#sinusoidal

Optical Phenomena and Applications

What causes the intensity of light to vary sinusoidally with angle when passing through two polaroids?

**Polarization requires transverse waves** because only transverse can have orientation perpendicular to propagation, longitudinal cannot be polarized, wave theory requires light transverse to explain polarization, polaroids transmit only component along pass-axis, unpolarized has random transverse orientations, after polaroid polarized. The intensity follows Malus’ law, where it varies as the square of the cosine of the angle between the polaroids’ axes, producing a sinusoidal pattern. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a cos(φ/2)

Ref: NCERT > Physics Book > Wave Optics > Optical Phenomena and Applications

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Rotational EMF and AC Generator

In an AC generator, the emf varies sinusoidally with time. What is the primary reason for this variation?

**Rotational emf** when coil area A rotates with angular speed ω in uniform field B, flux Φ = B A cos ωt, emf e = -N dΦ/dt = N B A ω sin ωt, maximum e₀ = N B A ω, frequency = ω/2π. For square side 22 cm area 0.0484 m² N=1 ω=14 rad/s B=0.15 T, e₀=1×0.15×0.0484×14=0.1016 V, sinusoidal. The emf varies sinusoidally because the coil rotates in a uniform magnetic field, causing the angle between the magnetic field and the coil’s area vector to change continuously, following a sine function. Using Φ = B A cosθ, e = -N dΦ/dt = -N A

Ref: NCERT > Physics Book > Electromagnetic Induction > Rotational EMF and AC Generator

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AC Through Capacitor - Capacitive Reactance

An AC voltage \( v = 141.4 \sin (100\pi t) \) V is applied across a \( 50 \, \Omega \) resistor. What is the rms current

**AC through capacitor** voltage lags current by 90°, Q= C V, I= dQ/dt = C dV/dt, V(t)=V_peak sin ωt, I(t)=I_peak sin(ωt+90°), average power zero because energy stored in electric field ½ C V² returned each cycle, capacitor blocks DC but passes AC, X_C decreases with f. Given: v = v_m sin ω t , where v_m = 141.4 V , R = 50 Ω . RMS voltage: V = (v_m/√(2)) = (141.4/√(2)) = 100 V . RMS current: I = (V/R) = (100/50) = 2 A . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P

Ref: NCERT > Physics Book > Alternating Currents > AC Through Capacitor - Capacitive Reactance

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Displacement Current and Ampere-Maxwell Law

An electromagnetic wave in vacuum has an electric field given by \( E_x = 150 \sin(5 \times 10^3 z - 1.5 \times 10^{12}

**Ampere-Maxwell law** ∮ B·dl = μ₀(I_c + ε₀ dΦ_E/dt) generalizes Ampere's law, displacement current arises from time-varying electric field, source of magnetic field like conduction current. For rate of change of flux 2×10¹¹ V·m/s, I_d = ε₀×2×10¹¹ =8.85×10⁻¹²×2×10¹¹=1.77 A. Comparing with E_x = E₀ sin(kz - ω t) , we have ω = 1.5 × 10¹² rad/s . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 1.5 × 10¹² rad/s, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Displacement Current and Ampere-Maxwell Law

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Damped Oscillations

Which function represents SHM? (\( \omega \) is a positive constant)

**Effect of damping** is gradual amplitude reduction while period remains nearly constant for light damping. Mechanical energy decreases as work done against damping force, E(t) = ½ k A(t)² decaying exponentially, and motion ceases without external energy input, distinguishing from ideal undamped SHM. SHM requires a = -ω² x : (a) 6 sin (2ω t - (π/4)) : a = -6 (2ω)² sin (2ω t - (π/4)) = -ω² x , SHM. (b) sin ω t + cos 3ω t : Not SHM (mixed frequencies). (c) eω t : Not periodic. (d) cos³ ω t : Periodic, not SHM. Applying x = A cos(ωt +

Ref: NCERT > Physics Book > Oscillations > Damped Oscillations

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Damped Oscillations

Which function represents simple harmonic motion? (\( \omega \) is a positive constant)

**Effect of damping** is gradual amplitude reduction while period remains nearly constant for light damping. Mechanical energy decreases as work done against damping force, E(t) = ½ k A(t)² decaying exponentially, and motion ceases without external energy input, distinguishing from ideal undamped SHM. SHM requires a = -ω² x : (a) cos ω t + sin 2ω t : Not SHM (mixed frequencies). (b) 4 sin (ω t + (π/6)) : a = -4ω² sin (ω t + (π/6)) = -ω² x , SHM. (c) eω t : Not periodic. (d) sin² ω t : Periodic, not SHM. Applying x = A cos(ωt + φ),

Ref: NCERT > Physics Book > Oscillations > Damped Oscillations

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