भौतिकशास्त्र Physics

थोडक्यात सारांश

The kinetic theory explains the behaviour of gases using the random motion of their molecules. From simple assumptions it gives the pressure of a gas, links temperature to molecular kinetic energy, and explains gas laws, degrees of freedom, mean free path and the Maxwell distribution of speeds.

महत्त्वाचे मुद्दे

• Assumptions for an ideal gas: molecules are tiny hard spheres; their size is negligible compared with distances between them; no forces except during collisions; collisions are perfectly elastic; motion is random; time of collision is negligible.
• Pressure is due to momentum transferred by molecules hitting the walls.
• Temperature is a measure of the average translational KE of molecules: average KE per molecule = (3/2)kT.
• The ideal gas equation PV = nRT follows from kinetic theory.
• Degrees of freedom (f): number of independent ways a molecule can store energy. Monatomic f = 3; rigid diatomic f = 5 (3 translational + 2 rotational).
• Law of equipartition of energy: each degree of freedom gets ½kT of energy on average.
• Mean free path λ: average distance a molecule travels between two collisions.
• Maxwell's distribution: molecular speeds are spread out; the curve rises, reaches a peak and falls; at higher temperature the peak moves to the right and flattens.
• Real gases deviate from ideal behaviour at high pressure and low temperature. Van der Waals' equation corrects for molecular size (b) and attraction (a).

महत्त्वाच्या संज्ञा

• Boltzmann constant k = R/N_A ≈ 1.38 × 10⁻²³ J K⁻¹.
• rms speed – square root of the mean of squares of molecular speeds.
• Most probable speed – speed possessed by the largest number of molecules.
• Avogadro number N_A ≈ 6.022 × 10²³ per mole.
• Ideal gas (आदर्श वायू) – a gas that obeys PV = nRT exactly.

सूत्रे

• Pressure: P = (1/3)ρc² = (1/3)(mN/V)c² (c = rms speed)
• PV = nRT = NkT
• Average KE per molecule = (3/2)kT; per mole = (3/2)RT
• c_rms = √(3RT/M) = √(3kT/m)
• c_avg = √(8RT/(πM)); c_mp = √(2RT/M); so c_mp < c_avg < c_rms
• Energy per mole = (f/2)RT; Cv = (f/2)R; Cp = Cv + R; γ = Cp/Cv = 1 + 2/f
• γ: monatomic 5/3 ≈ 1.67; diatomic 7/5 = 1.4
• Mean free path: λ = 1/(√2 π d² n) (d = molecular diameter, n = molecules per unit volume)
• Van der Waals: (P + a/V²)(V − b) = RT (for one mole)

५ महत्त्वाचे प्रश्न

Q1. State the law of equipartition of energy.

In thermal equilibrium, the total energy of a molecule is shared equally among all its degrees of freedom, each getting an average of ½kT.

Q2. Show that γ = 1.4 for a rigid diatomic gas.

For a diatomic gas f = 5. Cv = (5/2)R, Cp = Cv + R = (7/2)R. γ = Cp/Cv = 7/5 = 1.4.

Q3. Derive the expression for pressure exerted by a gas.

1. Cube of side l contains N molecules of mass m. A molecule has velocity components (cx, cy, cz).
2. On hitting wall perpendicular to x, momentum change = 2mcx.
3. Time between successive hits on same wall = 2l/cx, so force from one molecule = 2mcx ÷ (2l/cx) = mcx²/l.
4. Total force = (m/l) Σcx²; pressure = force/l² = (m/l³)Σcx².
5. By randomness Σcx² = (1/3)Σc² = (1/3)N c_rms².
6. P = (1/3)(mN/V)c_rms² = (1/3)ρc_rms².

Q4. Find the rms speed of oxygen molecules at 27 °C (M = 32 × 10⁻³ kg mol⁻¹, R = 8.31 J mol⁻¹ K⁻¹).

1. T = 27 + 273 = 300 K.
2. c_rms = √(3RT/M) = √(3 × 8.31 × 300/0.032).
3. 3 × 8.31 × 300 = 7479; 7479/0.032 = 233719.
4. c_rms = √233719 ≈ 483 m/s.

Q5. At what temperature will the rms speed of a gas be double its value at 27 °C?

1. c_rms ∝ √T, so c₂/c₁ = √(T₂/T₁).
2. 2 = √(T₂/300) → T₂/300 = 4 → T₂ = 1200 K.
3. In Celsius: 1200 − 273 = 927 °C.

परीक्षेसाठी टीप

Always convert °C to kelvin and molar mass from g to kg before using √(3RT/M). Remember the order c_mp < c_avg < c_rms — it is a favourite short question.

थोडक्यात सारांश

Real gases do not obey the ideal gas equation PV = RT at high pressure and low temperature. Van der Waals corrected the ideal gas equation for the finite size of molecules and the attraction between them. This topic covers the deviation of real gases, Andrews' experiment on CO₂, Van der Waals' equation, critical constants and the reduced equation of state.

महत्त्वाचे मुद्दे

• Ideal gas assumptions that fail for real gases: (1) molecules have no volume; (2) there are no intermolecular forces.
• Andrews' experiment (CO₂ isotherms): above 31.1 °C CO₂ cannot be liquefied by pressure alone; below it, the isotherm has a flat part where liquid and vapour coexist.
• Critical temperature Tc: the temperature above which a gas cannot be liquefied however large the pressure.
• Volume correction: molecules occupy space, so the free volume is (V − b). b ≈ 4 × the actual volume of the molecules.
• Pressure correction: attraction by inner molecules reduces the force with which molecules hit the wall, so the real pressure is P + a/V².
• Van der Waals equation (one mole): (P + a/V²)(V − b) = RT.
• At the critical point, the isotherm has a point of inflection: dP/dV = 0 and d²P/dV² = 0.
• Critical constants: Vc = 3b, Pc = a/27b², Tc = 8a/27Rb, and RTc/(PcVc) = 8/3 ≈ 2.67 (critical coefficient).
• Real gases give critical coefficients of about 3.4 to 3.7, so Van der Waals' equation is only approximately correct.
• Boyle temperature T_B = a/Rb: near it, a real gas behaves almost ideally over a wide range of pressure.

