dr. a louro - physics 225 notes

8/14/2019 Dr. A Louro - Physics 225 Notes

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Physics 225 Lecture Notes

A. A. Louro

Winter 2002


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2


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Contents

1 Rotations 1

1.1 Rotation kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.1.1 Angular displacement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.1.2 Angular velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.1.3 Angular acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.1.4 Are angular variables vectors? . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2 Relation between rotational and translational variables . . . . . . . . . . . . . . . . . 3

1.3 Rotational inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.3.1 Rotational kinetic energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.3.2 General procedure for calculating moments of inertia . . . . . . . . . . . . . . 6

1.3.3 Dependencies of the moment of inertia . . . . . . . . . . . . . . . . . . . . . . 6

1.3.4 The parallel-axis theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.4 Newtons 2nd. law for rotation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.5 Rolling motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.6 Conservation of energy, including rotation . . . . . . . . . . . . . . . . . . . . . . . . 9

1.7 General definition of torque . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.8 Angular momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.8.1 Extension to systems of particles . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.9 Conservation of angular momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2 Oscillations 15

2.1 Oscillators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.1.1 Simple harmonic motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.1.2 Hookes law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.1.3 Period and frequency of the oscillation . . . . . . . . . . . . . . . . . . . . . . 18

2.2 The ideal pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.3 Elastic potential energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

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ii CONTENTS

3 Waves 21

3.1 Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.1.1 Definitions and terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

3.1.2 Why study waves? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3.1.3 A general description of a sine wave . . . . . . . . . . . . . . . . . . . . . . . 23

3.1.4 Transverse and longitudinal waves . . . . . . . . . . . . . . . . . . . . . . . . 23

3.2 Waves in a string . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3.2.1 The speed of propagation of a wave in a string . . . . . . . . . . . . . . . . . 23

3.2.2 Energy transport by a wave . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3.3 The superposition principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.4 Interference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3.5 Standing waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3.6 Resonant modes of a string . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.7 Intensity of sound and sound level . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

3.7.1 The inverse square law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

3.7.2 Intensity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

3.7.3 The decibel scale . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

3.7.4 Hearing curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

3.8 The Doppler effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

3.8.1 Moving detector, stationary source . . . . . . . . . . . . . . . . . . . . . . . . 29

3.8.2 Moving source, stationary detector . . . . . . . . . . . . . . . . . . . . . . . . 30

3.8.3 The general Doppler effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

4 Fluids 33

4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

4.1.1 Solids, liquids, and gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

4.1.2 Density and pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

4.2 Fluid statics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

4.2.1 Variation of pressure with depth . . . . . . . . . . . . . . . . . . . . . . . . . 35

4.2.2 Buoyancy and Archimedes Principle . . . . . . . . . . . . . . . . . . . . . . . 36

4.3 Fluid dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

4.3.1 The velocity field in a fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

4.3.2 Streamlines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

4.3.3 The continuity equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374.3.4 Bernoullis law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

A Derivations for the final exam 43

A.1 Derivations for the final exam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44


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CONTENTS iii

B The vector cross product 45

B.1 The vector cross product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46B.1.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46B.1.2 The cross product is non-commutative . . . . . . . . . . . . . . . . . . . . . . 46B.1.3 The x, y, and z components of a cross product . . . . . . . . . . . . . . . . . 46

C Dimensional analysis 49

C.1 The magic of dimensional analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50C.1.1 Step 1: What does the quantity of interest depend on? . . . . . . . . . . . . . 50C.1.2 Step 2: Narrow down the possible functions by demanding that the dimen-

sions are right . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

D Average value of cos2(x) 53

D.1 Average value ofcos2(x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

E Work and potential energy 55

E.1 Work and potential energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56


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iv CONTENTS


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Chapter 1

Rotations

1


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2 CHAPTER 1. ROTATIONS

1.1 Rotation kinematics

We begin our study of rotation by considering a rigid body rotating about an axis fixed in space.Figure 1.1 shows an example, captured at two instants in time. To describe the rotational motion,

Rotation axis

s

r

t+ tt

Figure 1.1: Rotating rigid body.

we introduce new kinematic variables, analogous to displacement, velocity, and acceleration intranslational motion.

1.1.1 Angular displacement

During the interval [t, t + t] the body has rotated through an angle . This is the angulardisplacement. To make life simpler later, lets agree from the start to measure all angles inradians. Remember that one full turn is equivalent to an angle of 2 radians.

1.1.2 Angular velocity

The angular velocity is the rate at which the direction of the body changes as it spins:

=d

dt(1.1)

( is the Greek letter omega). Angular velocity is measured in radians per second.

1.1.3 Angular acceleration

The angular acceleration is the rate at which the angular velocity changes over time:

=d

dt(1.2)

and it is measured in (rad/s)/s .
http://www.translexis.demon.co.uk/alphabet.htm


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1.2. RELATION BETWEEN ROTATIONAL AND TRANSLATIONAL VARIABLES 3

1.1.4 Are angular variables vectors?

Are angular variables vectors? Sort of. For now, since we are only considering rotations about afixed axis, we dont have to worry about this. However, , , and are signed quantities. Theconvention is that if we are looking down on the body, so that the rotation axis is perpendicular tothe page, and the angle/rotation sense/angular acceleration is counterclockwise, then the quantityis negative.

1.2 Relation between rotational and translational variables

Consider any point on the rotating object at a distance r from the rotation axis. (For example,the tip of the triangular object in Figure 1.1). As the body rotates through an angle , the pointdescribes a circular arc of length s. Since is measured in radians we can write

s = r (1.3)

The speed of the point is

v =ds

dt(1.4)

and substituting s from equation (1.3), we find the following simple relation between the ordinaryspeed v of the point, and the angular speed of the rotating body, :

v =d(r)

dt= r

d

dt= r (1.5)

since r is of course constant. (The body is rigid, so the distance from the point to the rotation axisdoesnt change).

If the body does not rotate at a constant rate, but rather has an angular acceleration , thepoint has an acceleration in the direction it is moving, i.e. tangent to its circular path. Thistangential acceleration is easily found:

at =dv

dt=

d(r)

dt= r (1.6)

Recall that at the same time, because the point is turning in a circle, it has a centripetal, orradial acceleration

ar =

v2

r (1.7)which we can also write

ar = 2r (1.8)

See Figure 1.2.


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a_t

a_r

a

Figure 1.2: Radial and tangential components of the acceleration

1.3 Rotational inertia

Its becoming apparent that there is a close parallel between rotational and translational quantities.Theres more; for example, just as mass measures the inertia of an object, that is how hard it is to

accelerate given a certain applied force, there is an equivalent concept in the case of rotation.Since mass is defined in the context of Newtons 2nd. law, we would look for the rotational

equivalent of Newtons 2nd. law. There is indeed such a law, but we have to wait before weintroduce it. To discover what the rotational inertia is, well try a different approach: Well use thekinetic energy.

