(a) (i) Find the conjunctive normal form (CNF) of the following Boolean function: f(x, y, z, t) = x · y · z + x̄ · y · (t + z̄)…

(a) Boolean functions: CNF, DNF and truth table

Part (i): conjunctive normal form of f(x, y, z, t).

f = x·y·z + x̄·y·(t + z̄) = y·[x·z + x̄·(t + z̄)].

Let g = x·z + x̄·(t + z̄). The function g is 0 in exactly two situations: x = 1 with z = 0, or x = 0 with z = 1 and t = 0. Each situation is excluded by one sum term: x = 1, z = 0 makes (x̄ + z) false, and x = 0, z = 1, t = 0 makes (x + z̄ + t) false. So g = (x̄ + z)·(x + z̄ + t).

Therefore the conjunctive normal form is f = y·(x̄ + z)·(x + z̄ + t).

Part (ii): disjunctive normal form and truth table of f(x, y, z) = x + (x̄ȳ + x̄z)′ + z.

The bar in the question is over the whole bracket, so use De Morgan's law: (x̄ȳ + x̄z)′ = (x̄ȳ)′·(x̄z)′ = (x + y)(x + z̄) = x + y z̄.

Then f = x + (x + y z̄) + z = x + z + y z̄. Since z + y z̄ = z + y, f = x + y + z.

This is the simplified DNF. Its canonical DNF has the seven minterms where f = 1: f = x̄ȳz + x̄yz̄ + x̄yz + xȳz̄ + xȳz + xyz̄ + xyz.

Truth table:

x y z | f 0 0 0 | 0 0 0 1 | 1 0 1 0 | 1 0 1 1 | 1 1 0 0 | 1 1 0 1 | 1 1 1 0 | 1 1 1 1 | 1

The function is 0 only when x = y = z = 0, as expected for x + y + z.

(b) Ball rolling round inside a roller

Let R = b − a be the radius of the circle described by the ball's centre. The roller is drawn along the level path with uniform velocity V, so it rolls on the ground and its axis moves uniformly. Use axes moving with the axis, which is an inertial frame, with the roller's inner surface moving at speed V relative to them.

Initially the ball is at rest relative to the ground, at the lowest point. In the moving axes its centre has speed V backwards, with no spin, and this matches the surface speed of the roller at the contact. So rolling without slipping holds from the start, and no impulse acts.

Let θ be the angle between the line from the roller's axis to the ball's centre and the downward vertical. Let ω be the ball's spin. Rolling without slipping on the moving surface gives R dθ/dt + aω = −V. The ball's moment of inertia is I = (2/5)ma². The Lagrangian, after using the constraint to eliminate ω, is L = ½mR²θ̇² + (1/5)m(V + Rθ̇)² + mgR cosθ.

The Euler-Lagrange equation d/dt(∂L/∂θ̇) = ∂L/∂θ gives (7/5)mR²θ̈ = −mgR sinθ, and multiplying by θ̇ and integrating, (7/10)R²θ̇² − gR cosθ = constant.

At θ = 0, Rθ̇ = −V, so the constant is (7/10)V² − gR. Therefore R²θ̇² = V² − (10/7)gR(1 − cosθ).

Normal reaction N of the roller on the ball, towards the axis: N − mg cosθ = mRθ̇², so N = m[V²/R − (10/7)g(1 − cosθ)] + mg cosθ = m[V²/R − 10g/7 + (17/7)g cosθ].

This decreases as θ goes from 0 to π, so the minimum is at the top, θ = π: N_top = m[V²/R − 27g/7].

The ball stays in contact and goes completely round if N > 0 throughout, that is when N_top > 0: V² > (27/7)gR = (27/7)g(b − a).

At the top the ball's centre speed is then Rθ̇ with R²θ̇² = V² − (20/7)gR > gR, which is the condition for staying on the circle. Because the ball is perfectly rough, the friction needed for rolling is available, so it rolls completely round the inside of the roller.

(c) Vibrating string with given initial displacement and velocity

For a²u_xx = u_tt with u(0, t) = u(L, t) = 0, separation of variables gives u(x, t) = Σₙ [Aₙ cos(nπat/L) + Bₙ sin(nπat/L)] sin(nπx/L).

Initial displacement: u(x, 0) = Σ Aₙ sin(nπx/L) = x, so Aₙ = (2/L)∫₀ᴸ x sin(nπx/L) dx = 2L(−1)^(n+1)/(nπ).

Initial velocity: uₜ(x, 0) = Σ (nπa/L)Bₙ sin(nπx/L) = 1, so (nπa/L)Bₙ = (2/L)∫₀ᴸ sin(nπx/L) dx = 2(1 − (−1)ⁿ)/(nπ), and Bₙ = 2L(1 − (−1)ⁿ)/(a n²π²). This is 0 for even n and 4L/(a n²π²) for odd n.

Therefore u(x, t) = Σₙ₌₁^∞ [2L(−1)^(n+1)/(nπ) cos(nπat/L) + 2L(1 − (−1)ⁿ)/(a n²π²) sin(nπat/L)] sin(nπx/L).

Equivalently, u(x, t) = (2L/π) Σₙ [(−1)^(n+1)/n] cos(nπat/L) sin(nπx/L) + (4L/(aπ²)) Σₙ odd (1/n²) sin(nπat/L) sin(nπx/L).