Materials

Waves and Oscillations

Definition

A wave is a disturbance that travels through space and time, transferring energy without transferring matter. Oscillation is the repetitive variation, typically in time, of some measure about a central value.

Key Properties

Wavelength (λ): Distance between two consecutive identical points
Frequency (f): Number of oscillations per unit time
Period (T): Time for one complete oscillation
Wave speed (v): Speed at which the wave propagates

Important Formulas

v = fλ
T = 1/f
ω = 2πf = 2π/T

Example Problem

Problem: A wave has a frequency of 50 Hz and wavelength of 0.2 m. Find the wave speed.

Solution:

Given: f = 50 Hz, λ = 0.2 m
Using: v = fλ
v = 50 × 0.2 = 10 m/s

Periodic Motion

Definition

Periodic motion is motion that repeats itself at regular intervals of time. The motion returns to the same state after a fixed time period.

Key Characteristics

Particular Point: The motion passes through the same position repeatedly
Particular Time Interval: The motion repeats after a fixed time period (T)
Same Direction: At corresponding points in different cycles, velocity has the same direction

Mathematical Description

For periodic motion: x(t + T) = x(t) for all t

Example

Simple pendulum swinging back and forth
Mass on a spring oscillating up and down
Earth's rotation around the sun

Simple Harmonic Motion

Definition

Simple Harmonic Motion (SHM) is a type of periodic motion where the restoring force is directly proportional to the displacement from equilibrium position and acts in the opposite direction.

Mathematical Form

F = -kx
a = -ω²x

Where:

F = restoring force
k = force constant
x = displacement
ω = angular frequency

General Solution

x(t) = A cos(ωt + φ)

Where:

A = amplitude
ω = angular frequency
φ = phase constant

Example Problem

Problem: A mass of 2 kg is attached to a spring with k = 50 N/m. Find the period and frequency of oscillation.

Solution:

Given: m = 2 kg, k = 50 N/m

ω = √(k/m) = √(50/2) = √25 = 5 rad/s

T = 2π/ω = 2π/5 = 1.26 s

f = 1/T = 5/(2π) = 0.796 Hz

Angular Velocity & Frequency

Definition

Angular velocity (ω) is the rate of change of angular displacement with respect to time. In Simple Harmonic Motion (SHM), it represents how fast the oscillator would move if it were moving in a circle with the same period.

Frequency (f) is the number of complete oscillations per unit time.

Mathematical Relationships

ω = 2πf = 2π/T
f = 1/T = ω/(2π)
T = 2π/ω = 1/f

Where:

ω = angular velocity (rad/s)
f = frequency (Hz or s⁻¹)
T = period (s)

Physical Significance

Angular velocity connects linear Simple Harmonic Motion (SHM) to circular motion
Higher frequency means faster oscillations
Angular velocity determines the rate of phase change in Simple Harmonic Motion (SHM)

For Different Systems

Spring-Mass System:

ω = √(k/m)
f = (1/2π)√(k/m)

Simple Pendulum:

ω = √(g/L)
f = (1/2π)√(g/L)

Physical Pendulum:

ω = √(mgd/I)
f = (1/2π)√(mgd/I)

Example Problem

Problem: A mass-spring system has a spring constant k = 100 N/m and mass m = 4 kg. Find the angular velocity, frequency, and period.

Solution:

Given: k = 100 N/m, m = 4 kg

Angular velocity:
ω = √(k/m) = √(100/4) = √25 = 5 rad/s

Frequency:
f = ω/(2π) = 5/(2π) = 0.796 Hz

Period:
T = 1/f = 2π/ω = 2π/5 = 1.26 s

Units and Conversions

Angular velocity: rad/s (radians per second)
Frequency: Hz = s⁻¹ = cycles per second
Period: s (seconds)

Conversion:

1 revolution = 2π radians
1 Hz = 2π rad/s of angular velocity

Velocity & Acceleration in Simple Harmonic Motion (SHM)

Velocity

If position: x(t) = A cos(ωt + φ)

Then velocity: v(t) = dx/dt = -Aω sin(ωt + φ)

Maximum velocity: v_max = Aω (at equilibrium position)

Acceleration

Acceleration: a(t) = dv/dt = -Aω² cos(ωt + φ) = -ω²x(t)

Maximum acceleration: a_max = Aω² (at extreme positions)

Key Relationships

v² = ω²(A² - x²)
a = -ω²x

Example Problem

Problem: A particle in Simple Harmonic Motion (SHM) has amplitude 0.1 m and angular frequency 5 rad/s. Find velocity and acceleration when x = 0.06 m.

Solution:

Given: A = 0.1 m, ω = 5 rad/s, x = 0.06 m

Velocity:
v² = ω²(A² - x²)
v² = 25(0.01 - 0.0036) = 25(0.0064) = 0.16
v = ±0.4 m/s

Acceleration:
a = -ω²x = -25 × 0.06 = -1.5 m/s²

Springs in Simple Harmonic Motion (SHM)

Definition

Springs in Simple Harmonic Motion refer to elastic systems where a mass attached to a spring oscillates about its equilibrium position under the restoring force provided by the spring.

Hooke's Law

F = -kx

Spring-Mass System

For a mass m attached to a spring with spring constant k:

ω = √(k/m)
T = 2π√(m/k)
f = (1/2π)√(k/m)

Energy Considerations

Total Energy = (1/2)kA²
Kinetic Energy = (1/2)mv²
Potential Energy = (1/2)kx²

Example Problem

Problem: A 0.5 kg mass on a spring oscillates with period 2 s. Find the spring constant.

Solution:

Given: m = 0.5 kg, T = 2 s

T = 2π√(m/k)
2 = 2π√(0.5/k)
1/π = √(0.5/k)
1/π² = 0.5/k
k = 0.5π² = 4.93 N/m

Displacement

Definition

Displacement in Simple Harmonic Motion (SHM) is the distance from the equilibrium position, measured along the line of motion.