1.3.1 Rotational kinetic energy

The kinetic energy of a mass m moving with speed v is

K =

1

2 mv

2

(1.9)

A rotating body also has kinetic energy, even though it isnt going anywhere. Every atom in thebody is moving around the rotation axis as the entire object spins, so its total rotational kineticenergy is simply the sum of the kinetic energies of all the particles that make up the object.


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1.3. ROTATIONAL INERTIA 5

Following the analogy with translational motion, we write the rotational kinetic energy as

Krot =12

I2 (1.10)

where I stands for the rotational equivalent of mass, as yet undetermined.

Exercise 1.3.1 Show that the units of I are kg m2.

Consider for example a thin ring of mass M spinning about an axis through its centre, perpendicularto the plane of the ring. (See Figure 1.3). If we isolate a tiny mass element m of the ring, its kinetic

R

Tiny mass element

Figure 1.3: Spinning ring.

energy is1

2mv2 =

1

2m2R2 (1.11)

Since all the mass elements that make up the ring are at the same distance R from the axis, androtate with the same , the total kinetic energy of the ring is

Krot =

all ms

1

2m2R2 =

1

22R2

all ms

m =1

2M 2R2 (1.12)

Comparing this with equation (1.10), we see that

I = M R2 (1.13)

I is called the moment of inertia of the ring about this particular rotation axis. In general, ifthe ring were rotating about any other axis, for example around a diameter, the moment of inertiawould be different because it depends on how the mass is distributed relative to the rotation axis.


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1.3.2 General procedure for calculating moments of inertia

Since the bodies we consider are continuous, the tiny mass elements that we isolate to calculatethe kinetic energy are really infinitesimals dm. The sum over tiny mass elements is in fact anintegral:

K =1

2

v2dm =

1

22

r2dm (1.14)

where r is the distance of each mass element dm from the axis. This last integral is the moment ofinertia:

I =

r2dm (1.15)

In most cases, this integral turns out to be beyond the scope of this course. Refer to Table 11-2,page 227 in the textbook for moments of inertia of some common bodies.

1.3.3 Dependencies of the moment of inertia

Unlike mass, which is a simple property of any object, independent of its shape, the moment ofinertia depends on several things:

the mass of the object, the rotation axis the shape of the object, that is, how its mass is distributed around the rotation axis

1.3.4 The parallel-axis theoremVery often we know the moment of inertia of an object when it rotates around an axis that goesthrough the centre of mass (CM). The following theorem enables us to use this information to findeasily the moment of inertia about an axis parallel to the first one. Figure 1.4 shows an example.The parallel axis is labelled 2, and it is a distance h away from the centre of mass CM. We willshow that the moment of inertia with respect to axis 2 is related to the moment of inertia withrespect to the axis through CM by

I2 = ICM + M h2 (1.16)

where M is the mass of the object. Before we proceed with the proof, we need to be aware ofthe definition of the centre of mass of an extended ob ject. To find the coordinates of the centreof mass, consider an isolated mass element dm with coordinates (x, y). A word equation for the

coordinates of the centre of mass goes something like this:


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1.3. ROTATIONAL INERTIA 7

CM 2h

Figure 1.4: The parallel axis theorem.

[sum over all mass elements of (mass position)] / (total mass) . This being a continuousbody, the sum is really an integral:

xCM =

x dm

M(1.17)

yCM =y dm

M (1.18)

Now were ready to tackle the parallel-axis theorem. As usual, to find a moment of inertia we isolatefirst a mass element dm. For notation, see Figure 1.5. Notice in particular that the coordinatesystem is chosen with CM at the origin, that is xCM = yCM = 0. We have

ICM =

r2dm =

(x2 + y2) dm (1.19)

and

I2 =

r22dm =

[(h x)2 + y2] dm =

(x2 + y2 + h2 2xh) dm (1.20)

so

I2 =

(x2 + y2) + h2

dm 2hx dm (1.21)The last integral is just the x coordinate of the centre of mass, which is zero by our choice ofcoordinate system. We get finally

I2 = ICM + M h2 (1.22)


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CM 2

dm

x

y

x

y

hx

rr_2

Figure 1.5: Geometry for the parallel-axis theorem

which is what we set out to prove.

1.4 Newtons 2nd. law for rotation

For a particle, or the centre of mass of a system, Newtons 2nd law holds: If a net force F is appliedto the object, it acquires an acceleration a = F /m, where m, the objects mass, is a measure of itsinertia, that is, its resistance to being accelerated.

In rotation, we have learned that a body may acquire an angular acceleration as a result ofhaving a torque applied. We have also seen that the role of rotational inertia is played by I, themoment of inertia of the body. Replacing F, m, and a with their rotational counterparts, we getNewtons 2nd. law for rotation:

= I (1.23)

1.5 Rolling motion

A particular case of rotation that is of some interest is rolling motion. Consider a wheel, rollingwithout slipping on a flat surface. As Figure 1.6 shows, when the wheel turns through an angle d,the centre of mass moves a distance equal to the arc length ds = Rd parallel to the surface. The


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1.6. CONSERVATION OF ENERGY, INCLUDING ROTATION 9

dr

R

d

ds

Figure 1.6: Rolling

speed of the centre of mass is

vCM =ds

dt= R

d

dt= R (1.24)

If the centre of mass is accelerating,

aCM = Rd

dt= R (1.25)

1.6 Conservation of energy, including rotationIn general, the motion of a body may be separated into

the translational motion of the centre of mass with velocity v, and a rotation with angular velocity about an axis through the centre of mass.

Each of these contributes to the bodys total kinetic energy:

K = Ktr + Krot =1

2mv2CM +

1

2I2 (1.26)

The work-kinetic energy theorem holds, with the full kinetic energy. In particular, if only conser-

vative forces act on a body, the total energy is conserved:

U + Ktr + Krot = constant (1.27)


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Exercise 1.6.1 Two cylinders of equal mass M and radius R roll without slipping down an inclined

plane, dropping through a height h. Cylinder 1 is solid, but cylinder 2 is a hollow shell. Whichcylinder wins the race?

1.7 General definition of torque

Consider a point mass m as shown in Figure 1.7. Its position relative to the point O which we havechosen as the origin of the coordinate system is r, which lies in the xy plane. A force F is actingon m, also in the xy plane. According to our established definition of torque, we would write the

x

y

O

r

F

Figure 1.7: Calculating the torque on a point mass

torque on m about an axis that passes through O as

= +rF sin (1.28)

where the + sign is due to the fact that F accelerates m in a counterclockwise direction.Compare this with the vector formed by the cross-product of r and F (see Appendix B):

r F = rF sin k (1.29)What we have called is the z component ofr F.


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1.8. ANGULAR MOMENTUM 11

This suggests that we should extend our definition of torque. We shall say that the torque about

a point O on a mass at a position r relative to O by a forceF is

= r F (1.30)

Some observations about this definition of torque:

Torque is defined relative to a point rather than a rotation axis. In general, the orientationof the rotation axis may not stay constant.

Torque is a vector. If we limit ourselves to a rotation about a fixed axis, then the only choicesfor the direction of are up or down; for that we had the sign convention mentionedearlier.

The torque is zero if r F. Interestingly, this is the case of the gravitational pull by theSun on the planets: If we consider the torque due to gravity about the Sun, it must be zerobecause the force acts along the line joining the Sun and the planet, which is also the directionof the planets position vector relative to the Sun.

