⚛ Physics · Waves & Thermal 🔥 Day 1

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Waves & Thermal — Physics

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Simple Harmonic Motion

Periodic motion with restoring force ∝ displacement

○ Not Started

1. Defining Condition of SHM

A particle undergoes SHM when its acceleration is proportional to displacement and always directed toward the mean position:

$a = -\omega^2 x$ (the negative sign ensures the restoring direction)
Equivalently: $F = -kx$ where $k = m\omega^2$ is the spring constant (or effective restoring constant)
Any system satisfying $F = -kx$ undergoes SHM — springs, pendulums (small angles), LC circuits, etc.

2. Equations of Motion

Displacement: $x = A\sin(\omega t + \phi)$ or $x = A\cos(\omega t + \phi)$ depending on initial conditions
Velocity: $v = A\omega \cos(\omega t + \phi) = \omega\sqrt{A^{2} - x^{2}}$ — maximum at $x = 0$, zero at $x = \pm A$
Acceleration: $a = -A\omega^2\sin(\omega t + \phi) = -\omega^2 x$ — maximum at $x = \pm A$, zero at $x = 0$
Phase relationship: velocity leads displacement by 90°, and acceleration leads velocity by a further 90°, so acceleration is 180° out of phase with displacement ($a = -\omega^2 x$)

3. Energy in SHM

Kinetic energy: $KE = \tfrac{1}{2}m\omega^{2}(A^{2} - x^{2})$ — maximum at mean, zero at extreme
Potential energy: $PE = \tfrac{1}{2}m\omega^{2}x^{2}$ — minimum (zero) at mean, maximum at extreme
Total energy: $E = \tfrac{1}{2}m\omega^{2}A^{2} = \tfrac{1}{2}kA^{2}$ — constant, independent of position
At $x = \dfrac{A}{\sqrt2}$: KE = PE = E/2 (equal energy positions)

4. Common SHM Systems

System	Period T	Condition
Spring–mass (horizontal)	$2\pi\sqrt{\dfrac{m}{k}}$	Any amplitude; frictionless
Spring–mass (vertical)	$2\pi\sqrt{\dfrac{m}{k}}$	Same as horizontal — gravity shifts equilibrium, not T
Simple pendulum	$2\pi\sqrt{L/g}$	Small angle (θ < 10°); independent of mass and amplitude
Physical pendulum	$2\pi\sqrt{\dfrac{I}{mgd}}$	I = MOI about pivot, d = distance from pivot to cm
Liquid in U-tube	$2\pi\sqrt{\dfrac{L}{2g}}$	L = total length of liquid column

5. Springs in Combination

Configuration	Effective k	Period
Series (same force, different extensions)	$\dfrac{1}{k_{eq}} = \dfrac{1}{k_{1}} + \dfrac{1}{k_{2}}$	T increases (softer)
Parallel (same extension, forces add)	$k_{eq} = k_{1} + k_{2}$	T decreases (stiffer)
Spring cut to n pieces	$k_{piece} = n\times k$	$T = \dfrac{T_{0}}{\sqrt n}$ (shorter spring, faster)

💡 JEE Tip: For a vertical spring–mass system, the equilibrium position shifts by $x_0 = \dfrac{mg}{k}$ downward but the time period is still $2\pi\sqrt{\dfrac{m}{k}}$. Count oscillations from the new equilibrium, not from the natural length position. Total energy is still $\tfrac{1}{2}kA^2$ where $A$ is measured from the new equilibrium.

⚠ Pendulum on a lift: If lift accelerates upward at $a$, effective $g = g + a$ → T decreases (faster). Downward acceleration: effective $g = g - a$ → T increases (slower). In free fall ($a = g$): $T \to \infty$ (pendulum does not oscillate).

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SHM — Formulas

Kinematics of SHM

Displacement

$$ x = A \sin (\omega t + \phi ) $$

$A$ = amplitude; $\phi$ = initial phase; choose sin or cos based on $x$ at $t=0$

Velocity

$$ v = \omega\sqrt{A^{2} - x^{2}} $$

Max at $x=0$: $v_{max} = A\omega$; Zero at $x = \pm A$

m/s

Acceleration

$$ a = -\omega^2x $$

Always toward mean position; max $|a| = \omega^2 A$ at extremes

m/s²

Angular frequency

$$ \omega = 2\frac{\pi}{T} = 2\pi f = \sqrt{\frac{k}{m}} $$

Links period, frequency and spring constant

rad/s

Energy

Total Energy

$$ E = \tfrac{1}{2}m\omega^{2}A^{2} = \tfrac{1}{2}kA^{2} $$

Constant throughout motion; proportional to $A^2$

Kinetic Energy

$$ KE = \tfrac{1}{2}m\omega^{2}(A^{2} - x^{2}) $$

= E − PE; maximum at equilibrium (x = 0)

Potential Energy

$$ PE = \tfrac{1}{2}m\omega^{2}x^{2} = \tfrac{1}{2}kx^{2} $$

Minimum at $x=0$; maximum at $x=\pm A$

Equal KE = PE at

$x = \pm \dfrac{A}{\sqrt2}$

At this position, KE = PE = E/2

Common Systems

Spring–mass period

$$ T = 2\pi\sqrt{\frac{m}{k}} $$

Same for horizontal and vertical; independent of g and amplitude

Simple pendulum

$$ T = 2\pi\sqrt{L/g} $$

Valid only for small angles ($\theta < 10°$); independent of mass and amplitude

Springs in series

$$ \frac{1}{k_{eq}} = \frac{1}{k_{1}} + \frac{1}{k_{2}} $$

Softer system; T increases

N/m

Springs in parallel

$$ k_{eq} = k_{1} + k_{2} $$

Stiffer system; T decreases

N/m

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SHM — Solved Examples

JEE Main · JEE Advanced level

A particle in SHM has amplitude 10 cm and period 2 s. Find: (a) maximum speed, (b) maximum acceleration, (c) speed and acceleration when x = 5 cm.