महत्त्वाच्या संज्ञा

• Critical pressure (Pc) – pressure needed to liquefy the gas at Tc.
• Critical volume (Vc) – volume of one mole at Tc and Pc.
• Gas vs vapour – a vapour is the gaseous state below its critical temperature (can be liquefied by pressure); a gas is above Tc.
• Reduced variables – Pr = P/Pc, Vr = V/Vc, Tr = T/Tc.
• Law of corresponding states – gases with the same reduced pressure and temperature have the same reduced volume.

सूत्रे

• Van der Waals: (P + a/V²)(V − b) = RT (a in Pa m⁶ mol⁻², b in m³ mol⁻¹)
• For n moles: (P + an²/V²)(V − nb) = nRT
• Vc = 3b; Pc = a/(27b²); Tc = 8a/(27Rb)
• a = 3PcVc², b = Vc/3; also a = 27R²Tc²/(64Pc), b = RTc/(8Pc)
• Critical coefficient RTc/(PcVc) = 8/3
• Reduced equation: (Pr + 3/Vr²)(3Vr − 1) = 8Tr
• Boyle temperature T_B = a/(Rb) = (27/8) Tc

५ महत्त्वाचे प्रश्न

Q1. Define critical temperature, critical pressure and critical volume.

Tc is the highest temperature at which a gas can be liquefied by pressure alone. Pc is the pressure needed to liquefy it at Tc. Vc is the volume of one mole at Tc and Pc. For CO₂, Tc ≈ 31.1 °C (304 K) and Pc ≈ 73 atm.

Q2. Explain the two corrections made by Van der Waals.

1. Volume: molecules have finite size, so the space available to move is V − b, not V.
2. Pressure: a molecule about to hit the wall is pulled back by the other molecules. The reduction is proportional to (number striking) × (number pulling) ∝ ρ² ∝ 1/V², so the ideal pressure is P + a/V².

Q3. Derive the critical constants in terms of a and b.

1. Write P = RT/(V − b) − a/V².
2. At the critical point dP/dV = 0: RTc/(Vc − b)² = 2a/Vc³ … (i)
3. d²P/dV² = 0: 2RTc/(Vc − b)³ = 6a/Vc⁴ … (ii)
4. Dividing (i) by (ii): (Vc − b)/2 = Vc/3, so Vc = 3b.
5. Put in (i): RTc = 2a(2b)²/(27b³) = 8a/(27b), so Tc = 8a/(27Rb).
6. Put in P: Pc = R(8a/27Rb)/(2b) − a/(9b²) = 4a/(27b²) − 3a/(27b²) = a/(27b²).
7. Then RTc/(PcVc) = (8a/27b) ÷ (3b × a/27b²) = 8/3.

Q4. For CO₂, a = 0.364 Pa m⁶ mol⁻² and b = 4.27 × 10⁻⁵ m³ mol⁻¹. Find Tc (R = 8.314 J mol⁻¹ K⁻¹).

Tc = 8a/(27Rb) = (8 × 0.364)/(27 × 8.314 × 4.27 × 10⁻⁵) = 2.912/(9.585 × 10⁻³) ≈ 304 K, which matches the measured 31 °C.

Q5. Describe Andrews' experiment and its result.

CO₂ was compressed in a capillary tube at fixed temperatures, with air in a similar tube as a pressure gauge. Isotherms (P vs V) were plotted. At 13.1 °C and 21.5 °C the isotherm shows a gas region, a horizontal part (liquefaction at constant pressure) and a steep liquid part. The horizontal part shrinks as temperature rises and becomes a point at 31.1 °C — the critical point. At 48.1 °C the curve is close to a rectangular hyperbola (nearly ideal). Result: every gas has a critical temperature above which it cannot be liquefied by pressure alone.

परीक्षेसाठी टीप

Learn the three critical constants and the 8/3 ratio by heart — they come every year. Always state that the experimental ratio (about 3.5) differs, showing the limits of the equation.

थोडक्यात सारांश

Thermodynamics studies heat, work and internal energy and the rules that connect them. The zeroth law defines temperature, the first law is energy conservation, and the second law sets the direction of natural processes and limits engine efficiency. Entropy and the Carnot cycle are central ideas.

महत्त्वाचे मुद्दे

• Zeroth law: if A and B are each in thermal equilibrium with C, then A and B are in equilibrium with each other. This defines temperature.
• First law: heat given to a system = increase in internal energy + work done by the system: dQ = dU + dW.
• Internal energy U of an ideal gas depends only on temperature. U is a state function; Q and W depend on the path.
• Isothermal process: T constant, ΔU = 0, so Q = W.
• Adiabatic process: no heat exchange, Q = 0, so W = −ΔU; PVᵞ = constant.
• Isobaric (P constant) and isochoric (V constant, W = 0) processes.
• Second law (Kelvin–Planck): no engine working in a cycle can turn all the heat taken from a source into work.
• Second law (Clausius): heat cannot flow by itself from a colder body to a hotter body.
• Carnot engine: ideal reversible engine with two isothermal and two adiabatic steps. No engine working between the same two temperatures can be more efficient (Carnot's theorem).
• Entropy (एन्ट्रॉपी) measures disorder. In a reversible cycle total entropy change is zero; in an irreversible process the entropy of the universe increases.
• Third law: entropy of a perfect crystal approaches zero as T → 0 K.

महत्त्वाच्या संज्ञा

• System and surroundings – the part studied and everything outside it.
• Reversible process – can be retraced exactly, passing through equilibrium states.
• Heat engine – converts heat into work in a cycle; a refrigerator does the reverse using work.
• Coefficient of performance (COP) – heat removed from cold body ÷ work done (refrigerator).

सूत्रे

• dQ = dU + dW; dW = P dV
• Isothermal work (ideal gas): W = nRT ln(V₂/V₁) = 2.303 nRT log₁₀(V₂/V₁)
• Adiabatic: PVᵞ = const; TVᵞ⁻¹ = const; W = (P₁V₁ − P₂V₂)/(γ − 1)
• Mayer's relation: Cp − Cv = R
• Efficiency η = W/Q₁ = 1 − Q₂/Q₁; Carnot: η = 1 − T₂/T₁ (kelvin)
• Refrigerator COP β = Q₂/W = T₂/(T₁ − T₂)
• Entropy change: dS = dQ/T (reversible); for melting ΔS = mL/T

५ महत्त्वाचे प्रश्न

Q1. State the Kelvin–Planck and Clausius statements of the second law.