Exercise 1.7.1 A point mass M is atr = ( + + 2k) m from point O. Two forces act on M:

F1 = (2 + ) NF2 = ( + k) N

Calculate the net torque on M about O.

1.8 Angular momentum

Pursuing the analogy with translational motion, we know that Newtons 2nd law may be writtenin terms of the momentum of a particle as

F =dp

dt(1.31)

where F is the total external force on the system, and p is the particles momentum. We mayconjecture that there is a similar quantity for rotations, and write

= dLdt

(1.32)

where L is known as the angular momentum of the particle. (Immediately after this, we willdiscuss the angular momentum of a system of particles).


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We will show that the angular momentum is linked to the linear momentum; in fact,

L = r p (1.33)where r is the particles position vector. (Notice that the value of L depends on where the originis taken). To see this, take the derivative with respect to time of equation (1.33):

dL

dt=

dr

dt p + r dp

dt(1.34)

The first term vanishes, sincedr

dt= v p (1.35)

and in the second term we recognize dp/dt = F, so

dL

dt= r F = (1.36)

1.8.1 Extension to systems of particles

The angular momentum of a system of particles is simply the sum of the angular momenta of eachparticle.

In particular, the system may be a rigid body. We will limit ourselves once again to rotationsabout a fixed axis, so that the body has a moment of inertia I about that axis. By analogy withthe linear momentum, the angular momentum of the body is

L = I (1.37)

where we have dropped the vector arrows since L only has a component along the rotation axis,but the usual sign convention still holds.

1.9 Conservation of angular momentum

From Newtons 2nd. law for rotation, we find immediately that if there is no net external torqueon a system of particles, dL/dt = 0, and the angular momentum is conserved.

For example, consider a collapsing star. Stars like our Sun end their lives as white dwarves,essentially the core of the star laid bare. More massive stars can continue to collapse right downto the density of nuclear matter. A neutron star with 1.4 times the mass of the Sun may havea radius of the order of 10 km! (See for example http://www.astro.umd.edu/ miller/nstar.html).Since there are no external torques on the star as it collapses, it follows that

Iii = Iff IiTi

=IfTf

(1.38)
http://www.astro.umd.edu/~miller/nstar.html


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1.9. CONSERVATION OF ANGULAR MOMENTUM 13

where Ti,f are the initial and final periods of revolution of the star. The moment of inertia of

a sphere about an axis through its centre scales with the square of its radius. Let us assumeconservatively that the initial radius of our doomed star is of the same order as the Suns, 10 9 m,and Ti is also similar to the Suns, about 2 106 s. If the final radius is 104 m,

Tf = TiR2fR2i

= 1010 2 106 = 2 104s (1.39)

See pp 261-263 in the textbook for many more interesting applications of the conservation of angularmomentum.


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Chapter 2

Oscillations

15


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16 CHAPTER 2. OSCILLATIONS

2.1 Oscillators

An oscillator is a physical system where the physical variables that describe it are periodic func-tions of time. There are well-known instance of mechanical oscillators, of course, like the pendulum,or a loaded spring; there are also electrical circuits that produce an output voltage or a currentthat are periodic functions of time, and qualify as oscillators equally well. All oscillators may bedescribed formally in the same way. So even though we will often refer to pendulums or springs asillustrations, we should not lose sight of the generality of the properties of oscillating systems thatwe shall study here.

2.1.1 Simple harmonic motion

Position as a function of time

A simple harmonic oscillator is one where the variable of interest varies as a simple trigonometricfunction. For example, consider a mass on the end of a massless spring, as shown in Figure 2.1.Here, x is the position of the mass on the end of the spring, measured from its equilibrium positionwhen the mass is not oscillating1.

If the mass is pulled down to an initial position x0 and released, it will oscillate, so that itsposition as a function of time is given by

x = x0 cos t (2.1)

The argument of the cosine, t, is an angle, expressed as usual in radians; (Gr.: omega) iscalled the angular frequency of the oscillator, and it is measured in rad/s .

Of course, we could start measuring t at any instant, so that the most general form of x(t) is

x = x0 cos (t + 0) (2.2)

A little jargon:

x0 is the amplitude of the oscillation. Since the cosine function oscillates between 1, theextreme values of x are x0.

The argument of the cosine function, t + 0, is called the phase. Since 0 is the phase att = 0, it is called the initial phase. In our initial example, 0 = 0.

Exercise 2.1.1 Suppose the mass was pulled down and released from rest, but we didnt startmeasuring time until a while later, when it passed through x = 0. What would be the initial phaseof this oscillation?

1Strictly, we should be considering the effect of gravity as well. In fact, as we shall discuss later, it may be ignored,as the only effect is to shift the equilibrium position downwards. We will show this later.


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2.1. OSCILLATORS 17

x = 0 : Equilibrium

x

x

Figure 2.1: A mass on the end of a spring.

It is important to note that the choice of a cosine function to describe the oscillation is arbitrary;we could just as well have used a sine function.

Exercise 2.1.2 Write the position of the mass as a function of time in the conditions of Exercise2.1.1, but using a sine function instead of a cosine.

Velocity and acceleration as functions of time

Once the position as a function of time has been established, the velocity and acceleration are easilyfound by taking time derivatives (we may set 0 = 0 without loss of generality):

vx(t) = x0 sin t (2.3)ax(t) = 2x0 cos t (2.4)

To get a feel for the behaviour of an oscillator, you may want to look at the simulation athttp://www.phas.ucalgary.ca/physlets/shm.htm.
http://www.phas.ucalgary.ca/physlets/shm.htmhttp://www.phas.ucalgary.ca/physlets/shm.htm


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2.1.2 Hookes law

In 1660, Robert Hooke discovered his law of elasticity. With our notation, it stated that the elasticforce applied to the mass at the end of a spring is given by

Fx = kx (2.5)

where k is the spring constant, or elastic constant of the spring. It is a coefficient that mustdetermined experimentally, and it depends on characteristics of the particular spring used, such asthe material its made of, its shape and size.

Exercise 2.1.3 Sketch a plot of Fx(x). What is the effect of varying the value of k? Why do youthink a spring with a large value of k is called a stiff spring, and one with a small value of k iscalled a soft spring?

Since we postulated the form of the position of the mass as a function of time, it is nice to confirmthat it is consistent with Hookes empirical law:

Fx = max = m2x0 cos t = m2x (2.6)

so that we may identify

=

k

m(2.7)

2.1.3 Period and frequency of the oscillation

The period T of the oscillation is the time taken to complete one full cycle, that is for the phaseto go from to + 2. That is,

T =2

(2.8)

The frequency f is the number of cycles per second. A moments thought will convince the readerthat

f =1

T(2.9)

The units of f are inverse seconds, also known as Hertz (Hz).

Exercise 2.1.4 Find an expression for the frequency of the loaded spring, in terms of k and m.