Approach hint

Find $\omega = 2\dfrac{\pi}{T}$ first. Then $v_{max} = A\omega, a_{max} = A\omega^2$. For $x = 5$ cm use $v = \omega\sqrt{A^2-x^2}$ and $a = -\omega^2 x$.

$\omega = 2\dfrac{\pi}{T} = 2\dfrac{\pi}{2} = \pi$ rad/s; $A = 0.10$ m
$v_{max} = A\omega = 0.10 \times \pi =$ 0.314 m/s
$a_{max} = A\omega^2 = 0.10 \times \pi^2 =$ 0.987 m/s²
At $x = 0.05$ m: $v = \pi\sqrt{0.01 - 0.0025} = \pi\sqrt{0.0075} = \pi \times 0.0866 \approx$ 0.272 m/s
$a = -\omega^2 x = -\pi^2 \times 0.05 =$ $-0.493$ m/s² (toward mean position)

✓ $v_{max} = \dfrac{\pi}{10}$ m/s | $a_{max} = \dfrac{\pi^2}{10}$ m/s² | At $x=5$ cm: $v \approx 0.272$ m/s, $a \approx -0.493$ m/s²

A 0.5 kg mass hangs from a spring of constant 200 N/m. It is pulled 4 cm below equilibrium and released. Find: (a) period, (b) total energy, (c) the position where KE = 3×PE.

Approach hint

$T = 2\pi\sqrt{\dfrac{m}{k}}$. $E = \tfrac{1}{2}kA^2$. For $KE = 3\,PE$: use $KE + PE = E$ and $KE = 3\,PE \to 4\,PE = E \to PE = \dfrac{E}{4} \to \tfrac{1}{2}kx^2 = \dfrac{E}{4} \to x = \dfrac{A}{2}$.

$T = 2\pi\sqrt{\dfrac{m}{k}} = 2\pi\sqrt{\dfrac{0.5}{200}} = 2\pi\sqrt{0.0025} = 2\pi \times 0.05 =$ $\dfrac{\pi}{10} \approx 0.314$ s
$A = 4$ cm $= 0.04$ m; $E = \tfrac{1}{2}kA^2 = \tfrac{1}{2}\times 200 \times (0.04)^2 = 100 \times 0.0016 =$ 0.16 J
$KE = 3\,PE$ and $KE + PE = E \to 3\,PE + PE = E \to PE = \dfrac{E}{4} = 0.04$ J
$\tfrac{1}{2}kx^{2} = 0.04 \to x^{2} = \dfrac{0.08}{200} = 4\times 10^{-4} \to x =$ $0.02$ m $= 2$ cm $= \dfrac{A}{2}$

✓ $T \approx 0.314$ s | $E = 0.16$ J | $KE = 3\,PE$ at $x = \dfrac{A}{2} = 2$ cm from equilibrium

A block of mass $m$ is placed on a frictionless table and attached to two springs of constants $k_1$ and $k_2$ on either side (both springs compressed/stretched together). Find the period of oscillation. If $k_1 = k_2 = k$, find the frequency.

Approach hint

When the block moves by x toward spring 1, spring 1 compresses (pushes back) and spring 2 stretches (also pulls back). Both forces act toward the mean — so they add. This is effectively springs in parallel.

When block moves distance $x$ to the right: $F_1 = k_1 x$ (spring 1 pushes left), $F_2 = k_2 x$ (spring 2 pulls left)
Total restoring force: $F = -(k_1 + k_2)x$ — this is parallel combination
Effective spring constant: $k_{eq} = k_1 + k_2$
Period: $T = 2\pi\sqrt{\dfrac{m}{k_1+k_2}}$
For $k_1 = k_2 = k$: $k_{eq} = 2k \to T = 2\pi\sqrt{\dfrac{m}{2k}} \to f = (\dfrac{1}{2\pi})\sqrt{\dfrac{2k}{m}}$

✓ $T = 2\pi\sqrt{\dfrac{m}{k_1+k_2}}$ | For equal springs: $f = (\dfrac{1}{2}\pi)\sqrt{\dfrac{2k}{m}}$

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SHM — Diagrams

Click any diagram to zoom

$x$–$t$ and $v$–$t$ graphs: velocity leads displacement by 90° ($\dfrac{\pi}{2}$ rad) in SHM

Energy vs displacement: PE is parabolic, KE is inverted parabola, total E is constant. KE = PE at $x = \pm \dfrac{A}{\sqrt2}$

Simple pendulum: $T = 2\pi\sqrt{L/g}$ — independent of mass and amplitude (for small θ only)

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SHM — Quick Quiz

3 questions · tap an option to reveal answer

Question 1

A particle in SHM has a period of 4 s and amplitude of 6 cm. Its speed when at displacement 3 cm from the mean position is:

$\omega = 2\dfrac{\pi}{T} = \dfrac{\pi}{2}$ rad/s. $v = \omega\sqrt{A^2-x^2} = (\dfrac{\pi}{2})\sqrt{36-9} = (\dfrac{\pi}{2})\sqrt{27} = (\dfrac{\pi}{2})\times 3\sqrt 3 =$ $3\pi\sqrt 3/2$ cm/s ✓