Mathematical Expression

x(t) = A cos(ωt + φ)

Key Points

Displacement varies sinusoidally with time
Maximum displacement = Amplitude (A)
At equilibrium: x = 0
At extremes: x = ±A

Phase Relationships

Important phase differences in Simple Harmonic Motion (SHM):

Position	Displacement (x)	Velocity (v)	Acceleration (a)
Extreme position	Maximum (±A)	Zero	Maximum (opposite to x)
Equilibrium	Zero	Maximum	Zero
Intermediate	Between 0 and A	Between 0 and max	Between 0 and max

Phase differences:

Velocity leads displacement by π/2 (90°)
Acceleration leads velocity by π/2 (90°)
Acceleration opposes displacement (π or 180° out of phase)

Mathematical relationships:

x(t) = A cos(ωt + φ)           [displacement]
v(t) = -Aω sin(ωt + φ)         [velocity leads x by π/2]
a(t) = -Aω² cos(ωt + φ)        [acceleration opposes x]

Energy in Simple Harmonic Motion (SHM)

Total Mechanical Energy

E = (1/2)kA² = constant

Energy Distribution

At any instant:

E = KE + PE = (1/2)mv² + (1/2)kx²

Energy Variations

At equilibrium (x = 0): KE = maximum, PE = 0
At extremes (x = ±A): KE = 0, PE = maximum
At intermediate positions: Both KE and PE are non-zero

Example Problem

Problem: A mass oscillates with amplitude 0.05 m on a spring (k = 100 N/m). Find total energy and kinetic energy when x = 0.03 m.

Solution:

Given: A = 0.05 m, k = 100 N/m, x = 0.03 m

Total Energy:
E = (1/2)kA² = (1/2) × 100 × (0.05)² = 0.125 J

Potential Energy at x = 0.03 m:
PE = (1/2)kx² = (1/2) × 100 × (0.03)² = 0.045 J

Kinetic Energy:
KE = E - PE = 0.125 - 0.045 = 0.08 J

Amplitude

Definition

Amplitude (A) is the maximum displacement from the equilibrium position in Simple Harmonic Motion (SHM).

Characteristics

Always positive
Determines the energy of the system: E = (1/2)kA²
Determines maximum velocity: v_max = Aω
Determines maximum acceleration: a_max = Aω²

Determining Amplitude

From initial conditions:

A = √(x₀² + (v₀/ω)²)

Where x₀ and v₀ are initial position and velocity.

Kinetic Energy

Definition

Kinetic Energy is the energy due to motion of the oscillating particle.

Formula

KE = (1/2)mv²

In Simple Harmonic Motion (SHM):

KE = (1/2)m[Aω sin(ωt + φ)]²
KE = (1/2)mA²ω² sin²(ωt + φ)

Alternative Form

KE = (1/2)ω²(A² - x²) × m
KE = (1/2)k(A² - x²)

Maximum KE

KE_max = (1/2)mA²ω² = (1/2)kA²

This occurs at equilibrium position (x = 0).

Potential Energy

Definition

Potential Energy is the energy stored due to displacement from equilibrium.

Formula

PE = (1/2)kx²

Characteristics

Zero at equilibrium position
Maximum at extreme positions: PE_max = (1/2)kA²
Varies as square of displacement

Graph

PE vs displacement is a parabola with vertex at origin.

Simple Pendulum

Definition

A simple pendulum consists of a point mass suspended by a massless, inextensible string from a fixed point.

Small Angle Approximation

For small angles (θ < 15°): sin θ ≈ θ

Period Formula

T = 2π√(L/g)

Where:

L = length of pendulum
g = acceleration due to gravity

Key Points

Period is independent of mass and amplitude (for small angles)
Period depends only on length and gravitational acceleration

Example Problem

Problem: A simple pendulum has length 1 m. Find its period on Earth (g = 9.8 m/s²).

Solution:

Given: L = 1 m, g = 9.8 m/s²

T = 2π√(L/g) = 2π√(1/9.8) = 2π√(0.102) = 2π × 0.319 = 2.01 s

Pendulum in a Lift

Effective Gravity

When a pendulum is in a lift, the effective gravity changes:

Lift at rest or constant velocity: g_eff = g

Lift accelerating upward: g_eff = g + a

Lift accelerating downward: g_eff = g - a

Free falling lift: g_eff = 0 (pendulum doesn't oscillate)

Modified Period

T = 2π√(L/g_eff)

Example Problem

Problem: A pendulum has period 2 s when lift is at rest. Find period when lift accelerates upward at 2 m/s².

Solution:

Given: T₀ = 2 s, a = 2 m/s² (upward)

At rest: T₀ = 2π√(L/g) = 2 s
Therefore: L/g = 1/π²

In accelerating lift:
g_eff = g + a = 9.8 + 2 = 11.8 m/s²

T = 2π√(L/g_eff) = 2π√(L/11.8)
T = 2π√((1/π²) × (9.8/11.8)) = 2√(9.8/11.8) = 1.82 s

Center of Mass (Control of Mass)

Definition

The center of mass (also referred to as "control of mass" in some contexts) is the point where the total mass of a system can be considered to be concentrated for the purpose of analyzing translational motion. It's the average position of all mass elements in a system.

For Discrete Masses

x_cm = (m₁x₁ + m₂x₂ + ... + mₙxₙ)/(m₁ + m₂ + ... + mₙ)

For Continuous Mass Distribution

x_cm = (1/M) ∫ x dm

Properties

Center of mass moves as if all external forces act on it
Internal forces don't affect center of mass motion
For symmetric objects, center of mass is at geometric center

Example Problem

Problem: Two masses m₁ = 3 kg at x₁ = 2 m and m₂ = 2 kg at x₂ = 8 m. Find center of mass.

Solution:

x_cm = (m₁x₁ + m₂x₂)/(m₁ + m₂)
x_cm = (3×2 + 2×8)/(3+2) = (6+16)/5 = 4.4 m

Torque and Moment of Inertia

Torque

Torque (τ) is the rotational equivalent of force.

τ = r × F = rF sin θ
τ = Iα

Where:

r = distance from axis of rotation
F = applied force
θ = angle between r and F
I = moment of inertia
α = angular acceleration

Moment of Inertia

Moment of inertia (I) is the rotational equivalent of mass.

For point masses:

I = Σ mᵢrᵢ²

For continuous distributions:

I = ∫ r² dm

Common Moments of Inertia

Solid sphere about center: I = (2/5)MR²
Solid cylinder about axis: I = (1/2)MR²
Rod about center: I = (1/12)ML²
Rod about end: I = (1/3)ML²

Example Problem

Problem: A solid cylinder (M = 5 kg, R = 0.2 m) rotates about its axis. Find moment of inertia.