2.2 The ideal pendulum

An ideal pendulum is simply a mass of negligible size (so that its moment of inertia in anydirection through its centre of mass is zero), suspended from a string of negligible mass. Therestoring force that plays the role of the elastic force by a spring is the component of gravity
http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Hooke.html


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2.2. THE IDEAL PENDULUM 19

L

m

F_g

Figure 2.2: An ideal pendulum.

perpendicular to the string. As we shall see, for small amplitude oscillations, the pendulum alsoexecutes simple harmonic motion.

Consider the angle that the pendulum forms with the vertical, . (See Figure 2.2). The torqueon the mass about the suspension point is

= mgL sin (2.10)and since its moment of inertia about the same point is simply

I = mL2 (2.11)

we may also write = I mL2 = mgL sin (2.12)

where is the angular acceleration of the swinging mass. Therefore

= gL

sin (2.13)

Now, it is an interesting fact that if is measured in radians, for small-amplitude oscillations(


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Exercise 2.2.1 Verify this. Use your calculator to find sin(0.1), sin(0.01), andsin(0.001). In each

case, what is the percentage error incurred by approximating sin as ?So, for small-amplitude oscillations, we may approximate equation (2.13) as

= gL

(2.15)

which we may recognize as the fundamental relation between position and acceleration for a simpleharmonic oscillator:

acceleration = (positiveconstant) position (2.16)From this, we may write with confidence

= 0 cos t (2.17)

(where again well assume the initial phase is zero). The angular frequency is

=g

L(2.18)

(compare with equation (2.7)).

Exercise 2.2.2 What is the pendulums period T in terms of g and L?

2.3 Elastic potential energy

Lets return to the loaded spring. One conclusion we may draw from Hookes law is that the elastic

force only depends on the position of the mass. This implies that the elastic force is conservative,and we may associate a potential energy with it.

To find the potential energy of the mass when its at a certain position x, we have to calculatethe work done by the elastic force as the mass moves from say x = 0 to x. Using equations (??)and (??) (see Appendix C), and setting U = 0 at x = 0, we get

U(x) = W(x = 0 x) = x0

(kx)dx = 12

kx2 (2.19)

Exercise 2.3.1 If the principle of conservation of energy holds true, the total energy

E = 12

mv2 + 12

kx2 (2.20)

should remain constant over time. Verify this for the mass on the end of a spring, bearing in mindthat it is doing simple harmonic motion.


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Chapter 3

Waves

21


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22 CHAPTER 3. WAVES

3.1 Waves

3.1.1 Definitions and terminology

A wave is a periodic perturbation in a quantity that is a function of position in space. Consider,for example, a sound wave in air travelling in the x direction. The simplest form of wave is asine wave, so called because the pressure as a function of x and the time t has the form

p = p0 sin(kx t) (3.1)

Here, p0 is the amplitude of the wave: The pressure oscillates between p0. The argument ofthe sine function, (kx t), is called the phase of the wave. To understand the meaning of theparameters k, consider the form of the wave at a specific time, which we can call t = 0 without lossof generality. In that case,

p = p0 sin(kx) (3.2)

which is a periodic function ofx alone. The spatial period is the wavelength , which is clearlyequal to 2/k. The parameter k is called the wave number, defined as

k =2

(3.3)

If instead of looking at a snapshot of the wave in time, we look at the wave at a fixed position, whichagain we can call x = 0 without loss of generality, then we see that the pressure varies periodicallywith time, with angular frequency . If T is the period of the oscillation at a fixed point, we haveagain that

= 2T (3.4)

A sound wave travels. Look at a peak in the wave, for example where the phase is equal to 2.The peak will occupy different positions x over time, given by

kx t = 2 x = k

t (3.5)

This equation says that the position of the peak moves with velocity

vx =

k(3.6)

along x. Because it is the velocity of a certain value of the phase, it is called the phase velocity.

Exercise 3.1.1 Show that

p = p0 sin(kx + t) (3.7)

represents a sound wave travelling in the negative x direction.


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3.2. WAVES IN A STRING 23

3.1.2 Why study waves?

Many different phenomena have the same wave behaviour. The pressure in a sound wave, the shapeof a string under tension, the height of a liquid surface, the electric and magnetic fields in the caseof a light wave. In all these different contexts, the mathematics of wave motion is the same.

3.1.3 A general description of a sine wave

In all the types of waves mentioned above, there is a perturbation, which if it takes the formof a sine wave may be written as

y = y0 sin(kx t + 0) (3.8)

where y0 is the amplitude and k and have the same meaning as before. There is a new parameterin this equation, which we didnt include before not to complicate matters unnecessarily: 0 iscalled the initial phase, which is simply the value of the phase at x = 0 and t = 0. If we are onlyanalyzing a single wave, it may be set equal to zero without restriction, but if we consider two ormore sine waves interacting, as we shall do later, then the difference in initial phase between themis significant. More about this later.

3.1.4 Transverse and longitudinal waves

A sound wave is an example of a longitudinal wave. This means that the perturbation is alongthe same direction as that in which the wave travels. In the case of surface waves in a liquid, forexample, the perturbation the height of the surface is perpendicular to the direction in which

the wave propagates. These are called transverse waves.

3.2 Waves in a string

As our first concrete application, we shall study waves in a string under tension. These are trans-verse waves. The undisturbed string is straight; when the string is plucked, it is pulled sideways ata point, and this sideways displacement travels along the string. A sine wave in a string has theform of equation (3.8), with y representing the transverse displacement of points on the string.

3.2.1 The speed of propagation of a wave in a string

We shall apply the technique of dimensional analysis (see Appendix C) to find an expression forthe propagation speed of a wave in terms of relevant variables.

A wave in a string is generated by displacing a point sideways and releasing it. The restoringforce is the tension T applied to both ends of the string. The ability of the string to accelerate isaffected by its inertia, which may be measured by the density of the string. If the string is very


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24 CHAPTER 3. WAVES

thin, it may be represented by a one-dimensional line, so the relevant density is the linear density,

or mass per unit length. It is usually represented by the Greek letter .Let us settle on these two factors, then, as affecting the speed v, and write

v = CTab (3.9)

where C is a dimensionless constant and a and b are exponents to be determined.Substituting SI units in this equation,

ms1 = (kgms2)a(kgm1)b = kga+bmabs2a (3.10)

Both sides of the equation have the same units if

a + b = 0 (3.11)

a b = 1 (3.12)2a = 1 (3.13)

from which we find a = 1/2, b = 1/2. Therefore

v = C

T

(3.14)

Experimentally we find that C = 1 and the speed of a wave in a string is

v =

T

(3.15)

3.2.2 Energy transport by a wave

Continuing with the example of a wave on a string, although each element of the string only movesup and down, with its height as a function of time given by

y = y0 sin(kx t) (3.16)the wave excites new parts of the string as it travels. This means that the wave is transportingenergy along the string. At what rate is this energy transported?

The kinetic energy of a certain element of the string of mass dm = dx is

dK =

1

2 u2

dx (3.17)

where u is the speed of the oscillating element:

u =y

t= y0 cos(kx t) (3.18)


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3.3. THE SUPERPOSITION PRINCIPLE 25

Consequently,

dK =1

2 2

y20 cos

2

(kx t)dx (3.19)We find that the rate at which kinetic energy is transmitted by the wave as

dK

dt=

1

2v2y20 cos

2(kx t) (3.20)where we have replaced the speed of the wave v = dx/dt.

