(a) A body moves about a point 'O' under no force, the principal moments of inertia at 'O' being 3A, 5A and 6A. The components of…

(a)

For torque-free motion about a fixed point, Euler’s equations are I₁ dω₁/dt = (I₂ − I₃)ω₂ω₃, I₂ dω₂/dt = (I₃ − I₁)ω₃ω₁, I₃ dω₃/dt = (I₁ − I₂)ω₁ω₂.

Here I₁ = 3A, I₂ = 5A, I₃ = 6A. Hence dω₁/dt = −(1/3)ω₂ω₃, dω₂/dt = (3/5)ω₁ω₃, dω₃/dt = −(1/3)ω₁ω₂.

Initially ω₁ = n, ω₂ = 0, ω₃ = n. Since ω₁ = ω₃ initially, and d(ω₁ − ω₃)/dt = (ω₂/3)(ω₁ − ω₃), the equality ω₁ = ω₃ is preserved for all t. Put ω₁ = ω₃ = x, ω₂ = y. Then dx/dt = −(1/3)xy, dy/dt = (3/5)x².

Use conservation of kinetic energy. Since the body is torque-free, T = 1/2(3A x² + 5A y² + 6A x²) = constant. At t = 0, T = 1/2(3A n² + 6A n²) = (9/2)A n². Therefore 9x² + 5y² = 9n², so x² = n² − (5/9)y².

Substitute this in dy/dt: dy/dt = (3/5)[n² − (5/9)y²] = (3/5)n² − (1/3)y². This is separable: ∫₀ʸ dy/((3/5)n² − (1/3)y²) = t.

Using ∫ du/(a − bu²) = (1/√(ab)) artanh(u√(b/a)), with a = (3/5)n², b = 1/3, √(ab) = n/√5, √(b/a) = √5/(3n), we get t = (√5/n) artanh(√5 y/(3n)). Thus y = (3n/√5) tanh(n t/√5). Then x = √(n² − (5/9)y²) = n sech(n t/√5).

Hence the exact components are ω₁ = n sech(n t/√5), ω₂ = (3n/√5) tanh(n t/√5), ω₃ = n sech(n t/√5). For large t, tanh(n t/√5) → 1 and sech(n t/√5) → 0. Therefore ω₁ → 0, ω₂ → 3n/√5, ω₃ → 0. The components have units of s⁻¹. Physically, the initial conditions put the motion exactly on the separatrix (L² = (3An)² + (6An)² = 45A²n², and 2T·I₂ = 9An² × 5A = 45A²n², so L² = 2T·I₂), so the body approaches rotation about the intermediate axis (I₂ = 5A) asymptotically, taking infinite time; that axis is itself unstable.

(b)

Write the equation as d²x/dt² + 2β dx/dt + ω₀²x = 0, where β = γ/(2m) = 0.07/(2 × 0.25) = 0.14 s⁻¹, ω₀² = k/m = 85/0.25 = 340 s⁻². Since β < ω₀, the oscillator is underdamped. The damped angular frequency is ω_d = √(ω₀² − β²) = √(340 − 0.0196) = √339.9804 = 18.4386 s⁻¹.

(i) The period of oscillation is T = 2π/ω_d = 2π/18.4386 = 0.3408 s. Exactly, T = 100π/√849951 s ≈ 0.3408 s.

(ii) The amplitude decays as A(t) = A₀ e^−βt. For half amplitude, e^−βt = 1/2 ⇒ t = ln2/β = 0.693147/0.14 = 4.9511 s. Number of oscillations is N = t/T = 4.9511/0.3408 = 14.53. Exactly, N = (√849951 ln2)/(14π) ≈ 14.53. Thus the amplitude becomes half after about 14.5 oscillations, i.e. during the 15th complete oscillation.

(c)

In source-free vacuum, Maxwell’s equations give the electromagnetic wave equation ∇²E − (1/c²) ∂²E/∂t² = 0, and the same for B. Define the d’Alembertian operator □ = ∂²/∂x² + ∂²/∂y² + ∂²/∂z² − (1/c²) ∂²/∂t².

Consider a Lorentz boost along x with speed v: x' = γ(x − vt), y' = y, z' = z, t' = γ(t − vx/c²), where γ = 1/√(1 − v²/c²).

By the chain rule, ∂/∂x = γ(∂/∂x' − (v/c²) ∂/∂t'), ∂/∂t = γ(∂/∂t' − v ∂/∂x').

Therefore ∂²/∂x² = γ²(∂²/∂x'² − 2(v/c²) ∂²/∂x'∂t' + (v²/c⁴) ∂²/∂t'²), ∂²/∂t² = γ²(∂²/∂t'² − 2v ∂²/∂x'∂t' + v² ∂²/∂x'²).

Now compute ∂²/∂x² − (1/c²) ∂²/∂t² = γ²[(1 − v²/c²) ∂²/∂x'² − (1/c²)(1 − v²/c²) ∂²/∂t'²] = ∂²/∂x'² − (1/c²) ∂²/∂t'².

Since y and z coordinates are unchanged, ∂²/∂y² + ∂²/∂z² = ∂²/∂y'² + ∂²/∂z'². Hence □ = □'. Thus □E = 0 ⇒ □'E' = 0, and similarly for B. Therefore the electromagnetic wave equation has the same form in all inertial frames; it is invariant under Lorentz transformations.