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PH1420: Fields and Waves
- January 2, James Nicholls, Physics Department, Royal Holloway, University of London
- 1 Oscillations Contents
- 1.1 Some definitions
- 1.2 Model system: mass on a spring
- 1.3 SHO, complex numbers, & phasors
- 1.4 Examples of SHO
- 1.5 Coupled oscillators & normal modes
- 2 Damped mechanical oscillations
- 2.1 Viscous damping
- 2.2 Energy loss in a damped oscillator
- 2.3 Driven oscillator
- 2.4 Q-factor and resonance
- 3 Waves
- 3.1 The wave equation
- 3.2 Transverse waves on a string
- 3.3 Representations of travelling wave
- 3.4 Transmission and reflection of waves
- 3.5 Standing waves and normal modes
- 3.6 Longitudinal sound waves
- 3.7 Phase and group velocity
- 4 Electrostatics
- 4.1 Coulomb’s Law
- 4.2 Conservation and quantization of charge
- 4.3 Electric field E⃗ and potential V
- 4.4 Electric flux and Gauss’s law
- 4.5 Conductors
- 4.6 Parallel plate capacitor
- 4.7 Force in electric and magnetic fields
5 Magnetostatics 47 5.1 Electric Current........................... 47 5.2 Ampere’s experiments........................ 48 5.3 Biot-Savart Law........................... 49 5.4 Ampere’s Law............................. 50 5.5 Force between parallel wires..................... 52 5.6 Faraday’s law and magnetic flux.................. 53 5.7 Self-inductance............................ 54
6 Circuit theory 57 6.1 Resistors................................ 57 6.2 Capacitors............................... 58 6.3 Inductors............................... 62 6.4 DC and AC circuits......................... 62 6.4.1 Joule heating in AC circuits................. 65 6.4.2 LC and LCR circuits..................... 65
Introduction
These are summary notes that contain the key equations used in lectures. Many results are stated; the physics and mathematical steps behind the equations will be discussed in lectures. To get the most out of this course you need to: a. Attend lectures and make your own notes. This is an essential skill. b. Read appropriate chapters of Giancoli (14, 15, 21-30). Look also at chapters of the e-book by Pain - link on Moodle page. There are many other good books available online. c. Attempt all Mastering Physics assignments, problem sets and tu- torial question sheets. d. Discuss material with your Tutor and colleagues. Work you hand in should be your own. e. Try the animations used in the lectures; the web links are given on the Moodle page. References and links in the notes are given, but some use Flash animation software which is now unsupported. f. The material summarised in Secs. 1 and 2 is now taught in PH1320, hopefully using the same notation. These sections are included be- cause oscillations and damping are important in Waves (Sec. 3) and AC circuit theory (Sec. 6).
When solving a 2nd^ order ODE, there will always two unknowns (for example, A and B, C and ϕ, D and ϕ′, etc.) which will be determined by the boundary conditions. Later we will consider waves, which are periodic in both space and time. In space the repeat distance is
λ = wavelength where k = 2π/λ = wave number. (8)
1.2 Model system: mass on a spring
Revision of free oscillations
Consider a mass m attached to a horizontal spring, see Fig 1, which has spring constant K. Note the spring constant is normally written as k, but later this will be used for the wavevector. If the particle is moved from its equilibrium position at x = 0, there is a restoring force given by
Hooke′s law : F = −K x, (9)
where the minus sign ensures that the force restores the mass to its equilibrium position. Animation #1 demonstrating Hooke’s law. Applying Newton’s 2nd^ law to the mass-spring system we obtain the following
2 nd^ order ODE : m ¨x + K x = 0, (10)
and by comparing Eq. 10 and Eq. 5 we find that the mass oscillates with angular frequency ω =
p K/m. A possible solution for the displacement is x(t) = A cos(ωt + ϕ), where A is the amplitude. The velocity and acceleration are given by v = ˙x(t) = −Aω sin(ωt + ϕ) and a = ¨x(t) = −Aω^2 cos(ωt + ϕ). The potential energy U stored in a spring is calculated from the work done (= force × distance) as the spring is displaced from its equilibrium position x = 0 to a given position x
U = −
Z (^) x
0
F (x′) dx′^ =
Z (^) x
0
K x′^ dx′^ =
K x^2. (11)
The sum of the kinetic energy, mv^2 /2 = m x˙^2 /2, and the potential energy U = Kx^2 /2, gives the
total energy of spring system : E =
m x˙^2 +
K x^2. (12)
If there is no damping then the total energy is constant, E = mv max^2 /2 = mω^2 A^2 /2 = K A^2 /2. Note that the equation of motion in Eq. 10 can be obtained by differentiating Eq. 12 with respect to time t, and setting dE/dt = 0.
