one particular point in the oscillation. Two forces act on the bob: tension and gravity. The latter is decomposed into tangential and radial components. Since the radially-directed forces 𝑇and 𝐹𝐺 𝑟 cannot accelerate the mass, we only care about the tangential component 𝐹𝐺 𝑡.


14.6 The Pendulum

Slide 15

Applying Newton’s 2nd law to the tangential motion,

𝐹𝐺 𝑡 = −𝑚𝑔 sin 𝜃 = 𝑚𝑎𝑡

(we’ll review this in class…PHYS 211 was a while ago).

Since the tangential acceleration is 𝑎𝑡 =𝑑2𝑠

𝑑𝑡2, we

obtain the result𝑑2𝑠

𝑑𝑡2= −𝑔 sin 𝜃


14.6 The Pendulum

Slide 16

Next, we relate the arc length 𝑠 to the angle 𝜃. The geometry for an arc with radius 𝑟 is shown in the figure below, indicating that 𝑠 = 𝑟𝜃(𝜃 must be in radians when using this formula). In the case of the

pendulum, the radius of the bob’s arc is 𝐿, and thus 𝜃 =𝑠

𝐿.

Finally, we make the assumption that the angle 𝜃 is “small” such that sin 𝜃 ≈ 𝜃. The resulting equation of motion for the pendulum is

𝑑2𝑠

𝑑𝑡2= −

𝑔

𝐿𝑠

This is the same equation that we derived for a mass attached to a

spring, only with 𝑘

𝑚replaced with

𝑔

𝐿. Therefore, we can claim that

the pendulum undergoes SHM with angular frequency 𝝎 = 𝒈/𝑳.

Note that it is independent of the bob’s mass.


Problem #3: Pendulum

Slide 17

What is the period of a 1.0-meter-long pendulum on the Earth?What is the period of the same pendulum on Venus, where the free-fall acceleration has a magnitude of 8.87 m/s2?

Solution: in class

RDK EX 14.21


14.7 Damped Oscillations

Slide 18

In any realistic oscillating system, the amplitude will gradually diminish over time, as a result of dissipative forces such as friction or air resistance. These forces can be rather complex to analyze, so we will use a simplified linear drag model, in which the force is linearly proportional to speed:

𝐷 = −𝑏 𝑣The negative sign indicates that the drag force is directed opposite to the velocity. The damping constant 𝑏, with SI units of kg/s, depends both on the shape of the moving object and on the viscosity of the medium through which it moves.



Slide 19

The figure shows a mass oscillating on a spring in the presence of drag. Including drag, Newton’s 2nd law reads

𝐹net = −𝑘𝑥 − 𝑏𝑣 = 𝑚𝑎

With 𝑣 =𝑑𝑥

𝑑𝑡and 𝑎 =

𝑑2𝑥

𝑑𝑡2 , we have

𝑑2𝑥

𝑑𝑡2+

𝑏

𝑚

𝑑𝑥

𝑑𝑡+

𝑘

𝑚𝑥 = 0

Without proof, we’ll just present the solution to this differential equation:

𝑥 𝑡 = 𝐴𝑒−𝑏𝑡/2𝑚 cos(𝜔𝑡 + 𝜑0) with 𝜔 =𝑘

𝑚−

𝑏2

4𝑚2= 𝜔0

2 −𝑏

2𝑚

2



Slide 20

The first thing to notice is that the angular frequency is reduced from the case where there is no damping. When 𝑏/2𝑚 ≪ 𝜔0, this reduction is negligible, and we can make the approximation 𝜔 ≈ 𝜔0; this situation is known as weak or light damping.

A graph of the position vs. time for a lightly damped oscillator is shown here. The oscillation is modified by an

exponentially decaying envelope, indicated

by the term 𝑒−𝑏𝑡/2𝑚 in 𝑥 𝑡 . Larger values of b result in a faster decay.For a damped oscillation, mechanical energy is no longer conserved. It is relatively easy to

show that 𝐸 𝑡 = 𝐸0𝑒−𝑏𝑡/𝑚.


14.8 Driven Oscillations and Resonance

Slide 21

All of the systems that we have considered so far are undergoing free oscillation. That is, they are initially displaced from equilibrium and then allowed to oscillate until their energy dissipates.A more complex situation arises in the case of driven oscillation, in which a periodic force continuously performs work on the system. Some examples will be mentioned in class.

All oscillating systems have a natural frequency, 𝒇𝟎 – this is the frequency that we calculated earlier in this chapter. For a mass on a spring,

𝑓0 =1

2𝜋

𝑘

𝑚


14.8 Driven Oscillations and Resonance

Slide 22

The frequency of the periodic driving force is 𝑓ext. Using Newton’s 2nd law, it is possible to show that the oscillator’s amplitude is determined by the relationship between 𝑓ext and 𝑓0, and by the damping constant 𝑏. When the oscillator is driven “far from resonance” – that is, when 𝑓ext − 𝑓0 is fairly large – the resulting

oscillations are small in amplitude. When the oscillator is driven “on resonance”, the amplitude is maximized. Weaker damping allows for greater amplitudes.


Problem #4: Don’t Try This at Home

Slide 23

Vision is blurred if the head is vibrated at 29 Hz because the vibrations are resonant with the natural frequency of the eyeball in its socket. If the mass of the eyeball is 7.5 g, what is the effective spring constant of the musculature that holds the eyeball in its socket?

Solution: in class

RDK EX 14.29

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