Lateral vibration of a power line (string): 22-Mec-A2 Kinematics and Dynamics of Machines, December 2016, Q7

Question 7: Lateral vibration of a power line (string) (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A taut cable fixed at both ends: tension $T=2000$ N, length $L=100$ m, linear density $\rho=0.4$ kg/m. Lateral (transverse) vibration governed by the wave equation $\rho\,v_{tt}=T\,v_{xx}$ with $v(0,t)=v(L,t)=0$.

Find. (a) $f_1,f_2,f_3$; (b) the mode shapes; (c) the first-mode free response for the given initial-velocity profile.

Top: first three fixed–fixed mode shapes $Y_n=\sin(n\pi x/L)$. Bottom: the initial transverse-velocity profile $\dot v(x,0)=5\,(x/L)(1-x/L)$ m/s (parabolic, zero at both towers).

Approach. The wave speed sets the natural frequencies of a fixed–fixed string; the given velocity profile is expanded in the mode shapes and only the first term retained.

Wave speed. $$c=\sqrt{T/\rho}=\sqrt{2000/0.4}=\boxed{70.71\text{ m/s}}.$$
(a) Natural frequencies. $\omega_n=\dfrac{n\pi c}{L}$, $f_n=\dfrac{nc}{2L}$: $$\boxed{f_1=0.354\text{ Hz},\quad f_2=0.707\text{ Hz},\quad f_3=1.061\text{ Hz}}$$ ($\omega_1=2.221,\ \omega_2=4.443,\ \omega_3=6.664$ rad/s).
(b) Mode shapes. Fixed–fixed boundary conditions give sinusoidal modes $$Y_n(x)=\sin\!\frac{n\pi x}{L},\qquad n=1,2,3,\dots$$ i.e. a single half-wave (one antinode at mid-span), two half-waves (node at mid-span), three half-waves.
(c) First-mode free vibration. Expand $v(x,t)=\sum_n q_n(t)\sin(n\pi x/L)$. Zero initial displacement gives $q_n(t)=B_n\sin(\omega_n t)$. The initial-velocity coefficient is $$\dot q_n(0)=\frac{2}{L}\int_0^L \dot v(x,0)\sin\frac{n\pi x}{L}\,dx .$$ For $n=1$, with $\displaystyle\int_0^L x\bigl(1-\tfrac xL\bigr)\sin\frac{\pi x}{L}dx=\frac{4L^2}{\pi^3}$, $$\dot q_1(0)=\frac{2}{L}\cdot\frac{5}{L}\cdot\frac{4L^2}{\pi^3}=\frac{40}{\pi^3}=1.290\text{ m/s},\qquad B_1=\frac{\dot q_1(0)}{\omega_1}=\frac{1.290}{2.221}=0.581\text{ m}.$$ Hence the retained first-mode response $$\boxed{v(x,t)\approx 0.581\,\sin(2.221\,t)\,\sin\!\frac{\pi x}{100}\ \text{m}.}$$

Quantity	Value
Wave speed $c$	70.71 m/s
(a) $f_1,f_2,f_3$	0.354, 0.707, 1.061 Hz
(b) Mode shapes $Y_n$	$\sin(n\pi x/L)$
(c) First-mode response	$v\approx0.581\sin(2.221t)\sin(\pi x/100)$ m

Topic: Transverse vibration of a taut string (continuous system, wave equation).

Key relations: $c=\sqrt{T/\rho}$; fixed–fixed eigenpairs $\omega_n=n\pi c/L$, $Y_n=\sin(n\pi x/L)$; modal expansion with $\dot q_n(0)=\frac2L\int_0^L\dot v_0 Y_n\,dx$; $q_n(t)=B_n\sin\omega_n t$ when $v_0=0$.

Why this works: Separation of variables on $\rho v_{tt}=T v_{xx}$ with zero-end conditions forces sine mode shapes and the discrete frequency ladder $\omega_n\propto n$. The modes are orthogonal, so the given velocity profile projects onto each independently; a symmetric (parabolic) profile has a large first-mode content and zero even-mode content, which is why retaining only the first mode is a good approximation here.

Common pitfall: using cantilever/beam formulas (a power line has negligible bending stiffness — it is a string, governed by tension, not $EI$); or forgetting the $2/L$ normalising factor in the modal-velocity integral. Note even modes ($n=2,4,\dots$) are absent because the parabola is symmetric about mid-span.

Source: Rao, Mechanical Vibrations Ch. 8 (continuous systems, transverse vibration of a string); Meirovitch Ch. 8.

Question 7 of 7: Lateral vibration of a power line (string)

Question 7: Lateral vibration of a power line (string) (20 marks)