Figure 1: (a) At x = 0 the spring is at equilibrium and there is no force acting on the mass m. If the spring is (b) stretched or (c) compressed, there is a linear restoring force F = −Kx acting on m.
Figure 2: The addition of two phasors, A 1 ei(ωt+ϕ^1 )^ and A 2 ei(ωt+ϕ^2 ), which have the same angular frequency ω, but different amplitudes (A 1 , A 2 ) and phases (ϕ 1 , ϕ 2 ). The parallelogram shows the addition of the two at t = 0, A 1 eiϕ^1 + A 2 eiϕ^2 = AR eiϕR^ , to give a third phasor with amplitude AR and phase ϕR. With increasing time t, all three rotate in the anticlockwise direction with angular frequency ω, with the parallelogram preserving its shape.
Animation #2 demonstrates phasors for a simple harmonic oscillator.
Superposition of two phasors In general any two phasors can be added together; this is called superposition. For example, there are two sinusoidal voltages V 1 (t) = A 1 cos(ωt + ϕ 1 ) and V 2 (t) = A 2 cos(ωt + ϕ 2 ), which have the same angular frequency, but with dif- ferent amplitudes and phases. If we wish to look at the sum of the two voltages, V 1 (t) + V 2 (t) then we can look at their phasors z 1 = A 1 ei(ωt+ϕ^1 )^ and z 2 = A 2 ei(ωt+ϕ^2 ). Figure 2 shows the two phasors z 1 (t) = x 1 (t) + i y 1 (t) and z 2 (t) = x 2 (t) + i y 2 (t), and their sum zR(t) = z 1 (t) + z 2 (t) = xR(t) + i yR(t).
A 1 ei(ωt+ϕ^1 )^ + A 2 ei(ωt+ϕ^2 )^ = AR ei(ωt+ϕR)^ (20)
at time t = 0. With increasing time t, the three phasors z 1 (t), z 2 (t) and zR(t) rotate anticlockwise together at angular frequency ω, as does the parallelogram in Fig. 2. Applying the cosine rule to the triangle inside the parallelogram we find that
AR =
p (x 1 + x 2 )^2 + (y 1 + y 2 )^2 =
q A^21 + A^22 + 2 A 1 A 2 cos(ϕ 1 − ϕ 2 )
ϕR = tan−^1
y 1 + y 2 x 1 + x 2
= tan−^1
A 1 sin ϕ 1 + A 2 sin ϕ 2 A 1 cos ϕ 1 + A 2 cos ϕ 2
With AR and ϕR derived, the real and imaginary parts of Eq. 20 can be matched
to obtain the following two relations:
A 1 cos(ωt + ϕ 1 ) + A 2 cos(ωt + ϕ 2 ) = AR cos(ωt + ϕR) A 1 sin(ωt + ϕ 1 ) + A 2 sin(ωt + ϕ 2 ) = AR sin(ωt + ϕR).
The first expression being the voltage sum V 1 (t) + V 2 (t). Phasors will be used in the optics course PH2310 to calculate diffraction pat- terns.
1.4 Examples of SHO
p g/l
p g ρ A/m
p γ p/V
p 2 g/l
In these mechanical examples, when the systems are displaced from equilibrium, there is a linear restoring force. In Sec. 6.4.2 we will investigate SHO in an electrical circuit made from an inductor L and a capacitor C; charge oscillates on and off the plates of the capacitor with an angular frequency of ω = 1/
LC.