Solution:

For solid cylinder about its axis:
I = (1/2)MR² = (1/2) × 5 × (0.2)² = 0.1 kg⋅m²

Linear vs Circular Motion

Linear Motion - Force Changes Inertia

In linear motion:

Newton's 2nd Law: F = ma
Inertia: Resistance to change in linear motion (mass)
Force: Causes acceleration (change in velocity)

Circular Motion - Torque Changes Inertia

In rotational motion:

Rotational 2nd Law: τ = Iα
Rotational Inertia: Resistance to change in rotational motion (moment of inertia)
Torque: Causes angular acceleration (change in angular velocity)

Analogies

Linear Motion	Rotational Motion
Force (F)	Torque (τ)
Mass (m)	Moment of Inertia (I)
Acceleration (a)	Angular Acceleration (α)
Velocity (v)	Angular Velocity (ω)
Displacement (x)	Angular Displacement (θ)

Physical Pendulum

Definition

A physical pendulum is any rigid body that oscillates under gravity about a horizontal axis that doesn't pass through its center of mass.

Period Formula

T = 2π√(I/mgd)

Where:

I = moment of inertia about the pivot point
m = mass of the body
g = acceleration due to gravity
d = distance from pivot to center of mass

Parallel Axis Theorem

I = I_cm + md²

Where I_cm is the moment of inertia about center of mass.

Example Problem

Problem: A uniform rod of length L = 1 m and mass m = 2 kg oscillates about one end. Find the period.

Solution:

For rod about end: I = (1/3)mL²
Distance to center of mass: d = L/2

T = 2π√(I/mgd) = 2π√((1/3)mL²/(mg×L/2))
T = 2π√((1/3)L²/(g×L/2)) = 2π√(2L/3g)
T = 2π√(2×1/(3×9.8)) = 2π√(0.068) = 1.64 s

Thermodynamics

Definition

Thermodynamics is the branch of physics that deals with heat, work, and temperature, and their relation to energy, radiation, and physical properties of matter.

Laws of Thermodynamics

The fundamental laws governing energy transfer and transformation in thermodynamic systems are covered in detail in the First Law of Thermodynamics (Enhanced) section below.

Key Concepts

System: The part of universe under study
Surroundings: Everything else
State variables: P, V, T, U, S
Process: Path between two states

Types of Systems

Open: Mass and energy can cross boundary
Closed: Only energy can cross boundary
Isolated: Neither mass nor energy can cross boundary

Ideal Gas Laws

Individual Gas Laws:

Boyle's Law (constant T): PV = constant
Charles's Law (constant P): V/T = constant
Gay-Lussac's Law (constant V): P/T = constant

Combined Gas Law:

(P₁V₁)/T₁ = (P₂V₂)/T₂

Ideal Gas Equation:

PV = nRT

Where R = 8.314 J/(mol·K) = universal gas constant

Isothermal Process

Definition

An isothermal process is one where temperature remains constant throughout.

Characteristics

ΔT = 0
ΔU = 0 (for ideal gas)
Q = W (all heat goes into work)

Ideal Gas Law

PV = constant
P₁V₁ = P₂V₂

Work Done

W = nRT ln(V₂/V₁) = nRT ln(P₁/P₂)

Example Problem

Problem: 2 moles of ideal gas at 300 K expand isothermally from 10 L to 20 L. Find work done.

Solution:

Given: n = 2 mol, T = 300 K, V₁ = 10 L, V₂ = 20 L
R = 8.314 J/(mol⋅K)

W = nRT ln(V₂/V₁)
W = 2 × 8.314 × 300 × ln(20/10)
W = 4988.4 × ln(2) = 4988.4 × 0.693 = 3458 J

Adiabatic Process

Definition

An adiabatic process is one where no heat is exchanged with surroundings (Q = 0).

Characteristics

Q = 0
ΔU = -W
Temperature changes during process

Adiabatic Relations

PVᵞ = constant
TVᵞ⁻¹ = constant
TPᵞ⁻¹/ᵞ = constant

Where γ = Cp/Cv (adiabatic index)

Work Done

W = (P₁V₁ - P₂V₂)/(γ - 1)

Example Problem

Problem: An ideal gas (γ = 1.4) expands adiabatically from P₁ = 5 atm, V₁ = 2 L to V₂ = 8 L. Find final pressure.

Solution:

Given: γ = 1.4, P₁ = 5 atm, V₁ = 2 L, V₂ = 8 L

Using PVᵞ = constant:
P₁V₁ᵞ = P₂V₂ᵞ
P₂ = P₁(V₁/V₂)ᵞ = 5 × (2/8)^1.4 = 5 × (0.25)^1.4 = 5 × 0.0946 = 0.473 atm

Reversible Process

Definition

A reversible process is an ideal thermodynamic process that can be reversed without leaving any trace on the surroundings. It proceeds infinitely slowly through a series of equilibrium states.

Characteristics

Quasi-static: Proceeds very slowly
No friction: No energy lost to friction
No heat conduction: No spontaneous heat transfer
Pressure equilibrium: System pressure always equals external pressure

Examples

Slow compression/expansion of gas in a frictionless piston
Carnot cycle (theoretical)
Slow charging/discharging of a battery

Key Points

Maximum work output for expansion
Minimum work input for compression
Entropy of universe remains constant
Real processes are always irreversible

Mathematical Condition

For a reversible process:

dS_universe = dS_system + dS_surroundings = 0

Efficiency

Definition

Efficiency is the ratio of useful work output to the total energy input in any process or machine.

General Formula

η = Work Done (Output) / Energy Input × 100%
η = W_out / Q_in × 100%

For Heat Engines

η = (Q_H - Q_C) / Q_H = 1 - Q_C/Q_H
η = W / Q_H

Where:

Q_H = heat absorbed from hot reservoir
Q_C = heat rejected to cold reservoir
W = net work done

Maximum Efficiency (Carnot)

η_max = 1 - T_C/T_H

Example Problem

Problem: A heat engine absorbs 1000 J from a hot reservoir at 500 K and rejects 600 J to a cold reservoir at 300 K. Find actual and maximum possible efficiency.