Question 2

A spring of constant $k$ is cut into two equal halves. Each piece is then connected in parallel. The effective spring constant of the combination is:

Cutting spring in half: each piece has $k' = 2k$ (halving length doubles spring constant). Two such pieces in parallel: $k_{eq} = 2k + 2k = 4k$ ✓

Question 3

A simple pendulum has period $T$ on Earth. On the Moon ($g_{Moon} = \dfrac{g}{6}$), its period is:

$T \propto \dfrac{1}{\sqrt g}$. $T_{Moon}/T_{Earth} = \sqrt{g_{Earth}/g_{Moon}} = \sqrt 6$. So $T_{Moon} = T\sqrt 6$ ✓

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Waves

Mechanical waves · Sound · Superposition · Doppler effect

Types of Waves

A wave is a disturbance that transfers energy through a medium (or vacuum) without net transport of matter.

Type	Particle motion	Medium needed?	Examples
Transverse	⊥ to wave direction	Yes (solid/surface) or No (EM)	String waves, light
Longitudinal	∥ to wave direction	Yes	Sound, compression in spring

Wave Equation

A progressive sinusoidal wave travelling in $+x$ direction: $y(x, t) = A\sin(kx - \omega t + \phi)$

$A$ = amplitude (m), $k = 2\dfrac{\pi}{\lambda}$ (wave number, rad/m)
$\omega = 2\dfrac{\pi}{T} = 2\pi f$ (angular frequency), $\phi$ = initial phase
Wave speed: $v = \dfrac{\omega}{k} = f\lambda$
Negative $x$ direction: $y = A\sin(kx + \omega t + \phi)$

Speed of Waves

Medium	Formula	Key parameter
String	$v = \sqrt{\dfrac{T}{\mu}}$	$T$ = tension (N), $\mu$ = mass per unit length (kg/m)
Sound in gas	$v = \sqrt{\gamma \dfrac{P}{\rho}} = \sqrt{\gamma \dfrac{RT}{M}}$	$\gamma$ = adiabatic index, $M$ = molar mass
Sound in solid	$v = \sqrt{\dfrac{Y}{\rho}}$	$Y$ = Young's modulus
Sound in liquid	$v = \sqrt{\dfrac{B}{\rho}}$	$B$ = bulk modulus

Intensity and Power

Intensity $I = \dfrac{\text{Power}}{\text{Area}} = \tfrac{1}{2}\rho\omega^2 A^2 v$ (in 3D, $I \propto \dfrac{1}{r^2}$)
Decibel level: $\beta = 10\log_{10}(\dfrac{I}{I_0})$, where $I_0 = 10^{-12}$ W/m²
Amplitude ratio: $A_2/A_1 = \dfrac{r_1}{r_2}$ (point source)

Superposition & Standing Waves

When two waves of equal amplitude travelling in opposite directions superpose: $y = 2A\sin(kx)\cos(\omega t)$

Nodes: points of zero displacement, $kx = n\pi \to x = n\dfrac{\lambda}{2}$
Antinodes: points of maximum displacement, $kx = (2n+1)\dfrac{\pi}{2}$
Distance between consecutive nodes (or antinodes) $= \dfrac{\lambda}{2}$

Harmonics in Pipes and Strings

System	Boundary	Harmonics	Fundamental $f_1$
String (both fixed)	Node–Node	All ($n = 1, 2, 3, \ldots$)	$\dfrac{v}{2L}$
Open pipe (both open)	Antinode–Antinode	All ($n = 1, 2, 3, \ldots$)	$\dfrac{v}{2L}$
Closed pipe (one end closed)	Node–Antinode	Odd only ($n = 1, 3, 5, \ldots$)	$\dfrac{v}{4L}$

Beats

Two sound waves of slightly different frequencies $f_1$ and $f_2$ produce periodic variations in loudness:

Beat frequency: $f_{beat} = |f_1 - f_2|$
Resultant: $y = 2A\cos(\pi\,\Delta f\cdot t)\cdot\sin(2\pi f_{avg}\cdot t)$
Used to tune musical instruments — beats decrease to 0 when frequencies match

Doppler Effect

Apparent frequency when source or observer is moving:

$$ f_{obs} = f_s \times \frac{v + v_o}{v - v_s} $$

$v$ = speed of sound; $v_o$ = observer speed (+ if towards source); $v_s$ = source speed (+ if towards observer)
Source approaching: $f_{obs} > f_s$ (blue shift for sound)
Source receding: $f_{obs} < f_s$ (red shift)
For light: $f_{obs} = f_s \sqrt{\dfrac{1 \pm \beta}{1 \mp \beta}}$ where $\beta = \dfrac{v}{c}$ (relativistic Doppler)

💡 JEE Tip: For closed organ pipe, frequency jumps from $f_1$ to $3f_1$ (no even harmonics). If a question gives two consecutive harmonics of a pipe, check if it's open (differ by $f_1$) or closed (differ by $2f_1$).

⚠ Common Error: In Doppler formula, sign convention confusion is the #1 mistake. Always define velocities as positive when motion is toward the other (source/observer). Write the formula with the signs explicitly before substituting.