1.5 Coupled oscillators & normal modes
Consider a system of two equal masses (m) and three similar springs (K) shown in Fig. 3, with the two outer springs attached to fixed walls. Consider the two masses to be displaced by x 1 and x 2 in the horizontal direction (to the right) from their equilibrium positions.
Figure 3: Linear arrangement of two masses (both m) and three springs (all of spring constant K) arranged in a linear fashion, between two fixed outer walls. The coordinates x 1 and x 2 describe the horizontal displacement from equilibrium of each mass.
The equations of motion for the two masses are
mx¨ 1 = −Kx 1 − K(x 1 − x 2 ) and mx¨ 2 = −Kx 2 + K(x 1 − x 2 ). (23)
Assuming oscillatory solutions of the form
x 1 = X 1 eiωt^ and x 2 = X 2 eiωt, (24)
For a string, which is the limit of an infinite number of connected masses, the normal modes are the standing waves (fundamental plus the harmonics). The excitation of one mode will not excite motion in another mode; mathematically the modes are said to be orthogonal or normal to each other.
2 Damped mechanical oscillations
2.1 Viscous damping
When a body experiences a damping force that linearly depends on its velocity
Fdamping = −b v = −b x˙; (28)
this is known as viscous damping, where b is the damping coefficient. The minus sign shows that the damping force opposes the motion, and is slowing down the body. Adding this new term to Eq. 10 gives
the differential equation for a damped mechanical oscillator : m d^2 x dt^2
(29) where b, the mass m, and the spring constant K are real, positive constants. Equation 29 is a homogeneous 2 nd^ order linear differential equation, where because of the damping term the solutions are no longer sines and cosines. If we look for solutions of the form, x = A ept, and substitute this into Eq. 29, we obtain the auxiliary equation : m p^2 + b p + K = 0. (30) The quadratic in p has two solutions:
p± =
−b 2 m
b^2 − 4 m K 2 m
−b 2 m
D
2 m
where everything depends on the term inside the square root, D = b^2 − 4 m K. There are two solutions for the displacement
x(t) = Aep+t^ + Bep−t, (32)
where A and B are constants. The behaviour of the discriminant D gives three cases to consider:
i. Lightly damped D < 0 or 4mK > b^2 This is the most interesting case. There are two complex roots:
p± =
−b 2 m
b^2 − 4 mK 2 m
= −α±iω′^ where α = b/ 2 m and ω′^ =
p K/m − b^2 / 4 m^2. (33) ⇒ x(t) = e−αt(A eiω
′t
′t ) = C e | {z− αt} envelope
sin(ω′t + ϕ) | {z } oscillations
The exponentially damped sinusoidal solutions describe lightly damped oscil- lations. For the case of very light damping 4mK ≫ b^2 , then ω′^ ≈
p K/m, that is, the oscillations have almost the same frequency as the undamped case; however the envelope function e−αt^ makes the amplitude exponentially decay with time t. The amplitude of the oscillations is exponentially decreasing because energy is being dissipated by the damping term. Rewriting e−αt^ = e−bt/^2 m^ = e−t/τ^ , the characteristic decay time τ is how long it takes the amplitude to decrease to 1 /e of its original value. For a damped oscillator the decay time is τ = 2m/b. ii. Critically damped D = 0 or 4 m K = b^2. Repeated roots p± = −b/ 2 m, gives the solutions
x(t) = A e−bt/^2 m^ + B t e−bt/^2 m. (35)
Critical damping gives the quickest return to the equilibrium position, x = 0.
iii. Over damped D > 0 or b^2 > 4 m K. The two real roots p± are both negative, which gives two exponentially de- caying solutions.