Kelvin–Planck: it is impossible to build an engine that works in a cycle and converts all heat taken from a source into work with no other effect. Clausius: it is impossible for heat to flow from a colder body to a hotter body without external work.

Q2. Why is ΔU = 0 in an isothermal process of an ideal gas?

The internal energy of an ideal gas depends only on temperature. In an isothermal process T does not change, so U does not change and ΔU = 0. Hence all heat supplied is used as work.

Q3. Derive the work done in an isothermal expansion of an ideal gas.

1. For n moles at temperature T: P = nRT/V.
2. Work dW = P dV = (nRT/V) dV.
3. W = ∫ from V₁ to V₂ of nRT dV/V = nRT ln(V₂/V₁).
4. Since P₁V₁ = P₂V₂, W = nRT ln(P₁/P₂).
5. In log base 10: W = 2.303 nRT log₁₀(V₂/V₁).

Q4. A Carnot engine works between 527 °C and 127 °C. Find its efficiency. If it takes 1000 J from the source per cycle, how much work does it do?

1. T₁ = 527 + 273 = 800 K; T₂ = 127 + 273 = 400 K.
2. η = 1 − T₂/T₁ = 1 − 400/800 = 0.5 = 50%.
3. W = ηQ₁ = 0.5 × 1000 = 500 J.
4. Heat rejected Q₂ = 1000 − 500 = 500 J.

Q5. Find the entropy change when 1 kg of ice melts at 0 °C (latent heat L = 3.36 × 10⁵ J kg⁻¹).

1. Melting is at constant temperature T = 273 K.
2. Q = mL = 1 × 3.36 × 10⁵ J.
3. ΔS = Q/T = 3.36 × 10⁵/273 ≈ 1231 J K⁻¹.
4. Entropy increases because liquid water is more disordered than ice.

परीक्षेसाठी टीप

Always use kelvin in efficiency, COP and entropy formulas. Learn the four Carnot steps in order (isothermal expansion, adiabatic expansion, isothermal compression, adiabatic compression) with a labelled P–V diagram.

थोडक्यात सारांश

Maxwell's thermodynamic relations connect quantities that are easy to measure (P, V, T) with ones that are hard to measure (entropy S). They come from the four thermodynamic potentials. This topic covers the potentials, the four Maxwell relations, the TdS equations, the Clausius–Clapeyron equation and the Joule–Thomson (throttling) effect used to cool and liquefy gases.

महत्त्वाचे मुद्दे

• The four thermodynamic potentials: internal energy U, enthalpy H = U + PV, Helmholtz free energy F = U − TS, Gibbs free energy G = H − TS.
• Their differentials (using dU = TdS − PdV): dU = TdS − PdV, dH = TdS + VdP, dF = −SdT − PdV, dG = −SdT + VdP.
• Each is an exact differential, so the cross partial derivatives are equal. This gives the four Maxwell relations.
• Clausius–Clapeyron equation: dP/dT = L/[T(V₂ − V₁)]. Since ice contracts on melting, pressure lowers its melting point (regelation).
• Joule–Thomson effect: when a gas expands slowly through a porous plug from high to low pressure, with no heat exchange, its temperature changes. Enthalpy stays constant (isenthalpic process).
• For an ideal gas there is no JT effect. Real gases cool because work is done against intermolecular attraction.
• Every gas has an inversion temperature Ti: below it the gas cools on throttling, above it the gas warms. For hydrogen and helium Ti is very low, so they warm at room temperature and must be pre-cooled.
• The JT effect is used in the Linde process to liquefy air.

महत्त्वाच्या संज्ञा

• Throttling – slow expansion through a narrow opening or porous plug, at constant enthalpy.
• JT coefficient μ = (∂T/∂P) at constant H; positive μ means cooling on expansion.
• Inversion temperature – the temperature at which μ = 0 and the effect changes sign.
• Exact differential – a differential whose integral depends only on the end states.

सूत्रे

• (∂T/∂V)_S = −(∂P/∂S)_V
• (∂T/∂P)_S = (∂V/∂S)_P
• (∂S/∂V)_T = (∂P/∂T)_V
• (∂S/∂P)_T = −(∂V/∂T)_P
• First TdS equation: TdS = Cv dT + T(∂P/∂T)_V dV
• Cp − Cv = −T (∂V/∂T)_P² (∂P/∂V)_T; for an ideal gas this gives Cp − Cv = R
• Clausius–Clapeyron: dP/dT = L/[T(V₂ − V₁)]
• Van der Waals gas: inversion temperature Ti = 2a/(Rb); JT cooling ΔT ≈ (2a/RT − b) ΔP/Cp

५ महत्त्वाचे प्रश्न

Q1. Derive the Maxwell relation (∂S/∂V)_T = (∂P/∂T)_V.

Helmholtz function F = U − TS, so dF = dU − TdS − SdT = −SdT − PdV. Hence (∂F/∂T)_V = −S and (∂F/∂V)_T = −P. Since dF is exact, ∂²F/∂V∂T = ∂²F/∂T∂V, giving −(∂S/∂V)_T = −(∂P/∂T)_V, i.e. (∂S/∂V)_T = (∂P/∂T)_V.

Q2. Why does an ideal gas show no Joule–Thomson effect?

In throttling, H is constant. For an ideal gas H = U + PV = U + RT depends only on T (U depends only on T). If H is constant, T must be constant. So there is no cooling or heating. A real gas cools because its U also depends on the distance between molecules.

Q3. Derive the Clausius–Clapeyron equation and use it for ice.

Use (∂S/∂V)_T = (∂P/∂T)_V during a phase change, where T and P stay fixed. The entropy change is L/T for volume change (V₂ − V₁), so dP/dT = L/[T(V₂ − V₁)].
Example: ice at 273 K, L = 3.36 × 10⁵ J kg⁻¹, V(water) − V(ice) = 1.000 × 10⁻³ − 1.091 × 10⁻³ = −9.1 × 10⁻⁵ m³ kg⁻¹.
dT/dP = T(V₂ − V₁)/L = 273 × (−9.1 × 10⁻⁵)/(3.36 × 10⁵) ≈ −7.4 × 10⁻⁸ K Pa⁻¹. For 1 atm (1.013 × 10⁵ Pa) the melting point falls by about 0.0075 K.

Q4. Describe the porous plug experiment and give its results.