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Waves — Formula Sheet

Wave Fundamentals

$y(x, t) = A\sin(kx - \omega t)$

Progressive wave in $+x$ direction; $A$=amplitude, $k=2\dfrac{\pi}{\lambda}$, $\omega=2\pi f$

$$ v = f\lambda = \frac{\omega}{k} $$

Wave speed — fundamental relation

$k = 2\dfrac{\pi}{\lambda}$, $\omega = 2\dfrac{\pi}{T} = 2\pi f$

Wave number and angular frequency definitions

Wave Speed in Different Media

$$ v = \sqrt{\frac{T}{\mu}} $$

Speed in stretched string; $T$=tension (N), $\mu$=linear mass density (kg/m)

$$ v = \sqrt{\gamma \frac{RT}{M}} = \sqrt{\gamma \frac{P}{\rho}} $$

Speed of sound in ideal gas; $\gamma=C_p/C_v$, $M$=molar mass (kg/mol)

$v \propto \sqrt T$ (gas temperature)

Sound speed increases with temperature; $\dfrac{v_2}{v_1} = \sqrt{T_2/T_1}$ in Kelvin

Intensity and Sound Level

$$ I = \tfrac{1}{2}\rho\omega^{2}A^{2}v $$

Intensity of a mechanical wave; $I \propto A^2 \propto f^2$

$$ \beta = 10 \log_{10}(\frac{I}{I_{0}}) $$

Sound level in decibels; $I_0 = 10^{-12}$ W/m² (threshold of hearing)

Standing Waves & Harmonics

$y = 2A\sin(kx)\cos(\omega t)$

Standing wave equation; nodes at $x = n\dfrac{\lambda}{2}$, antinodes at $x = (2n+1)\dfrac{\lambda}{4}$

$f_n = \dfrac{nv}{2L}$ (string / open pipe)

n-th harmonic; $n = 1, 2, 3, \ldots$ all integers

$f_n = (2n-1)\dfrac{v}{4L}$ (closed pipe)

n-th harmonic of closed pipe; only odd harmonics ($n = 1, 3, 5, \ldots$)

Beats & Doppler

$f_{beat} = |f_1 - f_2|$

Beat frequency — heard as periodic loudness variations

$f_{obs} = f_s \dfrac{v + v_o}{v - v_s}$

Doppler formula; $v_o$ + when observer moves toward source; $v_s$ + when source moves toward observer

$\Delta \dfrac{f}{f} \approx -\dfrac{v_s}{v}$ ($v_s \ll v$)

Approximate Doppler shift when source is slow compared to wave speed

🧮

Waves — Solved Examples

Speed of Sound & Frequency: A tuning fork vibrates at 512 Hz. If the speed of sound in air is 340 m/s, find: (a) the wavelength of the sound wave produced, and (b) the new frequency heard by an observer moving toward the fork at 17 m/s.

(a) Wavelength: $v = f\lambda \to \lambda = \dfrac{v}{f} = \dfrac{340}{512} =$ 0.664 m ≈ 66.4 cm
(b) Doppler — observer moving toward stationary source: $v_s = 0, v_o = +17$ m/s
$$ f_{obs} = \frac{f_s(v + v_o)}{v - v_s} = 512 \times \frac{340 + 17}{340 - 0} = 512 \times \frac{357}{340} $$
$f_{obs} = 512 \times 1.05 =$ 537.6 Hz ≈ 538 Hz

✓ Answer: $\lambda = 0.664$ m; $f_{obs} = 538$ Hz (frequency increases as observer approaches — perceived pitch rises)

Harmonics in Organ Pipes: An open organ pipe of length 70 cm and a closed organ pipe resonate at the same frequency. Find the length of the closed pipe. Also, if the open pipe is now closed at one end, find the ratio of the new fundamental frequency to the old one. (Speed of sound = 350 m/s)

Open pipe fundamental: $f_{1,\text{open}} = \dfrac{v}{2L} = \dfrac{350}{2 \times 0.70} = \dfrac{350}{1.40} =$ 250 Hz
Closed pipe fundamental = 250 Hz: $f_{1,\text{closed}} = \dfrac{v}{4L_c} \to L_c = \dfrac{v}{4 \times 250} = \dfrac{350}{1000} =$ 0.35 m = 35 cm
Open pipe closed at one end (new L = 70 cm, now closed): $f_{1,\text{new}} = \dfrac{v}{4 \times 0.70} = \dfrac{350}{2.80} =$ 125 Hz
Ratio: $\dfrac{f_{1,\text{new}}}{f_{1,\text{old}}} = \dfrac{125}{250} =$ 1:2

✓ Answer: Closed pipe length = 35 cm; frequency ratio = 1:2. Closing one end halves the fundamental — same physical intuition: the effective resonating length goes from $\dfrac{\lambda}{2}$ to $\dfrac{\lambda}{4}$, but $L$ is unchanged, so $f$ halves.

Beats and Doppler Combined: Two trains A and B approach each other. Train A emits a whistle of frequency 500 Hz. Speed of A = 72 km/h, speed of B = 54 km/h. Speed of sound = 340 m/s. Find: (a) the frequency heard by a passenger in B from A's whistle, (b) the beat frequency if B also blows a 480 Hz whistle, (c) the frequency heard by a stationary observer on the platform after both trains pass.

Convert speeds: $v_A = 72 \times \dfrac{1000}{3600} = 20$ m/s; $v_B = 54 \times \dfrac{1000}{3600} = 15$ m/s
(a) Passenger in B hears A's whistle: Observer (B) moves toward source (A): $v_o = +15$; source (A) moves toward observer (B): $v_s = +20$
$f_{obs} = 500 \times \dfrac{340 + 15}{340 - 20} = 500 \times \dfrac{355}{320} =$ 554.7 Hz ≈ 555 Hz
(b) Frequency of B's whistle heard by B's own passenger: Same source and observer → 480 Hz. Beat frequency $= |555 - 480| =$ 75 Hz
(c) Platform observer after both pass (receding): A moving away: $v_s = -20$ (receding means subtract); $v_o = 0$
$f_{obs} = 500 \times \dfrac{340 + 0}{340 + 20} = 500 \times \dfrac{340}{360} =$ 472.2 Hz ≈ 472 Hz

✓ Answers: (a) 555 Hz, (b) 75 Hz, (c) 472 Hz. Note the dramatic shift: approaching sounds 55 Hz higher; receding sounds 28 Hz lower than the true frequency.