This is an oscillating function of x and t. The average kinetic energy transmitted isdK

dt

=

1

2v2cos2(kx t) (3.21)

where . . . denotes an average. In fact,

cos2(kx t) = 12 (3.22)(see Appendix D for the derivation of this result). Therefore the average kinetic energy transmittedis

dK

dt

=

1

4v2 (3.23)

This not all the energy that is transmitted by the wave, though. As each string element isperturbed, its elastic potential energy changes too, so potential energy is transmitted with thewave as well as kinetic energy. We will state without proof (see, however, Exercise ?? below) thatthe average rate at which the wave carries potential energy is the same as that of kinetic energy:

dUdt = dK

dt (3.24)

so that the average power, the rate at which energy is transmitted by the wave is

P = 2

dK

dt

=

1

2v2y20 (3.25)

A final note: Although we have derived this result for a wave in a string, it is a general propertyof waves that they transport energy at a rate proportional to the square of the amplitude, and thesquare of the frequency.

3.3 The superposition principle

So far we have only been concerned with waves that have a simple sine form, with a well-definedfrequency and a constant amplitude. In reality, of course, waves usually have far more complexforms. However, a complex wave can be broken down into a sum of simple sine waves, each onepropagating independently of the others. This is the superposition principle. Thanks to it, theanalysis of wave behaviour is greatly simplified.


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26 CHAPTER 3. WAVES

3.4 Interference

One typical wave phenomenon is neatly explained by the superposition principle: Interference.

Suppose that two waves, identical in every respect save for a constant phase difference, propagatealong a string. What is the resultant wave?

Call the two waves y1 and y2, with

y1 = A sin(kx t) (3.26)y2 = A sin(kx t + ) (3.27)

The resultant wave is

y = y1 + y2 = A[sin(kx

t) + A sin(kx

t + )] (3.28)

We use a standard trigonometric formula for the sum of two sine functions

sin + sin = 2sin

+

2

cos

2

(3.29)

to write y as

y = 2A cos(/2) sin(kx t + /2) (3.30)This is again a sine wave, but its amplitude depends on the phase difference between the twowaves. If = 0 so that the two waves coincide, then y is twice each of the two component waves.This situation is called constructive interference. At the other extreme, if = so that thetwo waves are half a cycle out of phase, y = 0: The waves cancel! This is called destructiveinterference.

3.5 Standing waves

Consider now two identical waves travelling in opposite directions along a string:

y1 = A sin(kx t) (3.31)y2 = A sin(kx + t) (3.32)

Exercise 3.5.1 Consider a peak in each wave, where the phase is equal to zero. What is x(t),the position of the peak as a function of time? From this, what is the direction of propagation of y1and y2?


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3.6. RESONANT MODES OF A STRING 27

To find the resultant wave, use equation (3.29) again:

y = 2A sin(kx) cos(t) (3.33)

This is an oscillating function of x sin(kx) with an amplitude that is a periodic function oftime 2A cos(t) . This is not a propagating wave, but rather a stationary wave whose amplitudeincreases and diminishes periodically: A standing wave.

Standing waves in a string are easily generated by tying down one end of the string and shakingthe other end. A wave travels down the string and is reflected at the fixed end (Newtons 3rd. law:The string pulls on the wall where the string is tied, and the wall pulls back, shaking the stringand sending a wave in the opposite direction!). Very quickly the reflected wave meets the incomingwave, and the two interfere to produce a standing wave.

3.6 Resonant modes of a string

If both ends are tied down and the string is plucked, waves travel in both directions, and arereflected in both directions. Under certain conditions, all these waves will interfere to produce astanding wave. Say the length of the string is L. If a standing wave of the form

y = ymax sin(kx)cos(t) (3.34)

is produced, y must be zero at x = 0 and at x = L at all times. Therefore

sin(kL) = 0 kL = n (3.35)

where n = 1, 2, . . . . Remembering that k = 2/, this means that

=2L

n(3.36)

Each of these possible standing waves is called a resonant mode of the string. The lowest frequencymode, with n = 1, is called the fundamental mode or first harmonic1. Higher frequency modesare called second, third, etc. harmonics.

3.7 Intensity of sound and sound level

3.7.1 The inverse square law

The energy emitted by a source of sound is carried away by the waves, as we have seen. If thesource is point-like, and the energy is spread uniformly in all directions, the wave fronts are spherical,

1The nomenclature differs between authors. Some call the first harmonic the mode above the fundamental. Here,however, we shall follow the same nomenclature as in the textbook.


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28 CHAPTER 3. WAVES

centered at the source. The surface of such a wave front is proportional to the square of its radius:

A = 4r2

. If a detector of collecting area a is located at a distance r from the source, the fractionof energy that the detector collects is a/A, which is proportional to 1/r2. Thus, the power (energyper second) collected by the detector is inversely proportional to its distance from the source.

3.7.2 Intensity

Since the power received by a detector is also proportional to its area, it is useful to have a measureof power per unit area at the detector. This is the intensity of the sound, and it is measuredin W/m2. If the power emitted by a point source is P, the intensity received by a detector at adistance r from the source is

I =P

4r2(3.37)

3.7.3 The decibel scale

The dynamic range of human hearing is surprisingly large: From about 1012 W/m2 to about1 W/m2 at the threshold of pain. Therefore a convenient measure of intensity is a logarithmicone.

The sound level of a signal is defined as

= (10 dB) log10I

I0(3.38)

where dB stands for decibel. I0 is arbitrarily chosen to be 1012 W/m2, approximately the lower

limit of human hearing. Thus the dynamic range of human hearing goes from approximately 0 dBto 120 dB.

3.7.4 Hearing curves

In fact, the sensitivity of the human ear depends on frequency: Two sounds at different frequenciesmay be perceived as being equally loud, although their sound level is quite different. Equal loud-ness curves may be drawn (see http://hyperphysics.phy-astr.gsu.edu/hbase/sound/eqloud.html),by comparing the perceived loudness of a sound at a certain frequency with the sound level in dBat 1000 Hz. For example a sound of 120 dB at 20 Hz (about the lowest frequency that we can hear)is equally loud to a sound of only 90 dB at 1000 Hz. As a general rule, we may say that at the

frequency extremes of hearing the sensitivity drops. But its also interesting to observe that we areexceptionally sensitive to frequencies around 3 - 4 kHz. This coincides with a resonant frequency ofthe auditory canal (see http://hyperphysics.phy-astr.gsu.edu/hbase/sound/maxsens.html). It hasbeen suggested that the evolutionary significance of this resonance is that this frequency coincideswith a babys cry!
http://hyperphysics.phy-astr.gsu.edu/hbase/sound/maxsens.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/sound/eqloud.html


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3.8. THE DOPPLER EFFECT 29

3.8 The Doppler effect

When either the source of sound, or the observer, or both, are moving relative to the medium, theperceived frequency is not the same as the frequency emitted by the source. This phenomenon iscalled the Doppler effect, after its discoverer Christian Doppler. (See http://www-groups.dcs.st-and.ac.uk/ history/Mathematicians/Doppler.html). Because of the precise relation between theemitted frequency and the detected frequency, it is possible to infer the relative velocity of sourceand observer from measurements of the frequency shift. Although we shall only deal here with theclassical Doppler effect, which applies to sound waves, the Doppler effect also applies to light wavesin a modified form. It is this application that has yielded extremely useful results in astrophysics,notably determining the expansion of the universe.