A gas is forced at constant high pressure P₁ through a porous plug (cotton or silk) in an insulated tube to a region of lower pressure P₂. Thermometers on both sides record the temperatures. Results: (1) at room temperature most gases (air, CO₂) cool; (2) hydrogen and helium warm up; (3) the change is proportional to the pressure drop; (4) every gas cools if its initial temperature is below its inversion temperature. Work done on the gas P₁V₁ − work done by it P₂V₂ = change in U, so U₁ + P₁V₁ = U₂ + P₂V₂: enthalpy is constant.

Q5. Using the Van der Waals constants for nitrogen (a = 0.137 Pa m⁶ mol⁻², b = 3.87 × 10⁻⁵ m³ mol⁻¹), estimate its inversion temperature.

Ti = 2a/(Rb) = (2 × 0.137)/(8.314 × 3.87 × 10⁻⁵) = 0.274/(3.217 × 10⁻⁴) ≈ 852 K. (The measured value is about 620 K; the Van der Waals model gives only an estimate.) Room temperature is far below Ti, so nitrogen cools on throttling.

परीक्षेसाठी टीप

To remember Maxwell's relations, start from the potential and its natural variables (U: S, V; H: S, P; F: T, V; G: T, P) and take cross derivatives. The relations from U and G carry a minus sign; those from H and F do not — check signs by writing dU or dG first.

थोडक्यात सारांश

An oscillation is a to-and-fro motion repeated about a mean position. This topic covers simple harmonic motion (SHM), its differential equation and energy, and then real oscillations: damped oscillations, forced oscillations and resonance.

महत्त्वाचे मुद्दे

• SHM: restoring force is proportional to displacement and directed towards the mean position: F = −kx.
• Differential equation of SHM: d²x/dt² + ω²x = 0, with ω² = k/m.
• Solution: x = A sin(ωt + φ), where A is amplitude and φ the initial phase (epoch).
• Velocity is maximum at the mean position and zero at the extreme positions; acceleration is the opposite.
• Total energy in SHM = ½kA², constant; KE and PE keep exchanging.
• Examples: simple pendulum (small angles), mass on a spring, compound pendulum, LC circuit.
• Damped oscillation: amplitude decreases with time because of friction or resistance. Three cases: underdamped (oscillates with falling amplitude), critically damped (returns fastest without oscillating), overdamped (returns slowly without oscillating).
• Forced oscillation: a body is driven by an external periodic force; it finally oscillates at the driver's frequency.
• Resonance: when the driving frequency equals the natural frequency, amplitude becomes maximum. Sharpness depends on damping (less damping → sharper resonance).
• Quality factor Q measures how slowly the energy is lost; high Q means sharp resonance.

महत्त्वाच्या संज्ञा

• Time period (आवर्तकाल) T – time for one complete oscillation.
• Frequency n – number of oscillations per second, n = 1/T, unit hertz.
• Angular frequency ω – ω = 2πn = 2π/T.
• Phase – the state of motion at an instant, given by the angle (ωt + φ).
• Relaxation time τ – time in which the energy of a damped oscillator falls to 1/e of its value.

सूत्रे

• x = A sin(ωt + φ); v = Aω cos(ωt + φ) = ω√(A² − x²); a = −ω²x
• v_max = Aω; a_max = Aω²
• Spring: T = 2π√(m/k); simple pendulum: T = 2π√(l/g)
• Compound pendulum: T = 2π√((K² + l²)/(lg))
• KE = ½k(A² − x²); PE = ½kx²; E = ½kA² = ½mω²A²
• Damped: m d²x/dt² + b dx/dt + kx = 0; underdamped amplitude falls as A₀e^(−bt/2m)
• Damped angular frequency ω' = √(ω₀² − (b/2m)²)
• Q = ω₀/(2β) where β = b/2m (for light damping)
• Series springs: 1/k = 1/k₁ + 1/k₂; parallel: k = k₁ + k₂

५ महत्त्वाचे प्रश्न

Q1. Define SHM and write its differential equation.

SHM is periodic motion in which the restoring force is directly proportional to displacement from the mean position and acts towards it. Equation: d²x/dt² + ω²x = 0, where ω = √(k/m).

Q2. At what displacement are the KE and PE of a particle in SHM equal?

KE = PE means ½k(A² − x²) = ½kx². So A² = 2x², x = A/√2 ≈ 0.707A.

Q3. Show that the total energy of a particle in SHM is constant.

1. x = A sin ωt, v = Aω cos ωt.
2. KE = ½mv² = ½mA²ω² cos² ωt.
3. PE = ½kx² = ½mω²A² sin² ωt (since k = mω²).
4. E = KE + PE = ½mω²A²(cos² ωt + sin² ωt) = ½mω²A².
5. This does not depend on t, so E is constant.

Q4. A body of mass 0.5 kg on a spring of k = 200 N/m oscillates with amplitude 0.1 m. Find period, maximum velocity and total energy.

1. ω = √(k/m) = √(200/0.5) = √400 = 20 rad/s.
2. T = 2π/ω = 2 × 3.1416/20 ≈ 0.314 s.
3. v_max = Aω = 0.1 × 20 = 2 m/s.
4. E = ½kA² = ½ × 200 × 0.01 = 1 J.
5. Check: ½mv_max² = ½ × 0.5 × 4 = 1 J ✓.

Q5. Explain damped and forced oscillations and the condition of resonance.

1. Damped: a resistive force −b(dx/dt) acts; amplitude decays exponentially; frequency is slightly less than natural.
2. Cases: underdamped, critically damped, overdamped (with sketches of x vs t).
3. Forced: external force F₀ sin pt acts; after transients die out the body oscillates at frequency p.
4. Amplitude depends on (ω₀ − p) and damping; maximum near p = ω₀ — resonance.
5. Examples: tuning a radio, soldiers breaking step on a bridge, sonometer resonance.

परीक्षेसाठी टीप

Draw the three damping curves (x against t) and the resonance curves for low and high damping — diagrams fetch easy marks. In numericals, find ω first; everything else follows from it.

थोडक्यात सारांश

When a particle is acted on by two simple harmonic motions (SHMs) at the same time, its resulting motion is a combination of the two. This topic covers the superposition of two SHMs along the same line (with equal and slightly different frequencies, giving beats) and two perpendicular SHMs, whose paths are called Lissajous figures. It also explains how a CRO uses these figures to compare frequencies and phases.