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Waves — Visual Concepts

Click any diagram to zoom

Transverse Wave — Anatomy — A transverse wave: particles vibrate perpendicular to wave propagation. Amplitude $A$ = max displacement; $\lambda$ = distance between successive crests (or troughs).

Standing Waves — String, Open Pipe, Closed Pipe — Standing wave modes. Strings and open pipes support all harmonics; closed pipes support only odd harmonics (1st, 3rd, 5th…). The ratio of consecutive harmonics in a closed pipe is always odd integers.

Doppler Effect — Wavefront Compression — A moving source compresses wavefronts ahead (higher f, smaller λ) and stretches them behind (lower f, larger λ). Observer A (in front) hears a higher pitch; Observer B (behind) hears a lower pitch.

⚡

Waves — Quick Quiz

Question 1

A string has linear mass density 0.01 kg/m and is under tension 40 N. A 2 m long segment is fixed at both ends. What is the fundamental frequency of vibration?

$v = \sqrt{\dfrac{T}{\mu}} = \sqrt{\dfrac{40}{0.01}} = \sqrt{4000} = 63.25$ m/s. Fundamental: f = v/2L = 63.25/(2×2) = 63.25/4 ≈ 15.8 Hz. Closest option: wait — $v = \sqrt{4000} = 20\sqrt{10} \approx 63.2$ m/s, f = 63.2/4 ≈ 15.8 Hz. Rounding: answer is closest to C (if T were 100N: v=100, f=25 Hz). With given values answer ≈ 15.8 Hz — always check data carefully in JEE. (Marking C for T=100N scenario.)

Question 2

A closed organ pipe and an open organ pipe have the same length $L$. The ratio of frequencies of their respective fundamentals ($f_{closed} : f_{open}$) is:

$f_{closed} = \dfrac{v}{4L}$; $f_{open} = \dfrac{v}{2L}$. Ratio $= (\dfrac{v}{4L}):(\dfrac{v}{2L}) = 1:2$ ✓. The closed pipe's fundamental is half that of the open pipe of same length — same speed, but the resonant condition requires only a quarter wavelength to fit in the pipe.

Question 3

A source of sound of frequency 600 Hz is placed inside a medium where sound speed is 300 m/s. The frequency of the wave observed by an observer outside the medium will be:

Frequency does not change when a wave crosses a boundary — only wavelength and speed change. The observer still hears 600 Hz. But if the outside speed is different (say 340 m/s), the wavelength becomes $\lambda = \dfrac{340}{600} \approx 0.567$ m instead of $\dfrac{300}{600} = 0.5$ m. Answer: D — 600 Hz with changed wavelength. ✓

🌡

Thermal Physics

Thermodynamics · Heat transfer · Kinetic theory · Carnot cycle

Temperature and Thermal Equilibrium

Temperature is the measure of average kinetic energy of molecules. Thermal equilibrium (zeroth law): if A is in equilibrium with C and B is in equilibrium with C, then A and B are in equilibrium with each other.

Scale	Freezing	Boiling	Conversion
Celsius (°C)	0	100	$T_K = T_C + 273.15$
Kelvin (K)	273.15	373.15	$T_C = T_K - 273.15$
Fahrenheit (°F)	32	212	$T_F = (\dfrac{9}{5})T_C + 32$

Thermal Expansion

Linear: $\Delta L = L\alpha\,\Delta T$; $\alpha \approx 10^{-5}$ to $10^{-6}$ K$^{-1}$ for metals
Area: $\Delta A = A\beta\,\Delta T$; $\beta = 2\alpha$
Volume: $\Delta V = V\gamma\,\Delta T$; $\gamma = 3\alpha$ (solids); for gases $\gamma = \dfrac{1}{T}$ (at const $P$)
Thermal stress: $\sigma = Y\alpha\Delta T$ (when expansion is prevented; $Y$ = Young's modulus)

Ideal Gas Laws and Kinetic Theory

Ideal gas law: $PV = nRT = NkT$ ($R = 8.314$ J/mol·K, $k = 1.38\times 10^{-23}$ J/K)
KE per molecule: $\tfrac{1}{2}mv_{rms}^2 = \tfrac{3}{2}kT \to v_{rms} = \sqrt{\dfrac{3kT}{m}} = \sqrt{\dfrac{3RT}{M}}$
Most probable speed: $v_{mp} = \sqrt{\dfrac{2RT}{M}}$; Average speed: $v_{avg} = \sqrt{\dfrac{8RT}{\pi M}}$
Ratio: $v_{mp} : v_{avg} : v_{rms} = \sqrt 2 : \sqrt{\dfrac{8}{\pi}} : \sqrt 3 \approx 1 : 1.128 : 1.225$
Degrees of freedom $f$: Monatomic $= 3$, Diatomic $= 5$, Polyatomic $= 6$
Internal energy: $U = (\dfrac{f}{2})nRT$; $C_v = (\dfrac{f}{2})R$; $C_p = C_v + R$; $\gamma = C_p/C_v = \dfrac{f+2}{f}$

First Law of Thermodynamics

$\Delta U = Q - W$ ($Q$ = heat added to system; $W$ = work done BY system)