Solution:

Given: Q_H = 1000 J, Q_C = 600 J, T_H = 500 K, T_C = 300 K

Actual efficiency:
η_actual = (Q_H - Q_C)/Q_H = (1000 - 600)/1000 = 0.4 = 40%

Maximum efficiency (Carnot):
η_max = 1 - T_C/T_H = 1 - 300/500 = 0.4 = 40%

This engine is operating at maximum theoretical efficiency!

Cyclic Process

Definition

A cyclic process is a thermodynamic process where the system returns to its initial state after completing the process. All state variables return to their original values.

Characteristics

Net change in internal energy: ΔU = 0
First law becomes: Q = W (heat supplied equals work done)
Forms closed loop on P-V diagram
Repeatable process

Types of Cycles

Carnot Cycle: Most efficient theoretical cycle
Otto Cycle: Gasoline engines
Diesel Cycle: Diesel engines
Brayton Cycle: Gas turbines

Mathematical Relations

∮ dU = 0 (internal energy returns to initial value)
∮ dQ = ∮ dW (total heat = total work)

Work Done in Cycle

Work done = Area enclosed by the cycle on P-V diagram

Example

A gas undergoes a cyclic process: expansion → cooling → compression → heating

Net ΔU = 0
Net work = area of loop on P-V diagram

Constant Pressure (Isobaric) Process

Definition

An isobaric process is one where pressure remains constant throughout the process.

Characteristics

Pressure: P = constant
Work done: W = P × ΔV = P(V₂ - V₁)
First law: ΔU = Q - PΔV

Ideal Gas Relations

V₁/T₁ = V₂/T₂ (Charles's Law)
V ∝ T (at constant P)

Work Done

W = ∫ P dV = P ∫ dV = P(V₂ - V₁)

Heat Capacity

At constant pressure: Q = nCₚΔT

Where Cₚ = molar heat capacity at constant pressure

Example Problem

Problem: 2 moles of ideal gas at 2 atm expand isobarically from 10 L to 20 L. Find work done.

Solution:

Given: n = 2 mol, P = 2 atm = 2 × 10⁵ Pa, V₁ = 10 L, V₂ = 20 L

W = P(V₂ - V₁) = 2 × 10⁵ × (20 - 10) × 10⁻³
W = 2 × 10⁵ × 10 × 10⁻³ = 2000 J

P-V Diagram

Horizontal line (constant pressure)

Constant Volume (Isochoric) Process

Definition

An isochoric process is one where volume remains constant throughout the process.

Characteristics

Volume: V = constant
Work done: W = 0 (no volume change)
First law: ΔU = Q (all heat goes to internal energy)

Ideal Gas Relations

P₁/T₁ = P₂/T₂ (Gay-Lussac's Law)
P ∝ T (at constant V)

Heat Capacity

At constant volume: Q = nCᵥΔT

Where Cᵥ = molar heat capacity at constant volume

Key Points

No work done since dV = 0
All supplied heat increases internal energy
Pressure changes with temperature

Example Problem

Problem: Gas at 300 K and 1 atm is heated at constant volume to 600 K. Find final pressure.

Solution:

Given: T₁ = 300 K, P₁ = 1 atm, T₂ = 600 K, V = constant

Using P₁/T₁ = P₂/T₂:
P₂ = P₁ × (T₂/T₁) = 1 × (600/300) = 2 atm

P-V Diagram

Vertical line (constant volume)

Slopes and Work

Definition

This section covers the slopes of isothermal and adiabatic curves on P-V diagrams and the work done in adiabatic processes. Understanding these slopes helps differentiate between different thermodynamic processes graphically.

Slope of Isotherm

On a P-V diagram, the slope of an isothermal curve is:

dP/dV = -P/V = -nRT/V²

Slope of Adiabatic

For an adiabatic process:

dP/dV = -γP/V

Comparison

Adiabatic curves are steeper than isothermal curves
Ratio of slopes: (dP/dV)_adiabatic/(dP/dV)_isothermal = γ

Work Done in Adiabatic Process

Definition: Work done in an adiabatic process where no heat is exchanged (Q = 0).

Primary Formula:

W = (P₁V₁ - P₂V₂)/(γ - 1)

Alternative Forms:

W = nCᵥ(T₁ - T₂)  (using temperature)
W = (nRT₁ - nRT₂)/(γ - 1)  (using ideal gas law)

From First Law: Since Q = 0 in adiabatic process:

ΔU = -W
W = -ΔU = -nCᵥΔT = nCᵥ(T₁ - T₂)

Key Points:

Work comes from internal energy of the gas
Temperature decreases during adiabatic expansion
Temperature increases during adiabatic compression
More work is required for adiabatic compression than isothermal

Example Problem: Problem: 2 moles of ideal gas (γ = 1.4, Cᵥ = 5R/2) expand adiabatically from 300 K to 250 K. Find work done.

Solution:

Given: n = 2 mol, T₁ = 300 K, T₂ = 250 K, γ = 1.4
Cᵥ = 5R/2 = 5 × 8.314/2 = 20.785 J/(mol·K)

W = nCᵥ(T₁ - T₂)
W = 2 × 20.785 × (300 - 250)
W = 2 × 20.785 × 50 = 2078.5 J

Work done by the gas = 2078.5 J

First Law of Thermodynamics (Enhanced)

Definition

The First Law of Thermodynamics is the law of energy conservation applied to thermodynamic systems. Energy cannot be created or destroyed, only converted from one form to another.

Mathematical Statement

ΔU = Q - W

Where:

ΔU = change in internal energy of the system
Q = heat added to the system (+) or removed (-)
W = work done by the system (+) or on the system (-)

Alternative Forms

Q = ΔU + W  (heat supplied = change in internal energy + work done by system)
dU = δQ - δW  (differential form)

Sign Conventions

Heat (Q): Positive if added to system, negative if removed
Work (W): Positive if done by system (expansion), negative if done on system (compression)
Internal Energy (ΔU): Positive if increased, negative if decreased

Applications to Different Processes

Isothermal Process (ΔT = 0):

ΔU = 0, so Q = W
All heat goes into work

Adiabatic Process (Q = 0):

ΔU = -W
Work comes from internal energy

Isochoric Process (W = 0):

ΔU = Q
All heat changes internal energy

Isobaric Process:

ΔU = Q - PΔV
Heat goes to both internal energy and work

Example Problem

Problem: A gas absorbs 500 J of heat and does 200 J of work during expansion. Find the change in internal energy.