महत्त्वाचे मुद्दे

• Principle of superposition: the resultant displacement is the vector sum of the individual displacements.
• Two collinear SHMs of the same frequency combine into one SHM of the same frequency with amplitude R = √(a₁² + a₂² + 2a₁a₂ cos δ).
• R is maximum (a₁ + a₂) when δ = 0 and minimum |a₁ − a₂| when δ = π.
• Two collinear SHMs of slightly different frequencies give beats: the amplitude rises and falls with beat frequency (n₁ − n₂).
• Two perpendicular SHMs of the same frequency give an ellipse in general. Special cases: δ = 0 or π → straight line; δ = π/2 with equal amplitudes → circle; δ = π/2 with unequal amplitudes → ellipse with axes along x and y.
• For a frequency ratio such as 1:2, 2:3 etc. the figure is more complex (figure-of-eight for 1:2) but stays a closed curve if the ratio is a whole-number ratio.
• Frequency ratio from the figure: draw a box around the figure; f_y/f_x = (points touching the top edge) ÷ (points touching a side edge).
• Lissajous figures are seen on a CRO by applying one signal to the X plates and another to the Y plates.

महत्त्वाच्या संज्ञा

• Phase difference (δ) – difference in the phase angles of two SHMs.
• Beats – periodic rise and fall of loudness when two close frequencies combine.
• Lissajous figure – path traced by a point under two perpendicular SHMs.
• CRO – cathode ray oscilloscope, an instrument that displays voltage signals.

सूत्रे

• Collinear, same ω: x = R sin(ωt + φ), R² = a₁² + a₂² + 2a₁a₂ cos δ, tan φ = a₂ sin δ/(a₁ + a₂ cos δ)
• Beat frequency = n₁ − n₂
• Perpendicular, same ω (x = a sin ωt, y = b sin(ωt + δ)): x²/a² + y²/b² − (2xy/ab) cos δ = sin²δ
• Frequency ratio from a Lissajous figure: f_y / f_x = (number of tangent points on a horizontal edge) ÷ (number of tangent points on a vertical edge)

५ महत्त्वाचे प्रश्न

Q1. What are Lissajous figures? Give two uses.

They are closed curves traced by a particle acted on by two perpendicular SHMs whose frequencies are in a simple whole-number ratio. Uses: (1) to compare or measure an unknown frequency against a known one on a CRO; (2) to measure the phase difference between two signals of the same frequency.

Q2. Two collinear SHMs of amplitudes 3 cm and 4 cm and the same frequency differ in phase by 90°. Find the resultant amplitude.

R = √(3² + 4² + 2 × 3 × 4 × cos 90°) = √(9 + 16 + 0) = 5 cm.

Q3. Derive the equation of the path for two perpendicular SHMs of the same frequency.

1. x = a sin ωt → sin ωt = x/a, cos ωt = √(1 − x²/a²).
2. y = b sin(ωt + δ) = b(sin ωt cos δ + cos ωt sin δ), so y/b − (x/a) cos δ = √(1 − x²/a²) sin δ.
3. Square both sides: y²/b² − (2xy/ab) cos δ + (x²/a²) cos²δ = sin²δ − (x²/a²) sin²δ.
4. Rearranging: x²/a² + y²/b² − (2xy/ab) cos δ = sin²δ, an ellipse.

Cases: δ = 0 → y = (b/a)x, a straight line; δ = π → y = −(b/a)x; δ = π/2 → x²/a² + y²/b² = 1; if also a = b, a circle.

Q4. Find the resultant of two collinear SHMs with the same frequency and derive the amplitude.

Let x₁ = a₁ sin ωt and x₂ = a₂ sin(ωt + δ). Then x = (a₁ + a₂ cos δ) sin ωt + (a₂ sin δ) cos ωt. Put R cos φ = a₁ + a₂ cos δ and R sin φ = a₂ sin δ. Then x = R sin(ωt + φ): an SHM of the same frequency. Squaring and adding: R² = a₁² + a₂² + 2a₁a₂ cos δ; dividing: tan φ = a₂ sin δ/(a₁ + a₂ cos δ).

Q5. A Lissajous figure on a CRO touches the horizontal edge at 2 points and the vertical edge at 1 point. The signal on the X plates is 100 Hz. Find the Y frequency.

f_y / f_x = (touches on the horizontal edge) ÷ (touches on the vertical edge) = 2/1, so f_y = 2 × 100 = 200 Hz. This is the horizontal figure-of-eight (∞) traced by x = a sin ωt, y = b sin 2ωt.

परीक्षेसाठी टीप

Draw the five standard figures for equal frequency (δ = 0, π/4, π/2, 3π/4, π). Examiners usually ask for the derivation of the ellipse equation and these special cases.

थोडक्यात सारांश

Real oscillators lose energy to friction and air resistance, so their amplitude dies away (damped oscillations). If a periodic external force keeps driving them, they settle into forced oscillations at the driver's frequency, and the amplitude becomes very large near the natural frequency (resonance). This topic covers the differential equations, the three cases of damping, logarithmic decrement, relaxation time, quality factor, forced oscillations and sharpness of resonance.

महत्त्वाचे मुद्दे

• Damping force is taken proportional to velocity: F_d = −b v (b = damping constant, kg s⁻¹).
• Equation of a damped oscillator: m d²x/dt² + b dx/dt + kx = 0, or d²x/dt² + 2γ dx/dt + ω₀²x = 0, where 2γ = b/m and ω₀² = k/m.
• Underdamped (γ less than ω₀): oscillates with decaying amplitude, x = A e^(−γt) cos(ω′t + φ), ω′ = √(ω₀² − γ²). Example: a pendulum in air.
• Critically damped (γ = ω₀): returns to rest fastest without oscillating. Example: moving-coil galvanometer needle, car shock absorbers.
• Overdamped (γ greater than ω₀): returns slowly without oscillating. Example: pendulum in thick oil.
• Energy of an underdamped oscillator decays as e^(−2γt).
• Forced oscillations: m d²x/dt² + b dx/dt + kx = F₀ sin ωt. After the transient dies, the body oscillates at the driving frequency ω.
• Amplitude resonance occurs at ω_r = √(ω₀² − 2γ²) ≈ ω₀ for small damping. Less damping → higher and sharper peak.
• At resonance in velocity (ω = ω₀), the driving force is in phase with velocity and power absorbed is maximum.
• Examples: tuning a radio (electrical resonance), soldiers breaking step on bridges, Tacoma Narrows bridge collapse (1940).