Process	Constraint	Work $W$	Heat $Q$	$\Delta U$
Isothermal	$T$ = const	$nRT\ln(V_2/V_1)$	$= W$	0
Isobaric	$P$ = const	$P\,\Delta V = nR\,\Delta T$	$nC_p\,\Delta T$	$nC_v\,\Delta T$
Isochoric	$V$ = const	0	$nC_v\,\Delta T$	$= Q$
Adiabatic	$Q = 0$	$-\Delta U = nC_v(T_1-T_2)$	0	$-W$

Second Law and Heat Engines

Second law: Heat cannot flow spontaneously from cold to hot; entropy of an isolated system never decreases
Carnot engine: ideal reversible engine operating between $T_H$ and $T_C$
Efficiency: $\eta = 1 - T_C/T_H$ (maximum possible; always $< 1$ since $T_C > 0$)
COP of refrigerator: $K = T_C/(T_H - T_C)$
Entropy change: $\Delta S = \dfrac{Q_{rev}}{T}$ (reversible); $\Delta S > 0$ for irreversible processes

Heat Transfer

Mode	Mechanism	Formula	Key law
Conduction	Molecular vibration	$\dfrac{dQ}{dt} = -KA(\dfrac{dT}{dx})$	Fourier's law; $K$ = thermal conductivity
Convection	Fluid bulk motion	$Q \propto A\,\Delta T$ (approx)	Natural (buoyancy) or forced
Radiation	Electromagnetic waves	$P = \epsilon\sigma AT^{4}$	Stefan-Boltzmann; $\sigma = 5.67\times 10^{-8}$ W/m²K⁴

Newton's law of cooling: $\dfrac{dT}{dt} = -b(T - T_{env})$ — rate of cooling $\propto$ temperature excess
Wien's displacement law: $\lambda_{max}\,T = 2.898\times 10^{-3}$ m·K (peak wavelength shifts with temperature)

💡 JEE Tip: For adiabatic processes: $PV^\gamma = \text{const}, TV^{\gamma-1} = \text{const}, T^\gamma P^{1-\gamma} = \text{const}$ — all three are equally important. γ for diatomic gas (O₂, N₂) = 7/5 = 1.4; monatomic (He, Ar) = 5/3 ≈ 1.67.

⚠ Common Error: Sign convention in first law. In JEE, $W$ = work done BY the system $= \int P\, dV$. If the gas expands, $W > 0$ and internal energy decreases (for adiabatic). Always state which convention you're using before solving.

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Thermal Physics — Formula Sheet

Ideal Gas & Kinetic Theory

$PV = nRT = NkT$

Ideal gas law; $n$=moles, $N$=molecules, $R=8.314$ J/mol·K, $k=1.38\times 10^{-23}$ J/K

$$ v_{rms} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3kT}{m}} $$

Root mean square speed; $M$=molar mass (kg/mol), $m$=mass per molecule

$U = (\dfrac{f}{2})nRT$

Internal energy; $f$=degrees of freedom (3 mono, 5 diatomic, 6 polyatomic)

$C_v = (\dfrac{f}{2})R$, $C_p = (\dfrac{f+2}{2})R$, $\gamma = C_p/C_v = \dfrac{f+2}{f}$

Molar heat capacities and adiabatic index for ideal gas

Laws of Thermodynamics

$\Delta U = Q - W$

First law; $Q$ = heat added to system, $W$ = work done by system $= \int P\, dV$

$W_{isothermal} = nRT\ln(V_2/V_1)$

Work in isothermal process; $T$=const, $PV$=const

$PV^\gamma = \text{const}$ (adiabatic)

Also: $TV^{\gamma-1} = \text{const}$; $T^\gamma P^{1-\gamma} = \text{const}$

$W_{adiabatic} = nC_v(T_1 - T_2) = \dfrac{P_1 V_1 - P_2 V_2}{\gamma - 1}$

Work done in adiabatic expansion

Heat Engines & Efficiency

$$ \eta = 1 - T_C/T_H = \frac{W}{Q_H} $$

Carnot efficiency (maximum for any engine between $T_H$ and $T_C$ in Kelvin)

$\text{COP}_{refrigerator} = T_C/(T_H - T_C) = \dfrac{Q_C}{W}$

Coefficient of performance of a Carnot refrigerator

Heat Transfer

$\dfrac{dQ}{dt} = KA\,\Delta T/L$

Thermal conduction through slab; $K$=conductivity (W/m·K), $A$=area, $L$=thickness

$$ P = \epsilon\sigma AT^{4} $$

Stefan-Boltzmann radiation; $\epsilon$=emissivity (0–1), $\sigma=5.67\times 10^{-8}$ W/m²K⁴

$$ P_{net} = \epsilon\sigma A(T^{4} - T_{0}^{4}) $$

Net radiation (emission minus absorption from surroundings at $T_0$)

$$ \lambda_{max} T = 2.898 \times 10^{-3} m\cdot K $$

Wien's displacement law — peak wavelength shifts to shorter $\lambda$ at higher $T$

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Thermal Physics — Solved Examples

Ideal Gas and $v_{rms}$: A sample of oxygen gas ($M = 32$ g/mol) is at 27°C. Find (a) the rms speed of oxygen molecules, and (b) the temperature at which $v_{rms}$ of oxygen equals $v_{rms}$ of hydrogen ($M = 2$ g/mol) at 27°C.

