 Solution

ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2018/paper-1/section-c/c12/torsional-waves/solution

The torque difference. Let

ϕ (x, t)

now be the local angular displacement, and take

G

and the section constant. The local twist per unit length is

\partial_{x} ϕ

, so the transmitted torque is

C (x, t) = G J_{p} \partial_{x} ϕ

. Across a short slice,

Δ C = C (x + Δ x) - C (x) = G J_{p} \frac{\partial ^{2} ϕ}{\partial x ^{2}} Δ x + O (Δ x^{2}) .

Thus

Δ C = \frac{π a ^{4} G}{2} \frac{\partial ^{2} ϕ}{\partial x ^{2}} Δ x

to first order in slice length.

The slice's angular acceleration. The slice's axial moment of inertia is

Δ I = ρ J_{p} Δ x

. Its net torque is the difference of the end torques, so

ρ J_{p} Δ x \partial_{t}^{2} ϕ = G J_{p} Δ x \partial_{x}^{2} ϕ .

\frac{\partial ^{2} ϕ}{\partial t ^{2}} = \frac{G}{ρ} \frac{\partial ^{2} ϕ}{\partial x ^{2}}, c_{t} = \frac{G}{ρ} .

For the given steel,

c_{t} ≃ 3.16 \times 1 0^{3} m s^{- 1}

. The radius cancels because both torsional stiffness and slice inertia contain the same

J_{p}

. This ideal one-dimensional wave model neglects dissipation and additional elastic modes.

 New to topics? Read the docs here!