महत्त्वाच्या संज्ञा

• Logarithmic decrement (λ) – natural log of the ratio of two successive amplitudes on the same side.
• Relaxation time (τ) – time in which energy falls to 1/e of its initial value; τ = 1/(2γ) = m/b.
• Quality factor (Q) – measure of sharpness of resonance; high Q means low damping.
• Sharpness of resonance – how fast the amplitude falls when ω moves away from resonance.

सूत्रे

• Damped: x = A₀ e^(−γt) cos(ω′t + φ), ω′ = √(ω₀² − γ²), γ = b/2m
• Logarithmic decrement λ = ln(A_n/A_(n+1)) = γT′
• Relaxation time τ = 1/(2γ); energy E = E₀ e^(−t/τ)
• Quality factor Q = ω₀τ = ω₀/(2γ) = 2π × (energy stored ÷ energy lost per cycle)
• Forced amplitude A = (F₀/m) ÷ √[(ω₀² − ω²)² + 4γ²ω²]
• Phase lag tan θ = 2γω/(ω₀² − ω²)
• Bandwidth (half-power): Δω = 2γ, so Q = ω₀/Δω

५ महत्त्वाचे प्रश्न

Q1. Distinguish between free, damped and forced oscillations.

Free: no damping, constant amplitude, natural frequency. Damped: amplitude falls exponentially due to resistive forces, frequency slightly less than natural. Forced: an external periodic force keeps the body oscillating at the force's frequency; amplitude depends on how close the driver is to the natural frequency.

Q2. The amplitude of a damped oscillator falls from 10 cm to 8 cm in one oscillation. Find the logarithmic decrement and the amplitude after 3 more oscillations.

λ = ln(10/8) = ln 1.25 = 0.223. Each oscillation multiplies the amplitude by 0.8, so after 3 more: 8 × 0.8³ = 8 × 0.512 = 4.1 cm.

Q3. Solve the equation of a damped oscillator and discuss the three cases.

1. Try x = e^(αt): α² + 2γα + ω₀² = 0, so α = −γ ± √(γ² − ω₀²).
2. γ greater than ω₀ (overdamped): both roots real and negative → x = e^(−γt)(C₁e^(βt) + C₂e^(−βt)), β = √(γ² − ω₀²); no oscillation, slow return.
3. γ = ω₀ (critical): repeated root → x = (C₁ + C₂t) e^(−γt); fastest non-oscillatory return.
4. γ less than ω₀ (underdamped): complex roots → x = A e^(−γt) cos(ω′t + φ), ω′ = √(ω₀² − γ²); oscillations with decaying amplitude.

Q4. Derive the amplitude of forced oscillations and the condition for amplitude resonance.

1. Equation: d²x/dt² + 2γ dx/dt + ω₀²x = (F₀/m) sin ωt.
2. Steady-state trial solution: x = A sin(ωt − θ). Substitute and equate coefficients of sin and cos.
3. This gives A = (F₀/m)/√[(ω₀² − ω²)² + 4γ²ω²] and tan θ = 2γω/(ω₀² − ω²).
4. A is maximum when the denominator is minimum: d/dω[(ω₀² − ω²)² + 4γ²ω²] = 0 → ω² = ω₀² − 2γ². So ω_r = √(ω₀² − 2γ²).
5. At ω = ω₀ the phase lag θ = 90°: displacement lags the force by a quarter cycle.

Q5. A tuning fork of frequency 256 Hz has its energy fall to 1/e in 4 s. Find its quality factor.

τ = 4 s, ω₀ = 2π × 256 = 1608 rad s⁻¹. Q = ω₀τ = 1608 × 4 ≈ 6.4 × 10³. A high Q means very low damping, so the fork rings for a long time.

परीक्षेसाठी टीप

Always state the three damping cases with one real example each. In numericals, check whether the question gives amplitude ratio (use λ) or energy (use τ); energy decays twice as fast as amplitude.

थोडक्यात सारांश

A wave carries energy from one place to another without carrying matter. This topic covers the differential equation of a wave, transverse and longitudinal waves, the speed of waves on a string and in a gas (Newton's formula and Laplace's correction), phase and group velocity, standing waves, and the Doppler effect for sound and light.

महत्त्वाचे मुद्दे

• Transverse wave: particles vibrate perpendicular to the direction of travel (waves on a string, light). Longitudinal wave: particles vibrate along the direction of travel (sound in air).
• A progressive wave moving along +x: y = a sin(ωt − kx), where k = 2π/λ and v = ω/k = nλ.
• Every travelling wave satisfies the wave equation ∂²y/∂t² = v² ∂²y/∂x².
• Speed on a stretched string v = √(T/μ), where T is tension and μ is mass per unit length.
• Newton assumed sound travels isothermally: v = √(P/ρ) ≈ 280 m s⁻¹ in air, which is too low.
• Laplace's correction: compressions and rarefactions are fast, so the process is adiabatic: v = √(γP/ρ) ≈ 332 m s⁻¹ at 0 °C.
• Speed of sound in a gas is proportional to √T and does not depend on pressure (at constant temperature).
• Phase velocity v_p = ω/k; group velocity v_g = dω/dk is the speed of the energy (wave packet). In a non-dispersive medium v_g = v_p.
• Standing waves form when two equal waves travel in opposite directions; nodes are λ/2 apart.
• Doppler effect: the apparent frequency changes when the source or the observer moves. Approach → higher pitch; recession → lower pitch.
• Uses: police radar speed guns, Doppler ultrasound of blood flow, red shift of galaxies (expanding universe).

महत्त्वाच्या संज्ञा

• Wavelength (λ) – distance between two nearest points in the same phase.
• Wave number (k) – 2π/λ, in rad m⁻¹.
• Dispersive medium – one where wave speed depends on frequency.
• Node / antinode – points of zero / maximum amplitude in a standing wave.
• Red shift – increase in wavelength of light from a receding source.