Solution:

Given: Q = +500 J (heat added), W = +200 J (work by system)

Using First Law: ΔU = Q - W
ΔU = 500 - 200 = 300 J

Internal energy increases by 300 J.

Physical Significance

Energy is conserved in all thermodynamic processes
Heat and work are different ways of transferring energy
Internal energy is a state function (path independent)
Heat and work are path functions (path dependent)

Carnot Cycle

Definition

The Carnot cycle is a theoretical thermodynamic cycle that provides maximum efficiency for any heat engine operating between two thermal reservoirs.

Four Processes

Isothermal expansion (A→B): Heat Qₕ absorbed from hot reservoir
Adiabatic expansion (B→C): Temperature drops from Tₕ to Tc
Isothermal compression (C→D): Heat Qc rejected to cold reservoir
Adiabatic compression (D→A): Temperature rises from Tc to Tₕ

Carnot Efficiency

η = 1 - Tc/Tₕ = 1 - Qc/Qₕ

Where:

Tₕ = temperature of hot reservoir (K)
Tc = temperature of cold reservoir (K)

Work Output

W = Qₕ - Qc = Qₕ(1 - Tc/Tₕ)

Example Problem

Problem: A Carnot engine operates between reservoirs at 500 K and 300 K. If it absorbs 1000 J from hot reservoir, find efficiency and work output.

Solution:

Given: Tₕ = 500 K, Tc = 300 K, Qₕ = 1000 J

Efficiency:
η = 1 - Tc/Tₕ = 1 - 300/500 = 1 - 0.6 = 0.4 = 40%

Work output:
W = ηQₕ = 0.4 × 1000 = 400 J

Heat rejected:
Qc = Qₕ - W = 1000 - 400 = 600 J

Refrigeration Cycle

Definition

A refrigeration cycle is the reverse of a heat engine - it removes heat from a cold reservoir and rejects it to a hot reservoir, requiring work input.

Coefficient of Performance (COP)

For refrigerator:

COP_R = Qc/W = Qc/(Qₕ - Qc)

For heat pump:

COP_HP = Qₕ/W = Qₕ/(Qₕ - Qc)

Carnot Refrigerator

Maximum theoretical COP:

COP_R = Tc/(Tₕ - Tc)
COP_HP = Tₕ/(Tₕ - Tc)

Relationship

COP_HP = COP_R + 1

Four Processes (Reverse Carnot)

Adiabatic compression: Work input, temperature rises
Isothermal compression: Heat rejected to hot reservoir
Adiabatic expansion: Temperature drops
Isothermal expansion: Heat absorbed from cold reservoir

Example Problem

Problem: A Carnot refrigerator operates between 250 K and 300 K. Find COP if it removes 500 J from cold reservoir.

Solution:

Given: Tc = 250 K, Tₕ = 300 K, Qc = 500 J

COP_R = Tc/(Tₕ - Tc) = 250/(300 - 250) = 250/50 = 5

Work required:
W = Qc/COP_R = 500/5 = 100 J

Heat rejected to hot reservoir:
Qₕ = Qc + W = 500 + 100 = 600 J

Quick Reference Formulas

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A. Waves and Oscillations (Section)

Basic Wave Properties:

v = fλ (wave speed)
T = 1/f (period)
ω = 2πf = 2π/T (angular frequency)

Periodic Motion: (Section)

x(t + T) = x(t) (condition for periodicity)

B. Simple Harmonic Motion (SHM) (Section)

Fundamental Relations:

F = -kx (Hooke's Law)
a = -ω²x (acceleration)
ω = 2πf = 2π/T = √(k/m) (angular frequency for spring-mass)

Motion Equations:

x(t) = A cos(ωt + φ) (displacement)
v(t) = -Aω sin(ωt + φ) (velocity leads displacement by π/2)
a(t) = -Aω² cos(ωt + φ) = -ω²x (acceleration opposes displacement)

Key Relationships: (Section)

v² = ω²(A² - x²) (velocity-displacement relation)
v_max = Aω (maximum velocity at equilibrium)
a_max = Aω² (maximum acceleration at extremes)
A = √(x₀² + (v₀/ω)²) (amplitude from initial conditions)

C. Energy in Oscillations

Energy Conservation: (Section)

E_total = (1/2)kA² = constant
E = KE + PE = (1/2)mv² + (1/2)kx²

Kinetic Energy: (Section)

KE = (1/2)mv²
KE = (1/2)k(A² - x²)
KE_max = (1/2)kA² (at equilibrium)

Potential Energy: (Section)

PE = (1/2)kx²
PE_max = (1/2)kA² (at extremes)

D. Pendulums

Simple Pendulum: (Section)

T = 2π√(L/g) (period, independent of mass for small angles)
ω = √(g/L) (angular frequency)
f = (1/2π)√(g/L) (frequency)

Pendulum in Lift: (Section)

g_eff = g + a (upward acceleration)
g_eff = g - a (downward acceleration)
T = 2π√(L/g_eff) (modified period)

Physical Pendulum: (Section)

T = 2π√(I/mgd) (period)
I = I_cm + md² (parallel axis theorem)

E. Rotational Mechanics

Center of Mass: (Section)

x_cm = (Σmᵢxᵢ)/(Σmᵢ) (discrete masses)
x_cm = (1/M) ∫ x dm (continuous distribution)

Torque and Moment of Inertia: (Section)

τ = r × F = rF sin θ (torque)
τ = Iα (rotational Newton's 2nd law)
I = Σ mᵢrᵢ² (discrete masses)
I = ∫ r² dm (continuous distribution)

Common Moments of Inertia:

Solid sphere (center): I = (2/5)MR²
Solid cylinder (axis): I = (1/2)MR²
Rod (center): I = (1/12)ML²
Rod (end): I = (1/3)ML²

Linear vs Circular Motion: (Section)

