--- title: "Rod longitudinal and shaft torsional frequencies" subject: "Theory of Mechanism and Machine II" chapter: 10 key: "semester_7/theory_of_mechanism_and_machine_ii:longitudinal-rod-torsional-shaft-frequencies" source: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/longitudinal-rod-torsional-shaft-frequencies site: BCT Engineering --- # Rod longitudinal and shaft torsional frequencies Theory of Mechanism and Machine II · Chapter 10: Vibration of Continuous Systems - Asked in: year not given, 4+4 marks - Video: [Lect 41 Frequency equation of longitudinal vibration of bar subjected to inertia force and spring fo](https://www.youtube.com/watch?v=Kn2vG5ekQ2k) — Amit Ramesh Bhende - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/longitudinal-rod-torsional-shaft-frequencies - Whole chapter: https://www.bctengineer.com/md/semester_7/theory_of_mechanism_and_machine_ii/chapter-10 ## Question (a) Derive the equation of motion for longitudinal vibration of a uniform rod and obtain the natural frequencies of a rod fixed at one end and free at the other. (b) A steel rod 2 m long is fixed at one end and free at the other ($E = 200$ GPa, $\rho = 7850$ kg/m³). Find the first three longitudinal natural frequencies. Also find the first two torsional natural frequencies of a steel shaft of length 1.2 m, fixed at one end and free at the other ($G = 80$ GPa). ## Short answer Longitudinal wave speed c = √(E/ρ); for fixed-free rod, f_n = (2n-1)c/(4l). With E=200 GPa, ρ=7850 kg/m³, l=2 m: f₁=630.9 Hz, f₂=1892.8 Hz, f₃=3154.7 Hz. Torsional wave speed c_t = √(G/ρ); same ρ and G=80 GPa, l=1.2 m gives f₁=665.1 Hz, f₂=1995.2 Hz. ## Answer #### (a) Longitudinal vibration of a rod For a rod of area $A$, modulus $E$ and density $\rho$, let $u(x,t)$ be the axial displacement. The axial force is $P = EA\dfrac{\partial u}{\partial x}$. For an element $dx$: $$ \left(P + \frac{\partial P}{\partial x}dx\right) - P = \rho A\,dx\,\frac{\partial^2 u}{\partial t^2} \;\Rightarrow\; \frac{\partial^2 u}{\partial t^2} = c^2\frac{\partial^2 u}{\partial x^2}, \qquad c = \sqrt{\frac{E}{\rho}} $$ With $u = U(x)\sin\omega t$, $\;U = A_1\cos\dfrac{\omega x}{c} + B_1\sin\dfrac{\omega x}{c}$. **Fixed at $x = 0$, free at $x = l$:** - Fixed end: $U(0) = 0 \Rightarrow A_1 = 0$. - Free end (no stress): $\dfrac{dU}{dx}(l) = 0 \Rightarrow \cos\dfrac{\omega l}{c} = 0$. $$ \frac{\omega_nl}{c} = \frac{(2n-1)\pi}{2} \;\Rightarrow\; f_n = \frac{(2n-1)c}{4l}, \quad n = 1, 2, 3, \ldots $$ Mode shape: $U_n = B\sin\dfrac{(2n-1)\pi x}{2l}$. (For fixed-fixed or free-free ends, $f_n = nc/(2l)$.) #### (b) Numerical values **Longitudinal:** $$ c = \sqrt{\frac{200\times10^9}{7850}} = 5047.5\ \text{m/s} $$ $$ f_1 = \frac{c}{4l} = \frac{5047.5}{4 \times 2} = 630.9\ \text{Hz}, \quad f_2 = 3f_1 = 1892.8\ \text{Hz}, \quad f_3 = 5f_1 = 3154.7\ \text{Hz} $$ **Torsional shaft:** The equation is the same with $E \to G$ and $u \to \theta$, so the wave speed is $c_t = \sqrt{G/\rho}$. Taking $\rho = 7850$ kg/m³ for steel: $$ c_t = \sqrt{\frac{80\times10^9}{7850}} = 3192.3\ \text{m/s} $$ $$ f_1 = \frac{c_t}{4l} = \frac{3192.3}{4 \times 1.2} = 665.1\ \text{Hz}, \qquad f_2 = 3f_1 = 1995.2\ \text{Hz} $$ The torsional frequencies do not depend on the shaft diameter, because $I_p$ cancels out of the equation. **Answer:** Longitudinal: $630.9$, $1892.8$, $3154.7$ Hz. Torsional: $665.1$ and $1995.2$ Hz. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The longitudinal wave equation for a rod has wave speed: A. sqrt(rho/E) B. E/rho C. sqrt(E/rho) D. sqrt(E A/rho) Answer: C. From EA u_xx = rho A u_tt, c = sqrt(E/rho). (Key point: Wave speed in rod) 2. For a rod fixed at x = 0 and free at x = l, the free-end condition and frequency equation are: A. U(l) = 0, sin(w l/c) = 0 B. dU/dx(l) = 0, cos(w l/c) = 0 C. dU/dx(l) = 0, sin(w l/c) = 0 D. U(l) = 0, cos(w l/c) = 0 Answer: B. A free end carries no stress so dU/dx = 0, giving cos(w l/c) = 0. (Key point: Fixed-free boundary conditions) 3. For the fixed-free rod, the higher frequencies are in the ratio: A. 1 : 2 : 3 B. 1 : 3 : 5 C. 1 : 4 : 9 D. 1 : 2 : 4 Answer: B. f_n = (2n-1) c/4l, so f2 = 3 f1 and f3 = 5 f1 (630.9, 1892.8, 3154.7 Hz). (Key point: Frequency sequence) 4. For the torsional shaft, the wave speed is c_t = sqrt(G/rho) and the frequencies: A. depend strongly on shaft diameter B. depend on diameter squared C. are independent of diameter because I_p cancels out D. need E instead of G Answer: C. The equation has the same form with E replaced by G; I_p cancels so diameter does not matter. 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