सूत्रे

• v = nλ = ω/k
• Wave equation: ∂²y/∂t² = v² ∂²y/∂x²
• String: v = √(T/μ); fundamental of a string fixed at both ends n = (1/2L)√(T/μ)
• Gas (Laplace): v = √(γP/ρ) = √(γRT/M); v_t = v₀√(T/273)
• Group velocity v_g = dω/dk = v_p − λ dv_p/dλ
• Doppler (sound, speed v in medium): n′ = n (v + v_o)/(v − v_s), with v_o positive when the observer moves towards the source and v_s positive when the source moves towards the observer
• Doppler (light, u much less than c): Δλ/λ = u/c

५ महत्त्वाचे प्रश्न

Q1. What is Laplace's correction to Newton's formula?

Newton took sound propagation as isothermal, giving v = √(P/ρ) ≈ 280 m s⁻¹ in air at STP, about 15% less than the measured 332 m s⁻¹. Laplace argued that compressions are too quick for heat exchange, so the change is adiabatic and the elasticity is γP. With γ = 1.4: v = √(1.4 × 1.013 × 10⁵/1.293) ≈ 331 m s⁻¹, which agrees with experiment.

Q2. Derive the relation between group velocity and phase velocity.

v_g = dω/dk. Since ω = v_p k: v_g = v_p + k dv_p/dk. With k = 2π/λ, dk = −(2π/λ²)dλ, so k dv_p/dk = −λ dv_p/dλ. Hence v_g = v_p − λ dv_p/dλ. In a non-dispersive medium dv_p/dλ = 0 and v_g = v_p.

Q3. Derive the expression for the Doppler effect when the source moves towards a stationary observer.

1. The source emits n waves per second; in 1 s the first wave travels v and the source moves v_s forward.
2. These n waves are squeezed into length (v − v_s), so the apparent wavelength λ′ = (v − v_s)/n.
3. The observer receives n′ = v/λ′ = n v/(v − v_s), which is greater than n.
4. If the observer also moves towards the source with v_o, he meets (v + v_o)/λ′ waves per second: n′ = n(v + v_o)/(v − v_s).

Q4. An ambulance siren of 500 Hz approaches a standing person at 20 m s⁻¹ and then moves away. Find the frequencies heard (v = 340 m s⁻¹).

Approaching: n′ = 500 × 340/(340 − 20) = 500 × 340/320 = 531 Hz.
Receding: n″ = 500 × 340/(340 + 20) = 500 × 340/360 = 472 Hz.
The drop in pitch as it passes is about 59 Hz.

Q5. Show that y = a sin(ωt − kx) satisfies the wave equation, and find the speed of a wave on a string with T = 100 N and μ = 0.01 kg m⁻¹.

∂²y/∂t² = −aω² sin(ωt − kx) and ∂²y/∂x² = −ak² sin(ωt − kx). So ∂²y/∂t² = (ω²/k²) ∂²y/∂x² = v² ∂²y/∂x² with v = ω/k.
For the string: v = √(T/μ) = √(100/0.01) = √10⁴ = 100 m s⁻¹.

परीक्षेसाठी टीप

In Doppler problems, write the sign convention first and check the answer: approach must raise the frequency, recession must lower it. Remember the Doppler effect for sound is not symmetric — a moving source and a moving observer at the same speed give different results.

थोडक्यात सारांश

Acoustics is the science of sound, especially how sound behaves inside halls and rooms. Ultrasonics deals with sound waves above 20 kHz that humans cannot hear. This topic covers intensity and decibels, reverberation and Sabine's formula, good acoustic design, and the production, detection and uses of ultrasonic waves.

महत्त्वाचे मुद्दे

• Sound is a longitudinal mechanical wave; it needs a medium.
• Audible range for humans: about 20 Hz to 20 kHz. Below 20 Hz is infrasonic; above 20 kHz is ultrasonic.
• Intensity of sound = energy flowing per second per unit area (W m⁻²). Loudness is our sensation of it and is roughly logarithmic (Weber–Fechner law).
• Reverberation: sound persisting in a hall after the source stops, due to repeated reflections.
• Reverberation time: time for sound intensity to fall to one-millionth (10⁻⁶) of its initial value, i.e. a drop of 60 dB.
• Sabine's formula relates reverberation time to hall volume and total absorption.
• A good hall has suitable reverberation time, uniform loudness, no echoes, no focusing, and low outside noise. Too little reverberation makes sound dull; too much makes it unclear.
• Absorbing materials: curtains, carpets, perforated boards, people in the audience.
• Production of ultrasonics: magnetostriction method (a ferromagnetic rod changes length in an alternating magnetic field; good for lower ultrasonic frequencies) and piezoelectric method (a quartz crystal vibrates under alternating voltage; for higher frequencies).
• Detection: piezoelectric detector, Kundt's tube, sensitive flames, thermal detectors.
• Uses: SONAR, echo sounding of sea depth, non-destructive testing of metals for flaws, medical scanning (sonography), cleaning, drilling, homogenising milk.

महत्त्वाच्या संज्ञा

• Decibel (dB) – unit of intensity level on a log scale.
• Absorption coefficient (a) – fraction of sound energy absorbed by a surface; open window = 1.
• Sabine – unit of absorption: 1 m² of open window (perfect absorber).
• Piezoelectric effect (दाब-विद्युत परिणाम) – voltage across a crystal when pressed; the inverse effect makes it vibrate under voltage.
• Magnetostriction – change in length of a ferromagnetic rod when magnetised.

सूत्रे

• Intensity level: L = 10 log₁₀(I/I₀) dB, I₀ = 10⁻¹² W m⁻²
• Sabine's formula: T = 0.161 V/A (SI; V in m³, A in m² sabine), with A = Σ aᵢSᵢ
• Average absorption coefficient: a_avg = Σ aᵢSᵢ / ΣSᵢ
• Piezoelectric oscillator frequency: n = (1/2t)√(Y/ρ) (t = crystal thickness)
• Magnetostriction oscillator: n = (1/2l)√(Y/ρ) (l = rod length)
• SONAR depth: d = vt/2 (t = time for echo)

५ महत्त्वाचे प्रश्न

Q1. Define reverberation time.

It is the time taken for the intensity of sound in a hall to fall to one-millionth of its original value after the source is stopped (a fall of 60 dB).

Q2. The intensity of a sound is 10⁻⁶ W m⁻². Find its intensity level.

L = 10 log₁₀(10⁻⁶/10⁻¹²) = 10 log₁₀(10⁶) = 10 × 6 = 60 dB.

Q3. Explain the factors affecting acoustics of buildings and their remedies.