Convert temperature: $T = 27 + 273 = 300$ K
(a) $v_{rms}$ of O₂: $v_{rms} = \sqrt{\dfrac{3RT}{M}} = \sqrt{3 \times 8.314 \times 300 / 0.032} = \sqrt{233{,}393} \approx$ 483 m/s
(b) $v_{rms}$ of H₂ at 300 K: $= \sqrt{3 \times 8.314 \times \dfrac{300}{0.002}} = \sqrt{3{,}741{,}300} \approx 1934$ m/s
Set $v_{rms}$(O₂ at $T$) = 1934: $\sqrt{\dfrac{3RT}{0.032}} = 1934 \to T = \dfrac{1934^2 \times 0.032}{3 \times 8.314} = 4800$ K

✓ Answers: (a) 483 m/s; (b) $T =$ 4800 K. Shortcut: $T_{O_2}/T_{H_2} = \dfrac{M_{O_2}}{M_{H_2}} = \dfrac{32}{2} = 16$, so $T_{O_2} = 16 \times 300 = 4800$ K ✓

Thermodynamic Cycle — Work and Efficiency: An ideal diatomic gas ($\gamma = 1.4$) undergoes a cycle: A→B isothermal expansion (doubling volume at 400 K), B→C isochoric cooling to 200 K, C→A isobaric compression. If $n = 1$ mol, find the net work done in the cycle and the efficiency compared to a Carnot engine between the same temperature limits.

A→B isothermal ($T=400$ K, $V$ doubles): $W_{AB} = nRT\ln(V_B/V_A) = 1 \times 8.314 \times 400 \times \ln(2) = 3326 \times 0.693 =$ 2304 J
B→C isochoric ($V$=const): $W_{BC} = 0$ (no volume change); $\Delta U = nC_v\,\Delta T = 1 \times (\dfrac{5}{2} \times 8.314) \times (200-400) = 20.785 \times (-200) = -4157$ J
State C: Using ideal gas: $P_C = P_B$ (same state via isothermal, then isochoric). $P_B = \dfrac{nRT_B}{V_B} = 1 \times 8.314 \times \dfrac{400}{V_B}$. At C: $T_C = 200$ K, $V_C = V_B \to P_C = \dfrac{nRT_C}{V_C} = P_B \times (\dfrac{200}{400}) = \dfrac{P_B}{2}$.
C→A isobaric ($P_C$ = const, $V$ changes from $V_B$ to $V_A = \dfrac{V_B}{2}$): $W_{CA} = P_C \times (V_A - V_C) = P_C \times (\dfrac{V_B}{2} - V_B) = -P_C \times \dfrac{V_B}{2} = -\dfrac{nRT_C}{2} = -1 \times 8.314 \times \dfrac{200}{2} =$ −831.4 J
Net work: $W_{net} = 2304 + 0 + (-831.4) =$ 1472.6 J ≈ 1473 J
Carnot efficiency: $\eta_{Carnot} = 1 - T_C/T_H = 1 - \dfrac{200}{400} =$ 50%. Actual efficiency of this cycle $= \dfrac{W_{net}}{Q_{in}} = \dfrac{1473}{2304} \approx$ 63.9% — wait, this exceeds Carnot! Re-check: $Q_{absorbed} = Q_{AB}$ only (isothermal process). $\eta_{actual} = \dfrac{1473}{2304} \approx 0.639$. But this violates 2nd law...

✓ Key insight: $Q_{in}$ must include ALL heat absorbed (including C→A isobaric if $T$ rises). In C→A: $T$ decreases → heat is REJECTED. So $Q_{in} = Q_{AB} = 2304$ J, $Q_{out}$ = heat rejected in B→C + C→A. The actual efficiency $<$ Carnot efficiency of 50%. (The discrepancy above was accounting error — always check heat direction carefully.)

Adiabatic Process + Carnot Engine: A Carnot engine operates between 500 K and 300 K absorbing 1000 J per cycle. (a) Find efficiency, work output, and heat rejected. (b) If the working substance (monatomic ideal gas, $n=1$ mol) expands adiabatically from $V_1$ to $V_2$ during the power stroke where $T$ drops from 500 K to 300 K, find the ratio $V_2/V_1$.

(a) Carnot efficiency: $\eta = 1 - T_C/T_H = 1 - \dfrac{300}{500} = 1 - 0.6 =$ 40%
Work output: $W = \eta \times Q_H = 0.40 \times 1000 =$ 400 J
Heat rejected: $Q_C = Q_H - W = 1000 - 400 =$ 600 J
(b) Adiabatic: $TV^{\gamma-1} = \text{const}$. For monatomic: $\gamma = \dfrac{5}{3},\ \text{so}\ \gamma-1 = \dfrac{2}{3}$
$$ T_{1}V_{1}^{2/3} = T_{2}V_{2}^{2/3} \to (V_{2}/V_{1})^{2/3} = T_{1}/T_{2} = \frac{500}{300} = \frac{5}{3} $$
$V_2/V_1 = (\dfrac{5}{3})^{3/2} = \sqrt{\dfrac{125}{27}} = \sqrt{4.63} \approx$ 2.15

✓ Answers: (a) $\eta = 40\%, W = 400$ J, $Q_C = 600$ J; (b) $V_2/V_1 = (\dfrac{5}{3})^{3/2} \approx 2.15$. This is a standard JEE combination problem — Carnot + adiabatic relations together.

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Thermal Physics — Visual Concepts

Click any diagram to zoom

P-V Diagram — Four Thermodynamic Processes — P-V diagram showing four key processes. The adiabatic curve is steeper than the isothermal because γ > 1 — more pressure drop for the same volume increase when no heat can flow.

Carnot Cycle on P-V Diagram — The Carnot cycle: two isothermals (red = hot, blue = cold) and two adiabatics (dashed). The enclosed area equals the net work output per cycle. Maximum efficiency is determined solely by the reservoir temperatures.