We will consider separately the cases of a moving detector and a moving source.

3.8.1 Moving detector, stationary source

The stationary source S emits waves with freqency f, and these travel towards the detector D withspeed v. Figure 3.1 is a time-position diagram (notice the unusual choice of axes), showing the

T

T

0

x_D=v_D tt

x

x_1 = v t

x_2 = v ( t T )

Figure 3.1: Moving detector, stationary source.

lines corresponding to two successive wave fronts,

x1 = vt (3.39)

x2 = v(t T) (3.40)
http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Doppler.htmlhttp://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Doppler.html


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30 CHAPTER 3. WAVES

where T = 1/f is the period of the signal. The diagram also shows the detector, whose position as

a function of time is xD = vDt (3.41)

For simplicity, we have chosen the detector to coincide with the source at t = 0, precisely when thefirst wave front is emitted.

The second wave front reaches the detector when x2 = xD, that is, at a time T given by

v(T T) = vDT (3.42)so that

T = Tv

v vD (3.43)

and the frequency measured by the detector is therefore

f = fv vD

v(3.44)

Exercise 3.8.1 Redo the calculation above for the case that the detector is moving towards thesource with speed vD, and verify that in this case

f = fv + vD

v(3.45)

3.8.2 Moving source, stationary detector

Figure 3.2 shows the time-position diagram for this situation.Now the sources position as a function of time is

xS = vSt (3.46)

where once again for simplicity we let the source coincide with the detector at the time when thefirst wave front is emitted. Then when the second wave front is emitted at time T, the sourcehas reached the position vST, and the second wave front moves away from the source towards thedetector with speed v, so that its position as a function of time is

x2 = vST v(t T) (3.47)This wave front will reach the stationary detector at a time T when x2 = 0, that is,

vST v(T T) = 0 (3.48)which gives

T = Tv + vS

v(3.49)


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3.8. THE DOPPLER EFFECT 31

T

0

t

x

T

x_S = v_S t

x_2 =

v S T

v_S T v( t T )

Figure 3.2: Moving source, stationary detector.

andf = f

v

v + vS(3.50)

Exercise 3.8.2 Like before, verify that if the source is moving towards the detector,

f = fv

v vS (3.51)

3.8.3 The general Doppler effect

If both the source and the detector are moving relative to the medium, the frequency measured bythe detector is the result of combining the two previous results:

f = fv vDv vS (3.52)

The correct signs are determined by bearing in mind the following simple rule (quoted here verbatim

from HRW):When the motion of detector or source is toward the other, the sign on its speed mustgive an upward shift in frequency. When the motion of detector or source is away fromthe other, the sign on its speed must give a downward shift in frequency.


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32 CHAPTER 3. WAVES


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Chapter 4

Fluids

33


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34 CHAPTER 4. FLUIDS

4.1 Introduction

4.1.1 Solids, liquids, and gases

Fluids a term that includes both liquids and gases can flow and change shape, unlike solids.This difference in basic properties has a simple microscopic explanation. As Figure 4.1 illustrates,atoms in solids are very closely bound in a rigid lattice; in a liquid, the atoms are a little furtherapart, and the bonds between them are weaker, so they can slide past each other; in a gas, theatoms are so far removed from each other that they hardly interact at all except when two happento collide. It follows that in a given volume, atoms will be most closely packed in a solid, a littleless so in a liquid, and far less in a gas. The typical densities of materials in the three states areconsistent with this picture. Typical densities of liquids and solids are on the order of 103 kg/m3,while gases at STP1 have typical densities on the order of 1 kg/m3.

4.1.2 Density and pressure

We shall be looking for the basic laws of motion that govern fluids, analogous to Newtons lawsin particle mechanics. However, we can no longer apply particle concepts like mass or force to acontinuous medium. Rather we speak of the density and the pressure at a point in a fluid.

The density is the amount of mass per unit volume at a point in the fluid. Since a point hasno volume, this may seem puzzling. What we really mean is a very small volume element of thefluid, much smaller than the size of the fluid sample, but large enough to contain a large numberof atoms. If the fluid is uniform, the density is the same at all points, so even a large volume of

the fluid has the same density.In a similar spirit we define the pressure at a point in the fluid as the magnitude of the force

per unit area on a tiny disk at that point. Interestingly, pressure is a scalar: The orientation of thedisk is irrelevant.

In the SI system of units, density is measured in kg/m3, and pressure is measured in N/m2, a unitthat also goes by the name of Pascal (Pa), after Blaise Pascal, philosopher and mathematician. (Seehttp://www-groups.dcs.st-and.ac.uk/ history/Mathematicians/Pascal.html for a short biography ofthis very interesting individual).

4.2 Fluid statics

We begin our study of fluids by considering fluids that are in equilibrium, and therefore static.

1STP means Standard Temperature and Pressure, defined as 0 C and 1 atmosphere of pressure (see section ).These are conditions typical of the Earths atmosphere at sea-level.
http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Pascal.html


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4.2. FLUID STATICS 35

4.2.1 Variation of pressure with depth

Consider first an incompressible fluid, typically a liquid. By definition, an incompressible fluidhas uniform density . Isolate a column in the fluid of cross-sectional area A extending from adepth h0 to a depth h1 (see Figure 4.2).

The pressure on the lower surface must be greater than the pressure on the top surface, tobalance the weight of the fluid in the column. Newtons 2nd. law in the vertical direction gives

A(h1 h0)g = p1Ap0A p1 = p0 + g(h1 h0) (4.1)Thus the pressure increases linearly with depth in an incompressible fluid.

Exercise 4.2.1 If the pressure on the surface of a body of water is 105 Pa, a typical value of theatmospheric pressure, and the density of water is 103 kg/m3, how far down would the pressure betwice the atmospheric pressure at the surface?

We get a different result when we consider a compressible fluid, such as the atmosphere.Since the density can now vary with height, we must consider a thin slice of the fluid at a height zwhere the density (z) can be assumed to be uniform. See Figure 4.3.

The balance of forces in the vertical direction gives

(z)Agdz = pA A(p + dp) dp = (z)gdz (4.2)In order to calculate the profile of p versus height z, we would have to know the form of (z).

For example if we assume the gas follows the ideal gas law, then2

p =

mkT (4.3)

where m is the mass of an atom, T is the temperature of the gas measured in degrees Kelvin (K),

and k is Boltzmanns constant, k = 1.38 1023 J/K. Substituting in equation (4.2) we getdp

p= mg

kTdz (4.4)

Notice that kT/mg has dimensions of length. This quantity is called the scale height H. To findthe pressure at any height z, we integrate between a base height z = 0 where the pressure is p0,and z: p

p0

dp

p= 1

H

z0

dz (4.5)

which gives

lnp

p0 =

z

H p = p0e

z/H (4.6)

The pressure decreases with height, not linearly but exponentially. The density, being proportionalto the pressure, also decreases linearly with height. This makes sense: The lower layers of the fluidshould be more compressed by the weight of the fluid on top, and therefore more dense.