⇒ x(t) = A ep+t^ + B ep−t^ (36)
2.2 Energy loss in a damped oscillator
The total energy stored in a mechanical spring system is proportional to the (amplitude)^2 , that is, E = Ka^2 /2. For a lightly damped oscillator there is an exponential decay e−t/τ^ of the amplitude, so the mechanical energy of the system goes as: E = E 0 e−^2 t/τ^ = E 0 e−b t/m, (37) where E 0 is the energy of the oscillator at t = 0. The decay time for the energy is τ /2 = m/b.
2.3 Driven oscillator
The differential equation for a driven oscillator is obtained by adding a time- dependent driving force f (t) to the right hand side of the free oscillator equation (29), giving: m
d^2 x dt^2
dx dt
This inhomogeneous 2 nd^ order linear differential equation can be rewritten as:
d^2 x dt^2
dx dt
frequency well below resonance, ω ≪ ω 0 , the amplitude remains constant and the displacement x(t) continues to be in phase with the driving force f (t). At resonance the amplitude is A(ω 0 ) = F 0 /(iω 0 b) = e−iπ/^2 F 0 /(ω 0 b). There- fore the displacement x(t) in phase ϕ = π/2 behind the driving force f (t). Note that at resonance the amplitude depends on damping as 1/b. Above resonance the amplitude is A(ω) = −F 0 /(m ω^2 ). The minus sign means the displacement x(t) is out of phase (lags by π) with the driving force f (t). The amplitude falls with driving frequency as 1/ω^2. Animation #4 shows the amplitude and phase in a damped oscillator.
2.4 Q-factor and resonance
i. Two quantities, both with units of inverse time, determine the behaviour of the general oscillator equation (Eq. 39). One is the damping term γ, and the other is the resonant frequency ω 0. The dimensionless parameter constructed from their ratio defines the Q-factor
quality factor : Q =
ω 0 γ
which can be used to characterise all possible behaviour, Q ≫ 1 describes light damping, Q ≪ 1 describes heavy damping, and Q = 1/ 2 is critical damping. For the damped spring γ = b/m and ω 0 =
p K/m:
⇒ Quality factor : Q =
ω 0 γ
p K/m b/m
m K b
ii. The amplitude and phase described by Eqs. 45 and 46 for Q = 1, 2 , 5 , 10 and 20 are shown in Fig. 4. As Q increases, the amplitude exhibits a resonance that becomes more peaked and symmetric. In the limit γ → 0 (Q → ∞) then the amplitude |A| → ∞ at ω = ω 0. Physically this is not possible.
iii. The Q-factor characterises the rate at which the system loses energy. It can be shown that: energy in system energy dissipated/cycle
E
−∆E
ω 0 2 πγ
Q
2 π
For light damping (high Q), the energy lost per cycle is very small.
iv. Consider light damping, where Q ≫ 1. In this limit the complex quantity A(ω) in Eq. 44 has a magnitude given by its modulus. One can use the following approximation for the denominator when ω is close to ω 0 :
(ω 02 − ω^2 ) + iωγ ≈ 2 ω 0 (ω 0 − ω) + iω 0 γ
to show that |A(ω)| = F 0 /(2 m ω 0 ) q (ω 0 − ω)^2 + γ 2 4
for Q ≫ 1. (50)
0
5
10
15
20
magnitude |
A
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1. / (^0)
(in units of
Q= Q= Q= Q= Q=
Figure 4: Top: The amplitude |A| of a driven oscillator as a function of driving frequency ω/ω 0 for Q = 1, 2 , 5 , 10 and 20. As Q decreases (more damping) the curves becomes less symmetric, with a maximum that shifts to lower frequency, in accordance with Eq. 55. The y-axis is in units of F/K, the displacement at ω = 0. Note that for light damping the amplitude at ω = ω 0 scales with Q. Bottom: The phase lag ϕ (in units of π) as a function of driving frequency ω/ω 0 for Q = 1, 2 , 5 , 10 and 20.