1. Reverberation time: keep it optimum by adding absorbing materials or audience seats.
2. Loudness: should be uniform; use reflecting boards and proper seating.
3. Focusing: avoid curved (concave) walls and domes, or cover them with absorbers.
4. Echoes: cover distant reflecting walls with absorbers.
5. Echelon effect (regular steps/railings): cover with carpet.
6. Noise: use double doors, sound-proof windows, and keep machinery isolated.

Q4. A hall has volume 2000 m³ and total absorption 200 m² sabine. Find the reverberation time. If 100 people (absorption 0.5 each) enter, what is the new time?

1. T = 0.161V/A = 0.161 × 2000/200 = 1.61 s.
2. Extra absorption = 100 × 0.5 = 50; new A = 250.
3. New T = 0.161 × 2000/250 = 322/250 ≈ 1.29 s.

Q5. Describe the piezoelectric method of producing ultrasonic waves.

1. A quartz crystal plate is placed between two metal plates connected to an oscillator circuit (tank circuit with L and C).
2. Alternating voltage applied to the crystal makes it expand and contract (inverse piezoelectric effect).
3. Adjust C so the circuit frequency equals the crystal's natural frequency n = (1/2t)√(Y/ρ).
4. At resonance the crystal vibrates strongly and emits ultrasonic waves.
5. Advantages: very high frequencies (up to hundreds of MHz), stable output.

परीक्षेसाठी टीप

Learn Sabine's formula with its constant 0.161 (SI). In acoustics long answers, list the defects and their remedies as a neat two-column table.

थोडक्यात सारांश

A blackbody absorbs all radiation falling on it and, when heated, emits radiation of all wavelengths. Classical physics could not explain the observed blackbody spectrum. Planck's quantum hypothesis (1900) solved this problem and began quantum physics. This topic covers Kirchhoff's law, Stefan–Boltzmann law, Wien's law, Rayleigh–Jeans law and Planck's law.

महत्त्वाचे मुद्दे

• A perfect blackbody absorbs all incident radiation (absorptivity = 1). A small hole in a hollow cavity with blackened inner walls is a good practical blackbody (Ferry's blackbody).
• Kirchhoff's law: at a given temperature and wavelength, the ratio of emissive power to absorptive power is the same for all bodies and equals the emissive power of a blackbody. Good absorbers are good emitters.
• Stefan–Boltzmann law: total energy radiated per unit area per second by a blackbody is proportional to T⁴.
• Energy distribution curves: for each temperature, emitted energy rises with wavelength, reaches a peak at λ_m and then falls. As T rises, the area under the curve increases and λ_m shifts to shorter wavelengths.
• Wien's displacement law: λ_m T = constant.
• Rayleigh–Jeans law (classical) fits only at long wavelengths. At short wavelengths it predicts infinite energy — the ultraviolet catastrophe.
• Wien's distribution law fits only at short wavelengths.
• Planck's hypothesis: oscillators in the cavity walls can have only discrete energies nhν; energy is emitted or absorbed in packets (quanta) of hν.
• Planck's law agrees with experiment at all wavelengths, and gives Wien's and Rayleigh–Jeans laws as limits.

महत्त्वाच्या संज्ञा

• Emissive power (eλ) – energy emitted per unit area per second per unit wavelength range.
• Absorptive power (aλ) – fraction of incident radiation absorbed.
• Quantum (पुंज) – smallest packet of energy, E = hν.
• Ultraviolet catastrophe – failure of classical theory at short wavelengths.

सूत्रे

• Stefan–Boltzmann: E = σT⁴, σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴; net loss E = σ(T⁴ − T₀⁴); for non-black body E = eσT⁴
• Wien's displacement: λ_m T = b, b ≈ 2.898 × 10⁻³ m K
• Energy of a quantum: E = hν = hc/λ, h = 6.63 × 10⁻³⁴ J s
• Planck's law: u_λ dλ = (8πhc/λ⁵) × 1/(e^(hc/λkT) − 1) dλ
• Rayleigh–Jeans: u_λ dλ = (8πkT/λ⁴) dλ (long λ limit)
• Wien's law: u_λ dλ = (8πhc/λ⁵) e^(−hc/λkT) dλ (short λ limit)
• Average energy of a Planck oscillator: ε = hν/(e^(hν/kT) − 1)

५ महत्त्वाचे प्रश्न

Q1. What is the ultraviolet catastrophe?

The Rayleigh–Jeans law says energy density grows as 1/λ⁴. So at very short (ultraviolet) wavelengths it predicts infinite energy, which is not seen. This failure of classical physics is called the ultraviolet catastrophe.

Q2. State Wien's displacement law and give one use.

The wavelength of maximum emission of a blackbody is inversely proportional to its absolute temperature: λ_m T = b. It is used to estimate the surface temperature of the Sun and stars from their spectrum.

Q3. State the assumptions of Planck's quantum theory and write Planck's radiation law.

1. A blackbody cavity has atomic oscillators of all frequencies.
2. An oscillator of frequency ν can have only energies 0, hν, 2hν, … nhν.
3. Energy is emitted or absorbed only in whole quanta of hν, not continuously.
4. Average energy of an oscillator: ε = hν/(e^(hν/kT) − 1), not kT.
5. Multiply by the number of modes 8π/λ⁴ per unit volume per unit wavelength: u_λ = (8πhc/λ⁵)/(e^(hc/λkT) − 1).

Q4. Show that Planck's law reduces to the Rayleigh–Jeans law at long wavelengths.

1. For long λ, x = hc/λkT is very small.
2. Then e^x ≈ 1 + x, so e^x − 1 ≈ hc/λkT.
3. u_λ ≈ (8πhc/λ⁵) × (λkT/hc) = 8πkT/λ⁴.
4. This is the Rayleigh–Jeans law. (For short λ, e^x ≫ 1 and we get Wien's law.)

Q5. The Sun's spectrum peaks near 500 nm. Estimate its surface temperature (b = 2.898 × 10⁻³ m K).

1. T = b/λ_m = 2.898 × 10⁻³/(500 × 10⁻⁹).
2. = 2.898 × 10⁻³/(5 × 10⁻⁷) = 0.5796 × 10⁴.
3. T ≈ 5796 K, i.e. about 5800 K.

परीक्षेसाठी टीप

Draw the energy-distribution curves for three temperatures with λ_m shifting left — it is asked almost every year. Know how Planck's law gives both Wien's and Rayleigh–Jeans laws as limits.