2This statement is offered without proof, purely for the purpose of illustration


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4.2.2 Buoyancy and Archimedes Principle

It is a common experience that objects sink or float in a fluid, depending on their density. Legendhas it that Archimedes of Syracuse discovered the exact mechanism for this. (Seehttp://www-groups.dcs.st-and.ac.uk/ history/Mathematicians/Archimedes.html)

Consider for example the fearsome beast in Figure 4.4 (Credit: D.J.,http://www.clay.k12.fl.us/wec/Animal.htm). If the animal was removed, it would be replaced bya portion of fluid of the same shape and volume which would be at rest, sustained by a force fromthe fluid equal to its weight. Since the fluid is unaware that this space is now occupied by a shark,it applies the same force to the shark. This is called the buoyancy force.

To put it mathematically, call the average density of the shark, f the density of the fluid, andV the volume of the shark, which is the same as that of the displaced fluid. The sharks weight isV g, and the buoyancy force, equal to the weight of the displaced fluid, is fV g. The net (upward)force on the shark is then

( f)V g (4.7)

which is positive if < f, in which case the shark floats upward, and negative if > f, in whichcase the shark sinks. If = f, the shark will neither sink nor swim; it is said to experience neutralbuoyancy.

Exercise 4.2.2 What is the weight of the shark under water? (Assume the shark is more densethan the surrounding water, which is hardly surprising). If you were able to measure the sharksweight outside the water, as well as under water, how would you calculate its density relative to

water from that information?

4.3 Fluid dynamics

4.3.1 The velocity field in a fluid

We turn now to the problem of a moving fluid. Our objective is to find conservation principlesanalogous to the known principles that apply to particle mechanics. To describe the motion ofthe fluid, we need to keep track of the velocity of the fluid at each point. Notice that this doesnot mean to follow a particular fluid element as it moves and record its velocity. It means that

we focus on a fixed point in the space occupied by the fluid, and record the velocity of each fluidelement that passes through that point. A map of the velocities at all points in the fluid is calledthe velocity field. It can be visualized approximately by drawing velocity vectors on a grid. Hereare some good examples: http://virga.sfsu.edu/crws/jetstream.html.

If the velocity field does not vary with time, the flow is said to be steady, or stationary.
http://virga.sfsu.edu/crws/jetstream.htmlhttp://www.clay.k12.fl.us/wec/Animal.htmhttp://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Archimedes.html


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4.3. FLUID DYNAMICS 37

4.3.2 Streamlines

If you looked at some of the wind velocity maps mentioned above, you could see that the arrowscould be quite easily joined by smooth curves. Such curves are called streamlines, defined ascurves which are tangent to the velocity vectors at each point. If the flow can be visualized usingstreamlines, it means that any fluid element that finds itself at some point between two givenstreamlines, remains between those streamlines at all future times. This is because of the definitionof streamlines: Since they are tangent to the velocity of the fluid at all points, it follows that nofluid element can cross a field line. The flow is said to be laminar, since the fluid moves in layers.The opposite of laminar flow is turbulent flow. There, streamlines are not defined. We shallrestrict our analysis to laminar flow only.

4.3.3 The continuity equation

The first conservation principle we apply is the conservation of mass. Figure 4.5 shows a portion ofa fluid flowing between two streamlines. At the left end, with cross-section area A1, the fluid moveswith speed v1. At the right end, with a different cross-section area A2, the speed is also different,v2. Since the fluid is contained within the streamlines, the amount of mass that enters from theleft during a time dt is the same that leaves at the right during the same time. Symbolically,

1v1A1dt = 2v2A2dt (4.8)

and therefore, within the same streamlines anywhere

vA = constant (4.9)

This is known as the continuity equation. The quantity vA is called the mass flow rate,measured in kg/s.

If the fluid is incompressible, = constant, and we can further say that

vA = constant (4.10)

Here, vA has dimensions of m3/s; its called the volume flow rate.In the case of an incompressible fluid, the continuity equation implies that if a fluid element is

forced through a constriction, it will speed up, or if it is allowed to widen, it will slow down. Inother words, the fluid is accelerated when the cross-section area changes, and that acceleration canonly be brought about by a pressure difference across the fluid element. The pressure differencethat produces a certain acceleration is determined by conservation of energy, to which we turn now.

4.3.4 Bernoullis law

Well consider now the principle of conservation of energy as applied to a fluid. In this section,we shall argue something quite surprising: The pressure in a fluid behaves mathematically as apotential energy function.


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As we show in Appendix E, if a particle moves along a direction x under the influence of a

conservative forceF, and the potential energy is U(x), there is a simple relation between thepotential energy and the component of the force in the direction of motion:

Fx = dUdx

(4.11)

Now consider instead a fluid element of width dx and cross-sectional area A moving between a pairof streamlines. Considering the pressure difference across the fluid element, the net force in the xdirection is

pA (p + dp)A = Adp (4.12)Now, the volume of the fluid element is Adx; so, interestingly, the force per unit volume along x is

Fx = dpdx (4.13)

so p mimics a potential energy!3

Conservation of energy per unit volume then implies that

p +1

2v2 = constant (4.14)

along a streamline. Furthermore, if the fluid is also subject to a conservative force, its associatedpotential energy should be included in the equation. For instance, if the fluid is moving in thegravitational field near the surface of the Earth, along any streamline

p + gh + 12 v2 = constant (4.15)

This is known as Bernoullis equation, after the talented 18th. century mathematical physicistDaniel Bernoulli.

3It is important to realize that although the pressure has the mathematical properties of a potential energy, itisnt really a p otential energy. A conservative force requires a source: The gravitational pull on the Moon is due tothe Earth (and other bodies). There is no source of the pressure.


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Solid

Liquid

Gas

Figure 4.1: A microscopic view of solids, liquids, and gases.


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h

0

Crosssectional area A

Weight

p_0

p_1

h_0

h_1

Figure 4.2: Variation of pressure with depth in an incompressible fluid.


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z

z

z+dz

p+dp

p

Crosssectional area A

Figure 4.3: Variation of pressure with depth in a compressible fluid.

Figure 4.4: D.J.s hammerhead shark.


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A_1

A_2

v 1 dt

v_2 dt

v_1 v_2

Figure 4.5: Conservation of mass.


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Appendix A

Derivations for the final exam

43


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44 APPENDIX A. DERIVATIONS FOR THE FINAL EXAM

A.1 Derivations for the final exam

The following is a list of derivations we have seen in the course, one of which will be included inthe final exam, with exactly the same wording as given here. Beside each one, a link is provided tothe page in these notes where the derivation may be found.

1. The parallel-axis theorem: See Section 1.3.4. The moment of inertia of an extended bodyabout a fixed axis is the integral

I =

r2 dm

where r is the distance from the mass element dm to the axis, and the integral is taken overthe whole body.