Linear: F = ma, Rotational: τ = Iα
v ↔ ω, a ↔ α, m ↔ I, F ↔ τ, x ↔ θ

F. Thermodynamics Fundamentals

Ideal Gas Laws: (Section)

PV = nRT (ideal gas equation, R = 8.314 J/(mol·K))
PV = constant (Boyle's Law, constant T)
V/T = constant (Charles's Law, constant P)
P/T = constant (Gay-Lussac's Law, constant V)
(P₁V₁)/T₁ = (P₂V₂)/T₂ (combined gas law)

First Law of Thermodynamics: (Section)

ΔU = Q - W (energy conservation)
Q = ΔU + W (alternative form)
Sign convention: Q > 0 (heat in), W > 0 (work by system)

Efficiency: (Section)

η = W_out/Q_in = (Q_H - Q_C)/Q_H (heat engine)
η_max = 1 - T_C/T_H (Carnot efficiency)

G. Thermodynamic Processes

Isothermal Process: (Section)

T = constant, ΔU = 0, Q = W
PV = constant, P₁V₁ = P₂V₂
W = nRT ln(V₂/V₁) = nRT ln(P₁/P₂)

Adiabatic Process: (Section)

Q = 0, ΔU = -W
PVᵞ = constant, TVᵞ⁻¹ = constant, TPᵞ⁻¹/ᵞ = constant
W = (P₁V₁ - P₂V₂)/(γ-1) = nCᵥ(T₁ - T₂)

Isobaric Process: (Section)

P = constant, V/T = constant
W = PΔV = P(V₂ - V₁)
Q = nCₚΔT, ΔU = Q - PΔV

Isochoric Process: (Section)

V = constant, P/T = constant, W = 0
ΔU = Q = nCᵥΔT

Slopes and Work: (Section)

Isothermal slope: dP/dV = -P/V
Adiabatic slope: dP/dV = -γP/V
Adiabatic curves are steeper by factor γ

Reversible Process: (Section)

Quasi-static, no friction, maximum efficiency
dS_universe = 0

H. Thermodynamic Cycles

Cyclic Process: (Section)

ΔU = 0 (returns to initial state)
Q_net = W_net (total heat = total work)
Work = area enclosed on P-V diagram

Carnot Cycle: (Section)

Four processes: isothermal expansion → adiabatic expansion → isothermal compression → adiabatic compression
η = 1 - T_C/T_H = 1 - Q_C/Q_H
Maximum theoretical efficiency

Refrigeration Cycle: (Section)

COP_R = Q_C/W = T_C/(T_H - T_C) (refrigerator)
COP_HP = Q_H/W = T_H/(T_H - T_C) (heat pump)
COP_HP = COP_R + 1

Important Constants

g = 9.8 m/s² (acceleration due to gravity)
R = 8.314 J/(mol·K) (universal gas constant)
γ = Cₚ/Cᵥ (adiabatic index: ~1.4 for diatomic gases, ~1.67 for monatomic)
Cᵥ = R/(γ-1) (molar heat capacity at constant volume)
Cₚ = γR/(γ-1) (molar heat capacity at constant pressure)

Final Exam Practice Questions

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Question 1: Simple Harmonic Motion and Energy

A mass of 2.0 kg is attached to a horizontal spring with spring constant k = 800 N/m. The mass is displaced 0.15 m from its equilibrium position and released from rest. The surface is frictionless.

(a) Calculate the angular frequency, frequency, and period of oscillation.

(b) Find the maximum velocity and maximum acceleration of the mass.

(c) When the displacement is 0.10 m from equilibrium, calculate:

(i) The velocity of the mass
(ii) The kinetic energy
(iii) The potential energy

(d) Write the equations for displacement x(t), velocity v(t), and acceleration a(t) as functions of time, assuming the mass starts from maximum displacement.

Solution 1:

(a) Angular frequency, frequency, and period:

Given: m = 2.0 kg, k = 800 N/m, A = 0.15 m

Angular frequency: ω = √(k/m) = √(800/2.0) = √400 = 20 rad/s

Frequency: f = ω/(2π) = 20/(2π) = 3.18 Hz

Period: T = 1/f = 2π/ω = 2π/20 = 0.314 s

(b) Maximum velocity and acceleration:

Maximum velocity (at equilibrium):
v_max = Aω = 0.15 × 20 = 3.0 m/s

Maximum acceleration (at extremes):
a_max = Aω² = 0.15 × 20² = 0.15 × 400 = 60 m/s²

(c) At x = 0.10 m:

(i) Velocity:
Using v² = ω²(A² - x²)
v² = 20²(0.15² - 0.10²) = 400(0.0225 - 0.01) = 400(0.0125) = 5
v = ±2.24 m/s

(ii) Kinetic energy:
KE = ½mv² = ½ × 2.0 × 5 = 5.0 J

(iii) Potential energy:
PE = ½kx² = ½ × 800 × 0.10² = 400 × 0.01 = 4.0 J

Check: Total energy = KE + PE = 5.0 + 4.0 = 9.0 J
E_total = ½kA² = ½ × 800 × 0.15² = 9.0 J ✓

(d) Equations of motion:

Starting from maximum displacement (x₀ = A, v₀ = 0):

x(t) = 0.15 cos(20t) m
v(t) = -3.0 sin(20t) m/s
a(t) = -60 cos(20t) m/s²

Question 2: Thermodynamic Cycle Analysis

An ideal gas undergoes a cyclic process consisting of four stages:

Process A→B: Isothermal expansion at 400 K from 2.0 L to 8.0 L
Process B→C: Adiabatic expansion (γ = 1.4) until temperature drops to 300 K
Process C→D: Isothermal compression at 300 K
Process D→A: Adiabatic compression back to initial state

The gas contains 0.5 moles.

(a) For process A→B, calculate:

(i) Initial and final pressures
(ii) Work done by the gas
(iii) Heat absorbed

(b) For process B→C, find the final volume and work done.

(c) Calculate the efficiency of this heat engine.

(d) Compare this efficiency with the maximum possible Carnot efficiency between the same temperature limits.

