--- title: "Natural frequency by energy method" subject: "Theory of Mechanism and Machine II" chapter: 6 key: "semester_7/theory_of_mechanism_and_machine_ii:rayleigh-energy-method-spring-mass" source: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/rayleigh-energy-method-spring-mass site: BCT Engineering --- # Natural frequency by energy method Theory of Mechanism and Machine II · Chapter 6: Vibration of Single Degree of Freedom Systems - Asked in: year not given, 5 marks - Video: [Natural Frequency Of Free Longitudinal Vibrations (Energy Method) | Theory Of Machines](https://www.youtube.com/watch?v=pMXeyu5xH_s) — LEARN AND GROW - 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/rayleigh-energy-method-spring-mass - Whole chapter: https://www.bctengineer.com/md/semester_7/theory_of_mechanism_and_machine_ii/chapter-6 ## Question Using the energy method, derive the natural frequency of a spring-mass system. Hence show the effect of the mass of the spring on the natural frequency. ## Short answer For massless spring, equating maximum kinetic energy ½ m ω_n² X² to maximum potential energy ½ k X² gives ω_n = √(k/m). Including spring mass m_s, assuming linear velocity along the spring, the effective mass becomes m + m_s/3, so ω_n = √[k/(m + m_s/3)]. ## Answer For a conservative system with no damping, the total energy stays constant: $$ T + U = \text{constant} $$ #### Spring-mass (massless spring) Let the mass $m$ be displaced by $x$ from the static equilibrium position and have velocity $\dot{x}$. $$ T = \tfrac12 m\dot{x}^2, \qquad U = \tfrac12 k x^2 $$ (The change of potential energy of gravity is cancelled by the static spring force, so only the extra spring energy $\tfrac12kx^2$ remains.) Differentiating $T + U = $ constant: $$ m\dot{x}\ddot{x} + kx\dot{x} = 0 \;\Rightarrow\; m\ddot{x} + kx = 0 $$ $$ \omega_n = \sqrt{\frac{k}{m}}, \qquad f_n = \frac{1}{2\pi}\sqrt{\frac{k}{m}} = \frac{1}{2\pi}\sqrt{\frac{g}{\delta_{st}}} $$ Alternatively, with $x = X\sin\omega_n t$, the maximum kinetic energy $\tfrac12 m\omega_n^2X^2$ equals the maximum potential energy $\tfrac12 kX^2$, which gives the same result. #### Effect of the spring's own mass Let the spring have mass $m_s$ and length $l$. A small element $dy$ at distance $y$ from the fixed end moves with velocity $\dfrac{y}{l}\dot{x}$ (assuming a linear variation). $$ T_{spring} = \int_0^l \tfrac12\left(\frac{m_s}{l}dy\right)\left(\frac{y\dot{x}}{l}\right)^2 = \tfrac12\cdot\frac{m_s}{3}\dot{x}^2 $$ Total kinetic energy $T = \tfrac12\left(m + \dfrac{m_s}{3}\right)\dot{x}^2$. Hence $$ \omega_n = \sqrt{\frac{k}{m + m_s/3}} $$ One-third of the spring mass is added to the suspended mass as an **effective mass**. The natural frequency decreases slightly; the correction is negligible when $m_s \ll m$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What principle gives the equation of motion by the energy method? A. T + U = constant for a conservative undamped system B. T - U = constant C. T = U at all times D. Force = zero Answer: A. Differentiating T + U = constant gives m x'' + k x = 0. (Key point: Energy method) 2. The resulting natural frequency of the spring-mass system is: A. sqrt(k/m) B. sqrt(m/k) C. k/m D. 2 pi sqrt(k/m) Answer: A. m x'' + k x = 0 gives w_n = sqrt(k/m). (Key point: Natural frequency) 3. Taking the spring's own mass m_s into account, the effective mass added to the suspended mass is: A. m_s/3 B. m_s/2 C. m_s D. m_s/4 Answer: A. Integrating the kinetic energy of a linearly varying velocity gives 1/2 (m_s/3) x'^2. (Key point: Spring mass effect) 4. Including the spring mass makes the natural frequency: A. Slightly lower: w_n = sqrt(k/(m + m_s/3)) B. Higher C. Unchanged D. Zero Answer: A. Effective mass increases, so frequency decreases slightly; negligible if m_s << m. 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