Show that for an arbitrary fixed axis, I is related to ICM, the moment of inertia with respect

to a parallel axis that passes through the centre of mass, byI = ICM + M h

2

where M is the mass of the body, and h is the separation between the two axes.

In your proof, you may use the definition of the coordinates of the centre of mass:

rCM =

r dm

where r is the position of a mass element dm, and the integral is again taken over the wholebody.

2. Torque and angular momentum: See Section 1.8. Show that for a point mass m with

angular momentumL and a net torque about a point P

=dL

dt

3. A standing wave on a string: See Sections ?? and ??. Show that if two waves of identicalangular frequency and wave number k travel in opposite directions along a string, theycombine to form a standing wave of the form

y = A(t) sin(kx)

where A(t) is a time-dependent amplitude. Give an expression for A(t).

Use the following trigonometrical relation:

sin( ) = 2 sin 2

cos

2

4. Variation with depth of pressure in an incompressible fluid: See Section 4.2.1. Derive

the pressure p as a function of depth h in an incompressible fluid of density .


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Appendix B

The vector cross product

45


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46 APPENDIX B. THE VECTOR CROSS PRODUCT

B.1 The vector cross product

B.1.1 Definition

The cross product is an operation between two vectors in 3 dimensions that gives a third vectoras a result. It is written as

ab (B.1)and its pronounced a cross b.

The direction of the vector ab is perpendicular to both a and b, and given by a right-handrule: Point the fingers of the right hand along the first vector a, and sweep them towards thesecond vector b. The thumb points automatically in the direction ofab.

The magnitude of the cross product is

|a| |b| = |a||b| sin (B.2)where is the (small) angle between a and b.

One important consequence of this definition is that the cross product of two vectors that areparallel is zero, since = 0 in that case.

B.1.2 The cross product is non-commutative

With the cross product, the order of the factors does matter! From the definition,

a b = b a (B.3)

B.1.3 The x, y, and z components of a cross productIf we are given

a = ax + ay + azk (B.4)

b = bx + by + bzk (B.5)

we can construct the cross product as a sum of terms containing the cross products of all possiblecombinations of , and k:

ab = axbx + axby + . . . (B.6)So, we need to construct all the possible cross products of unit vectors. Actually, the only ones we

need are:

= k (B.7) k = (B.8)k = (B.9)


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B.1. THE VECTOR CROSS PRODUCT 47

The cross products of each unit vector with itself are all zero, and a product like is obtainedas .Now we can write the components ofab. Well leave it as an exercise to show that

(a b)x = aybz azby (B.10)(a b)y = azbx axbz (B.11)(ab)z = axby aybx (B.12)

Exercise B.1.1 Make up a mnemonic rule to remember the components of a cross product. (Hint:Cyclic permutations of indices).

Exercise B.1.2 a = 1/2 + 1/2 andb = 1/2 + 1/2. Calculate ab.


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48 APPENDIX B. THE VECTOR CROSS PRODUCT


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Appendix C

Dimensional analysis

49


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50 APPENDIX C. DIMENSIONAL ANALYSIS

C.1 The magic of dimensional analysis

The technique of dimensional analysis may seem almost magical: It is possible to derive theform of physical laws without knowing any physics! Its a two-step technique: To illustrate it, weshall use it here to derive the form of the period of an ideal pendulum.

C.1.1 Step 1: What does the quantity of interest depend on?

What can the period of a pendulum depend on? By playing with a pendulum, we can concludethat it depends on the length of the pendulum L. We can also surmise that it depends on thegravitational field g, even if we cant leave the Earth to test the idea. And forgetting everythingweve learned about pendulums, we might imagine that it depends on the mass m of the pendulumbob. We will, however, make the strong assumption that the period is independent of the amplitude

of the oscillation. Summarizing, we can write

T = f(L,g ,m) (C.1)

C.1.2 Step 2: Narrow down the possible functions by demanding that the di-

mensions are right

The fundamental assumption that we make next is that the function f is in fact a power law:

T = CLagbmc (C.2)

where the exponents a, b, and c are yet to be determined, and C is a dimensionless constant.Necessarily, the dimensions of both sides of this equation must be the same. Using SI units, thismeans that

s = ma(ms2)bkgc = ma+bs2bkgc (C.3)

This leads to a set of equations for the exponents:

a + b = 0 (C.4)

2b = 1 (C.5)c = 0 (C.6)

Notice that we find straight away that the exponent of m is zero, so the period T is independentof the mass. Solving for the remaining exponents, we get b = 1/2 and a = 1/2.

Now we can write

T = C

L

g(C.7)

This is as far as we can go with dimensional analysis. We could find C experimentally, bymeasuring the period of a pendulum and varying the length L. Since we know that the general


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C.1. THE MAGIC OF DIMENSIONAL ANALYSIS 51

form ofT as a function of L is given by our last result, we could plot T2 vs. L, and from the slope

find C. We would find that it is very close to 2.Indeed, one of the useful features of dimensional analysis is that it suggests experiments. An-other extremely useful application of dimensional analysis is in modelling. To simulate how a 1meter long pendulum on the Moon might behave, we would observe that g on the Moon is about1/6 the value on Earth, so we would construct a pendulum of length

6 m here at home.


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52 APPENDIX C. DIMENSIONAL ANALYSIS


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Appendix D

Average value of cos2(x)

53


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54 APPENDIX D. AVERAGE VALUE OF COS2(X)

D.1 Average value ofcos2(x)

In general, the average of a function f(x) over a certain interval [a, b] is calculated as

f(x) = 1b a

ba

f(x)dx (D.1)

Since cos2(x) is a periodic function of x with period 2, we only need to find the average over onecycle, say [0, 2]:

cos2(x) = 12

2

0

cos2(x)dx (D.2)

Alls fair in love, war, and integration, so well use a dirty trick. A moments thought willconvince you that

cos2(x) = sin2(x) (D.3)since sin(x) is the same function as cos(x), but with a phase shift of , (remember, the phase isthe argument of a trig function), so the averages should be the same. Now,

cos2(x) = 1 sin2(x) cos2(x) = 1 sin2(x) cos2(x) = sin2(x) = 12

(D.4)


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Appendix E

Work and potential energy

55


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56 APPENDIX E. WORK AND POTENTIAL ENERGY

E.1 Work and potential energy

In this section, we revisit briefly the familiar topics of work and potential energy, now with thebenefit of calculus.

A particle moves in the x direction under the action of a force F. As the particle is displace bydx, the amount of work done on it is

dW = Fxdx (E.1)

and the total amount of work done in moving the particle from xi to xf is

W =

xfxi

Fxdx (E.2)

If F is a conservative force, we can associate a potential energy function U to it, defined simply as

U = W (E.3)

In this case, we may rewrite equation (E.1) as

dU = Fxdx (E.4)

so that there is a simple relation between the force and the potential energy:

Fx = dUdx

(E.5)

Here, for simplicity, we restricted the motion of the particle to one dimension. In general, thecomponents of the force along x, y, and z are found from the potential energy U as

Fx =U

x(E.6)

Fy =U

y(E.7)

Fz =U

z(E.8)

dr. a louro - physics 225 notes

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