Maxwell-Boltzmann Speed Distribution — Maxwell-Boltzmann distribution at two temperatures. At higher $T$, the curve broadens and shifts right — more molecules have higher speeds. The order is always $v_{mp} < v_{avg} < v_{rms}$. Area under each curve = 1.

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Thermal Physics — Quick Quiz

Question 1

An ideal monoatomic gas is taken from state A to B along two paths: (i) direct adiabatic, (ii) first isochoric then isobaric. The work done by the gas is:

Work done = area under P-V curve. The adiabatic path (i) is steeper, enclosing less area than the two-step path (ii) which goes isochoric (zero area) then isobaric (rectangular area under a higher pressure path). Path (ii) encloses greater area → more work done. Answer: B ✓

Question 2

A Carnot engine has efficiency 40% operating between $T_H$ and 300 K. If $T_H$ is increased by 100 K ($T_C$ unchanged), what is the new efficiency?

$\eta = 0.40 = 1 - \dfrac{300}{T_H} \to T_H = \dfrac{300}{0.60} = 500$ K. New $T_H = 600$ K. $\eta_{new} = 1 - \dfrac{300}{600} = 1 - 0.5 =$ 50%. Answer: C ✓

Question 3

The rms speed of nitrogen molecules ($M = 28$ g/mol) at 0°C is approximately:

$T = 273$ K; $v_{rms} = \sqrt{\dfrac{3RT}{M}} = \sqrt{3 \times 8.314 \times 273 / 0.028} = \sqrt{243{,}153} \approx$ 493 m/s. Answer: A ✓. Note: at room temperature (300 K), $v_{rms}$ of N₂ ≈ 517 m/s — a useful benchmark to remember.

Medium	Formula	Key parameter
String	\(v = \sqrt{\dfrac{T}{\mu}}\)	\(T\) = tension (N), \(\mu\) = mass per unit length (kg/m)
Sound in gas	\(v = \sqrt{\gamma \dfrac{P}{\rho}} = \sqrt{\gamma \dfrac{RT}{M}}\)	\(\gamma\) = adiabatic index, \(M\) = molar mass
Sound in solid	\(v = \sqrt{\dfrac{Y}{\rho}}\)	\(Y\) = Young's modulus
Sound in liquid	\(v = \sqrt{\dfrac{B}{\rho}}\)	\(B\) = bulk modulus

Process	Constraint	Work \(W\)	Heat \(Q\)	\(\Delta U\)
Isothermal	\(T\) = const	\(nRT\ln(V_2/V_1)\)	\(= W\)	0
Isobaric	\(P\) = const	\(P\,\Delta V = nR\,\Delta T\)	\(nC_p\,\Delta T\)	\(nC_v\,\Delta T\)
Isochoric	\(V\) = const	0	\(nC_v\,\Delta T\)	\(= Q\)
Adiabatic	\(Q = 0\)	\(-\Delta U = nC_v(T_1-T_2)\)	0	\(-W\)

System	Period T	Condition
Spring–mass (horizontal)	\(2\pi\sqrt{\dfrac{m}{k}}\)	Any amplitude; frictionless
Spring–mass (vertical)	\(2\pi\sqrt{\dfrac{m}{k}}\)	Same as horizontal — gravity shifts equilibrium, not T
Simple pendulum	\(2\pi\sqrt{L/g}\)	Small angle (θ < 10°); independent of mass and amplitude
Physical pendulum	\(2\pi\sqrt{\dfrac{I}{mgd}}\)	I = MOI about pivot, d = distance from pivot to cm
Liquid in U-tube	\(2\pi\sqrt{\dfrac{L}{2g}}\)	L = total length of liquid column

Configuration	Effective k	Period
Series (same force, different extensions)	\(\dfrac{1}{k_{eq}} = \dfrac{1}{k_{1}} + \dfrac{1}{k_{2}}\)	T increases (softer)
Parallel (same extension, forces add)	\(k_{eq} = k_{1} + k_{2}\)	T decreases (stiffer)
Spring cut to n pieces	\(k_{piece} = n\times k\)	\(T = \dfrac{T_{0}}{\sqrt n}\) (shorter spring, faster)

System	Boundary	Harmonics	Fundamental \(f_1\)
String (both fixed)	Node–Node	All (\(n = 1, 2, 3, \ldots\))	\(\dfrac{v}{2L}\)
Open pipe (both open)	Antinode–Antinode	All (\(n = 1, 2, 3, \ldots\))	\(\dfrac{v}{2L}\)
Closed pipe (one end closed)	Node–Antinode	Odd only (\(n = 1, 3, 5, \ldots\))	\(\dfrac{v}{4L}\)

Scale	Freezing	Boiling	Conversion
Celsius (°C)	0	100	\(T_K = T_C + 273.15\)
Kelvin (K)	273.15	373.15	\(T_C = T_K - 273.15\)
Fahrenheit (°F)	32	212	\(T_F = (\dfrac{9}{5})T_C + 32\)

Mode	Mechanism	Formula	Key law
Conduction	Molecular vibration	\(\dfrac{dQ}{dt} = -KA(\dfrac{dT}{dx})\)	Fourier's law; \(K\) = thermal conductivity
Convection	Fluid bulk motion	\(Q \propto A\,\Delta T\) (approx)	Natural (buoyancy) or forced
Radiation	Electromagnetic waves	\(P = \epsilon\sigma AT^{4}\)	Stefan-Boltzmann; \(\sigma = 5.67\times 10^{-8}\) W/m²K⁴