3 Waves
A wave is a disturbance that propagates from one region to another. This dis- turbance travels at a definite velocity c and transfers energy as it propagates. In UG physics you will meet three types of waves:
- Mechanical waves These waves travel through a material medium (air, string, solid, liquid, etc.) by vibrating particles, but the particle themselves do not propagate. Mechanical waves are classified as either transverse or longitu- dinal, depending on the direction of the vibrations relative to that of the wave propagation. The wave velocity c is determined by Newton’s laws, which in turn depend on the mechanical properties of the material.
- Electromagnetic waves These can travel either through a medium or a vacuum. The waves are governed by the equations of electricity and magnetism (Maxwell’s equations), which will be covered in PH2420.
- Matter waves These are quantum mechanical (non-classical) waves whose properties are governed by the equations of Quantum Mechanics (Schrodinger’s equation), which will be covered briefly in PH1920, and more thoroughly in PH2210. In this course we will focus mostly on the classical mechanical waves, because they are the easiest to visualise and understand. The mathematics and results we obtain will be applicable to the other types of wave.
A periodic classical wave can be thought of as a correlated collection of oscil- lators, hence an understanding of simple harmonic motion is required for an understanding of waves. For transverse waves travelling along a string each point in the string oscillates back and forth in the transverse direction (not along the direction of the string), and in sound waves, the molecules of air os- cillate back and forth in the longitudinal direction (the direction in which the sound is travelling). The molecules do not have any net motion in the direction of the sound propagation. In water waves, each water molecule also undergoes oscillatory motion, and again, there is no overall net motion.
3.1 The wave equation
The 2nd^ order partial differential equation (PDE) that describes how a wave propagates in space x and time t is called the wave equation.
The wave equation in one-dimension is :
∂^2 y ∂t^2
= c^2
∂^2 y ∂x^2
where c is the velocity of the wave. For a wave on a string, y(x, t) is the transverse displacement of the string (from equilibrium) at position x at some time t. The wave equation holds for mechanical waves (which can be transverse or longitudinal), but also for electromagnetic waves (transverse). In Sec. 3. we will show that for transverse waves on a string, Eq. 59 can be derived by applying Newton’s Second Law to an element of the string.
It can be shown that a function f 1 of the form y(x, t) = f 1 (x − c t) is a solution to the wave equation. This is a wave or pulse moving to the right; solutions moving to the left have the form y(x, t) = f 2 (x + c t). Therefore the general solution to the 1D wave equation can be written as:
y(x, t) = f 1 (x − c t) + f 2 (x + c t), (60)
where f 1 and f 2 are arbitrary functions, requiring only properties of continuity and differentiability when substituted into Eq. 59. We will show that plane waves
y(x, t) = A 1 ei^ (+kx−ωt)^ + A 2 ei^ (−kx−ωt)^ (61)
are also solutions, and consistent with the general form in Eq. 60. The solution A 1 ei^ (+kx−ωt)^ with wavevector +k is a travelling (progressive) wave that propa- gates in the positive x-direction (to the right), whereas the solution A 2 ei^ (−kx−ωt) with wavevector −k travels to the left. The wave equation (Eq. 59) and its solutions have important properties: i. The wave equation applies to sound waves, electromagnetic waves, waves on a string, plus many other systems. From everyday life we know that there are a transverse waves on a string, where the oscillations of the string are perpendicular to the direction of the wave. In a longitudinal wave, such as a sound wave, the oscillations are parallel to the direction of propagation. There are systems where both types of wave can exist, for example, in a solid. ii. The plane waves (Eq. 61), which are sinusoidal in both space and time, y = sin(k x − ωt) can be written as y = sin(k (x − c t)) = f 1 (x − c t). Waves of arbitrary shape moving to the right or left with speed c can also be described by f 1 (x − c t) and f 2 (x + c t). A Gaussian pulse moving to the right is shown in Fig. 5. Note that the pulse of form f (x ± c t) will retain its shape. It shows no dispersion.
iii. Equation 59 is a linear 2nd^ order partial differential equation. Linearity means that if y 1 (x, t) and y 2 (x, t) are solutions to Eq. 59, then a superposition of these two solutions is also a solution. More formally the principle of superposition states:
If y 1 (x, t) and y 2 (x, t) are solutions, then y = a 1 y 1 (x, t) + a 2 y 2 (x, t) (62) is another solution, where a 1 and a 2 are real constants. The principle in words applied to waves: if two or more waves meet in a region of space, then at each instant of time the net disturbance they cause at any point is equal to the sum of the disturbances caused by each of the waves individually. Note that particles do not show superposition properties. For example, two particles that occupy the same position at the same time will undergo a col- lision, and will be deflected from their original path.