Solution 2:

(a) Process A→B (Isothermal at 400 K):

Given: n = 0.5 mol, T = 400 K, V₁ = 2.0 L, V₂ = 8.0 L
R = 8.314 J/(mol·K)

(i) Pressures:
Initial: P₁ = nRT/V₁ = (0.5)(8.314)(400)/(2.0×10⁻³) = 831,400 Pa ≈ 8.31 atm
Final: P₂ = nRT/V₂ = (0.5)(8.314)(400)/(8.0×10⁻³) = 207,850 Pa ≈ 2.08 atm

(ii) Work done:
W = nRT ln(V₂/V₁) = (0.5)(8.314)(400) ln(8.0/2.0)
W = 1662.8 × ln(4) = 1662.8 × 1.386 = 2305 J

(iii) Heat absorbed:
For isothermal process: ΔU = 0, so Q = W = 2305 J

(b) Process B→C (Adiabatic expansion):

From 400 K to 300 K, starting at V = 8.0 L

Using TV^(γ-1) = constant:
T₁V₁^(γ-1) = T₂V₂^(γ-1)
400 × (8.0)^0.4 = 300 × V₂^0.4
V₂^0.4 = 400 × (8.0)^0.4 / 300 = 1.333 × 2.297 = 3.063
V₂ = (3.063)^2.5 = 12.67 L

Work done: W = nCᵥ(T₁ - T₂)
For ideal gas: Cᵥ = R/(γ-1) = 8.314/0.4 = 20.785 J/(mol·K)
W = (0.5)(20.785)(400 - 300) = 1039 J

(c) Efficiency calculation:

Heat absorbed (from hot reservoir) = Q_AB = 2305 J

For complete cycle, need to find heat rejected at low temperature.
Process C→D: Isothermal compression at 300 K
Process D→A: Adiabatic compression

By symmetry and energy conservation:
Q_CD = -nRT_cold ln(V_D/V_C)

For Carnot-like cycle efficiency:
η = 1 - T_cold/T_hot = 1 - 300/400 = 0.25 = 25%

(d) Carnot efficiency comparison:

Maximum Carnot efficiency:
η_Carnot = 1 - T_C/T_H = 1 - 300/400 = 0.25 = 25%

This cycle achieves the maximum theoretical efficiency because it operates
between the same temperature limits with reversible processes.

Question 3: Coupled Pendulum System

A physical pendulum consists of a uniform rod of length L = 1.2 m and mass M = 3.0 kg, pivoted at a point 0.3 m from one end. The pendulum is inside an elevator.

(a) Calculate the period of small oscillations when the elevator is:

(i) At rest
(ii) Accelerating upward at 2.0 m/s²

(b) A point mass m = 0.5 kg is attached at the free end of the rod. How does this change the period when the elevator is at rest?

(c) If the pendulum is displaced by 5° and released, find the maximum angular velocity and maximum angular acceleration for case (a)(i).

Solution 3:

(a) Period calculations:

Given: L = 1.2 m, M = 3.0 kg, pivot at 0.3 m from end
Distance from pivot to center of mass: d = 0.6 - 0.3 = 0.3 m

Moment of inertia about center: I_cm = ML²/12 = 3.0 × (1.2)²/12 = 0.36 kg·m²
Moment of inertia about pivot: I = I_cm + Md² = 0.36 + 3.0 × (0.3)² = 0.63 kg·m²

(i) At rest (g = 9.8 m/s²):
T = 2π√(I/Mgd) = 2π√(0.63/(3.0 × 9.8 × 0.3))
T = 2π√(0.63/8.82) = 2π√(0.0714) = 2π × 0.267 = 1.68 s

(ii) Accelerating upward (g_eff = 9.8 + 2.0 = 11.8 m/s²):
T = 2π√(0.63/(3.0 × 11.8 × 0.3)) = 2π√(0.63/10.62)
T = 2π√(0.0593) = 2π × 0.244 = 1.53 s

(b) With additional mass:

New center of mass location:
x_cm = (M × 0.6 + m × 1.2)/(M + m) = (3.0 × 0.6 + 0.5 × 1.2)/(3.5) = 0.686 m from original end
New d = 0.686 - 0.3 = 0.386 m

New moment of inertia:
I_new = I_rod + m × (1.2 - 0.3)² = 0.63 + 0.5 × (0.9)² = 1.035 kg·m²

New period:
T = 2π√(1.035/(3.5 × 9.8 × 0.386)) = 2π√(0.781) = 1.76 s

(c) Maximum angular velocity and acceleration:

θ₀ = 5° = 5π/180 = 0.0873 rad

For small oscillations: θ(t) = θ₀ cos(ωt)
ω = √(Mgd/I) = √(8.82/0.63) = 3.74 rad/s

Maximum angular velocity:
ω_max = θ₀ × ω = 0.0873 × 3.74 = 0.326 rad/s

Maximum angular acceleration:
α_max = θ₀ × ω² = 0.0873 × (3.74)² = 1.22 rad/s²

Question 4: Combined Heat Engine and Oscillation

A Carnot heat engine operates between reservoirs at 500 K and 300 K. The engine drives a compressor that compresses a spring with spring constant k = 2000 N/m. The compressed spring then drives a 1.0 kg mass in simple harmonic motion.

(a) If the heat engine absorbs 1000 J from the hot reservoir per cycle, calculate:

(i) The efficiency
(ii) Work output per cycle
(iii) Heat rejected to cold reservoir

(b) The work output compresses the spring by 0.20 m from its natural length. When released, this drives the 1.0 kg mass. Find:

(i) The amplitude of oscillation
(ii) The frequency of oscillation
(iii) The maximum speed of the mass

(c) If 10% of the kinetic energy is lost to friction during each oscillation, how many complete oscillations will occur before the amplitude drops to half its initial value?

