(a) A regular tetrahedron, formed of six light rods, each of length l, rests on a smooth horizontal plane. A ring of weight W and…

(a)

Set-up. Let ABC be the horizontal base on the smooth plane and O the vertex. All six rods have length l. The ring (radius r, weight W) stays horizontal with its centre on the vertical axis through O, and it rests on the three slant rods OA, OB, OC at the points where they are at horizontal distance r from the axis. Let T be the tension in each horizontal rod (equal by symmetry).

Virtual displacement. Replace the horizontal rods by the tensions T and let the base side change from l to a general value x, keeping the slant rods rigid and the figure symmetrical. The forces that do work are only the weight W and the three tensions. The reactions at A, B, C are vertical and these points stay on the plane, so the reactions do no work. The contacts between ring and rods are smooth, so they do no work either.

Geometry for base side x. The circumradius of the base is x/√3. If α is the angle of a slant rod with the vertical, then sin α = x/(√3 l). The height of O is h = √(l² − x²/3). The contact point P is at horizontal distance r from the axis, so it lies a depth r cot α below O, where cot α = h/(x/√3) = √3h/x. The height of the ring above the plane is therefore

z = h − r cot α = h(1 − √3 r/x).

Equation of virtual work. When the base side increases by δx, each tension T does work −T δx, and the weight does work −W δz. By the principle of virtual work the total is zero:

−W δz − 3T δx = 0, i.e. T = −(W/3)(dz/dx).

dz/dx = (dh/dx)(1 − √3 r/x) + √3 r h/x², with dh/dx = −x/(3h).

At the actual position x = l we have h = l√(2/3), so x/(3h) = 1/√6 and √3 r h/x² = √2 r/l. Then

dz/dx = −(1/√6)(1 − √3 r/l) + √2 r/l = −1/√6 + (r/l)(1/√2 + √2) = −1/√6 + 3r/(√2 l).

Hence

T = W/(3√6) − Wr/(√2 l) = W(l − 3√3 r)/(3√6 l).

Interpretation and check. The ring can rest on the slant sides only if r < l/√3. For r < l/(3√3) the horizontal rods are in tension. For l/(3√3) < r < l/√3 the formula gives a negative T, so the rods are then in thrust. An independent resolution of forces at the joints gives the same value. There, each slant rod takes a normal reaction W/(3 sin α) from the ring, the load is shared between the rod's end joints by the lever rule, and horizontal equilibrium at A is taken. The formula also has the correct dimensions, since T is a force proportional to W.

(b)

Let the centre of motion be the origin, a the amplitude and ω² the constant of the motion. Then velocity² = ω²(a² − x²) and acceleration = −ω²x. Let the two positions be x₁, x₂, with velocities u, v and (signed) accelerations f₁ = −ω²x₁ and f₂ = −ω²x₂.

Distance between the positions. u² = ω²(a² − x₁²), v² = ω²(a² − x₂²) ⇒ v² − u² = ω²(x₁² − x₂²) = ω²(x₁ − x₂)(x₁ + x₂).

Since ω²(x₁ + x₂) = −(f₁ + f₂), v² − u² = −(x₁ − x₂)(f₁ + f₂) ⇒ x₂ − x₁ = (v² − u²)/(f₁ + f₂).

So the distance between the positions is k(v² − u²) with k = 1/(f₁ + f₂) (signed accelerations). In terms of the magnitudes of the accelerations there are two cases:

• both positions on the same side of the centre: k = 1/(f₁ + f₂);
• positions on opposite sides of the centre: k = 1/(f₁ − f₂).

In the second case x₁ + x₂ is a difference of the distances from the centre. In either case k is chosen so that k(v² − u²) > 0.

Amplitude. From u² − v² = ω²(x₂² − x₁²) and x_i = −f_i/ω²:

u² − v² = (f₂² − f₁²)/ω² ⇒ ω² = (f₂² − f₁²)/(u² − v²).

Then a² = x₁² + u²/ω² = (f₁² + u²ω²)/ω⁴. Now

f₁² + u²ω² = [f₁²(u² − v²) + u²(f₂² − f₁²)]/(u² − v²) = (u²f₂² − v²f₁²)/(u² − v²),

and ω⁴ = (f₂² − f₁²)²/(u² − v²)². Therefore

a² = (u² − v²)(u²f₂² − v²f₁²)/(f₂² − f₁²)²,

a = [(u² − v²)(u²f₂² − v²f₁²)]^(1/2) / (f₂² − f₁²), as required.

Check. The expression is symmetric under interchanging (u, f₁) with (v, f₂). Dimensionally, (L²T⁻²)(L²T⁻²·L²T⁻⁴)^(1/2)/(L²T⁻⁴) = L. Also, ω² > 0 requires u² − v² and f₂² − f₁² to have the same sign. This holds because the particle moves faster where it is nearer the centre, which is where |f| is smaller.

(c)(i)

Given solution. For u = −e⁻ˣ we have u′ = e⁻ˣ and u″ = −e⁻ˣ. Then x(−e⁻ˣ) + (x − 1)e⁻ˣ + e⁻ˣ = 0, so u is a solution.

Reduction of order. In standard form y″ + P y′ + Q y = 0 with P = (x − 1)/x = 1 − 1/x. So

∫P dx = x − ln x, e^(−∫P dx) = x e⁻ˣ.

Put y = u·v. Then v′ = e^(−∫P dx)/u² = x e⁻ˣ / e⁻²ˣ = x eˣ, and

v = ∫x eˣ dx = (x − 1)eˣ.

Second solution: y₂ = u v = −e⁻ˣ(x − 1)eˣ = 1 − x, or equivalently y₂ = x − 1.

Verification. For y = x − 1, y′ = 1 and y″ = 0, so x·0 + (x − 1)·1 − (x − 1) = 0. The Wronskian of −e⁻ˣ and x − 1 is −x e⁻ˣ, which is not identically zero, so the two solutions are independent.

General solution: y = c₁e⁻ˣ + c₂(x − 1). A constant multiple of the second solution is immaterial, so 1 − x and x − 1 are the same solution. A simple polynomial solution of this kind is expected here: trying y = x + b in the equation gives (x − 1) − (x + b) = 0, i.e. b = −1, which agrees with the result of the reduction.

(c)(ii)

Complementary function. The homogeneous part is a Cauchy–Euler equation. Putting y = xᵐ: m(m − 1) − 2m + 2 = 0 ⇒ m² − 3m + 2 = 0 ⇒ m = 1, 2. So y_c = c₁x + c₂x², with y₁ = x and y₂ = x².

Standard form. Dividing by x²: y″ − (2/x)y′ + (2/x²)y = x sin x, so R(x) = x sin x.

Wronskian. W = y₁y₂′ − y₂y₁′ = x·2x − x²·1 = x².

Variation of parameters. y_p = A(x)·x + B(x)·x², where

A′ = −y₂R/W = −x²·x sin x/x² = −x sin x ⇒ A = ∫(−x sin x)dx = x cos x − sin x; B′ = y₁R/W = x·x sin x/x² = sin x ⇒ B = −cos x.

y_p = x(x cos x − sin x) + x²(−cos x) = −x sin x.

Check. For y_p = −x sin x: y_p′ = −sin x − x cos x and y_p″ = −2cos x + x sin x. Then x²y_p″ − 2xy_p′ + 2y_p = (−2x² cos x + x³ sin x) + (2x sin x + 2x² cos x) − 2x sin x = x³ sin x.

General solution: y = c₁x + c₂x² − x sin x, where c₁ and c₂ are arbitrary constants. The substitution x = eᵗ reduces the equation to one with constant coefficients and leads to the same general solution.