Linearity and superposition are important properties of most solutions that you will meet in UG physics, and can be useful in understanding electromag- netism, quantum mechanics, etc.
iv. Substituting a plane wave solution y(x, t) = A ei^ (±kx−ωt)^ into the wave equa- tion (Eq. 59) gives ω^2 = c^2 k^2 , (63) which is known as a dispersion relation ω(k). In this case ω = ± c k, where the positive solution can be rewritten in terms of wavelength and frequency as the well known equation
c = f λ. (64)
v. For the dispersion relation ω = c k, it can be shown that the phase velocity vp = ω/k and the group velocity vg = ∂ω∂k are the same: ω k
∂ω ∂k
= c. (65)
The dispersion ω = ck is therefore a special case where all the different fre- quencies move at the same speed, and waves and pulses preserve their shape as they propagate. In this case the solutions are said to be dispersionless.
vi. For a general dispersion relation ω(k), then ∂ω∂k ̸= ωk , and wavepackets will show dispersion.
3.2 Transverse waves on a string
A simple case of transverse waves is to consider their description on a uniform string with tension T and density μ (mass per unit length). We will assume that the string extends infinitely in both directions and is infinitesimally thin and completely flexible. Let x be the coordinate along the string, and let y(x, t) be the small transverse displacement of the string at position x and time t. Consider two nearby points separated by a displacement δx in the longitudinal direction along the string, and by a displacement δy in the transverse direction. If δy is small, see Fig. 6, then the length of the stretched segment of string is
δl =
p δx^2 + δy^2 = δx
s
1 +
δy δx
≈ δx 1 +
δy δx
If δy/δx ≪ 1 , then all points on the string move only in the transverse direction, and there is no longitudinal motion. i. We will assume that the tension T in the string has the same magnitude at all points along the string. This is an approximation because the passage of the wave will stretch the string slightly, from δx → δl, which will increase the tension. However, if the wave has small amplitude, this is a small effect that can be ignored.
a b
Figure 6: A segment of string of original length δx, with end points at x = a and x = b. The stretched segment has length δl =
p δx^2 + δy^2. Tangents to the string at a and b, make angles θ and θ + δθ with the positive x-axis. The tension throughout the string is T.
ii. We will also assume that θ and θ + δθ are small angles, allowing the use of the following approximations: cos θ ≈ 1 and sin θ ≈ tan θ.
iii. We will ignore the weight of the segment. The only forces that act on the segment are the tension T due to the neighbouring segments of string. Referring to Fig. 6, and noting that the angles in the figure have been exagger- ated, the tension at a and b is T , but acting in different directions. At a, the x and y components of the force are: Fx(a) = −T cos θ and Fy (a) = −T sin θ. At b, the corresponding components are: Fx(b) = T cos(θ+δθ) and Fy (b) = T sin(θ + δθ). Net horizontal force on the segment
Fx(a) + Fx(b) = T (cos(θ) − cos(θ + δθ)) ≈ 0. (67)
⇒ from Newton’s second law of motion the segment will not accelerate in the x-direction. Net vertical force on segment In the y-direction the forces on the segment are: Fy (a) + Fy (b) = −T sin(θ) + T sin(θ + δθ) Fy (a) + Fy (b) ≈ −T tan(θ) + T tan(θ + δθ),
where tan(θ) is the gradient at a, and tan(θ + δθ) is the gradient at b.
⇒ Fy (a) + Fy (b) = T
∂y ∂x
b
∂y ∂x
a
= T
∂^2 y ∂x^2
a
δx (69)