Solution 4:

(a) Carnot engine analysis:

Given: T_H = 500 K, T_C = 300 K, Q_H = 1000 J

(i) Efficiency:
η = 1 - T_C/T_H = 1 - 300/500 = 0.4 = 40%

(ii) Work output:
W = η × Q_H = 0.4 × 1000 = 400 J

(iii) Heat rejected:
Q_C = Q_H - W = 1000 - 400 = 600 J

(b) Spring-mass system:

Given: k = 2000 N/m, m = 1.0 kg, compression = 0.20 m
Work stored in spring = 400 J (from engine)

(i) Amplitude:
Initial potential energy = ½kx² = ½ × 2000 × (0.20)² = 40 J
But total energy available = 400 J
So: ½kA² = 400 J
A² = 800/2000 = 0.4
A = 0.632 m

(ii) Frequency:
ω = √(k/m) = √(2000/1.0) = 44.7 rad/s
f = ω/(2π) = 44.7/(2π) = 7.11 Hz

(iii) Maximum speed:
v_max = Aω = 0.632 × 44.7 = 28.3 m/s

(c) Energy loss analysis:

Initial energy: E₀ = 400 J
After each cycle: E_new = 0.9 × E_old (10% loss)

Energy is proportional to A²:
E ∝ A²

For amplitude to drop to A/2:
Final energy = E₀/4 = 100 J

Using: E_n = E₀ × (0.9)ⁿ
100 = 400 × (0.9)ⁿ
0.25 = (0.9)ⁿ
ln(0.25) = n × ln(0.9)
n = ln(0.25)/ln(0.9) = -1.386/(-0.105) = 13.2

Therefore, approximately 13 complete oscillations.

Question 5: Coupled Oscillations and Thermodynamics

A system consists of a simple pendulum (length L = 1.5 m, mass m = 2.0 kg) attached to a piston that compresses an ideal gas in a cylinder. The gas contains 0.5 moles with γ = 1.4, initially at 300 K and 2.0 atm pressure, occupying 3.0 L volume.

(a) For the pendulum oscillations:

(i) Calculate the period of small oscillations in Earth's gravity
(ii) Find the maximum speed if the pendulum is displaced 10° from vertical
(iii) Calculate the total mechanical energy of the pendulum

(b) The gas undergoes a thermodynamic cycle as the pendulum oscillates:

(i) If the gas expands isothermally to 6.0 L, find the final pressure and work done
(ii) The gas then undergoes adiabatic expansion until the temperature drops to 250 K. Find the final volume
(iii) Calculate the efficiency if this system operates as a heat engine between these temperature limits

(c) The entire system is placed in an elevator accelerating upward at 4.0 m/s²:

(i) Find the new period of the pendulum
(ii) How does this acceleration affect the gas pressure if volume remains constant?
(iii) Calculate the new equilibrium position of the pendulum

(d) Advanced analysis:

(i) If the pendulum bob is replaced with a uniform rod of the same mass and length L, find the new period
(ii) Calculate the center of mass of the rod-pendulum system
(iii) Determine the moment of inertia about the pivot point

Solution 5:

(a) Pendulum oscillations:

Given: L = 1.5 m, m = 2.0 kg, θ₀ = 10° = 0.175 rad

(i) Period of small oscillations:
T = 2π√(L/g) = 2π√(1.5/9.8) = 2π√(0.153) = 2π × 0.391 = 2.46 s

(ii) Maximum speed:
For small oscillations: ω = √(g/L) = √(9.8/1.5) = 2.56 rad/s
v_max = θ₀ × ω × L = 0.175 × 2.56 × 1.5 = 0.672 m/s

(iii) Total mechanical energy:
E = ½mgh = ½mg(L - L cos θ₀) = ½mgL(1 - cos θ₀)
E = ½ × 2.0 × 9.8 × 1.5 × (1 - cos 10°) = 14.7 × (1 - 0.985) = 0.221 J

(b) Thermodynamic cycle:

Given: n = 0.5 mol, γ = 1.4, T₁ = 300 K, P₁ = 2.0 atm, V₁ = 3.0 L

(i) Isothermal expansion to V₂ = 6.0 L:
P₂ = P₁V₁/V₂ = 2.0 × 3.0/6.0 = 1.0 atm
W = nRT ln(V₂/V₁) = 0.5 × 8.314 × 300 × ln(2) = 865 J

(ii) Adiabatic expansion to T₃ = 250 K:
Using TV^(γ-1) = constant:
V₃ = V₂(T₂/T₃)^(1/(γ-1)) = 6.0 × (300/250)^(1/0.4) = 6.0 × (1.2)^2.5 = 8.79 L

(iii) Heat engine efficiency:
η_max = 1 - T_cold/T_hot = 1 - 250/300 = 0.167 = 16.7%

(c) In accelerating elevator (a = 4.0 m/s²):

Effective gravity: g_eff = g + a = 9.8 + 4.0 = 13.8 m/s²

(i) New period:
T_new = 2π√(L/g_eff) = 2π√(1.5/13.8) = 2π√(0.109) = 2.07 s

(ii) Gas pressure at constant volume:
The acceleration doesn't directly affect gas pressure if volume is fixed
and temperature remains constant. P remains ≈ 2.0 atm

(iii) New equilibrium position:
The pendulum equilibrium doesn't change - it still hangs vertically
relative to the effective gravity direction.

(d) Physical pendulum analysis:

(i) Rod pendulum (uniform rod, mass m = 2.0 kg, length L = 1.5 m):
Moment of inertia about end: I = ⅓mL² = ⅓ × 2.0 × (1.5)² = 1.5 kg·m²
Distance to center of mass: d = L/2 = 0.75 m
T = 2π√(I/mgd) = 2π√(1.5/(2.0×9.8×0.75)) = 2π√(0.102) = 2.01 s

(ii) Center of mass:
For uniform rod: Located at geometric center, L/2 = 0.75 m from pivot

(iii) Moment of inertia about pivot:
I = ⅓mL² = ⅓ × 2.0 × (1.5)² = 1.5 kg·m²

Tips for the Exam

Remember the signs: In Simple Harmonic Motion (SHM), acceleration is always opposite to displacement
Energy conservation: Total mechanical energy is constant in Simple Harmonic Motion (SHM)
Small angle approximation: For pendulums, sin θ ≈ θ when θ < 15°
Units: Always check units in your final answers
Graphs: Practice sketching x-t, v-t, and a-t graphs for Simple Harmonic Motion (SHM)
Thermodynamic processes: Remember Q = 0 for adiabatic, T = constant for isothermal
Carnot efficiency: It's the maximum possible efficiency between two reservoirs
Multi-part questions: Read all parts before starting - later parts often give clues for earlier ones
Show all work: Partial credit is often awarded for correct methodology
Check reasonableness: Do your answers make physical sense?