Free Vibration of Undamped SDOF Systems

Natural frequency of SDOF systems

A single-degree-of-freedom (SDOF) system is described by one coordinate. An undamped linear SDOF system obeys m_{eq}\ddot{x} + k_{eq}x = 0, where m_{eq} is the equivalent mass and k_{eq} the equivalent stiffness, both referred to the chosen coordinate. Its natural circular frequency is \omega_n = \sqrt{k_{eq}/m_{eq}} in rad/s, its cyclic frequency is f_n = \omega_n/2\pi in Hz, and its period is T = 2\pi/\omega_n in seconds. Keep the units separate, because an answer requested in Hz must be divided by 2\pi from a value in rad/s.

Since \omega_n^2 = k/m depends only on the ratio of stiffness to mass, any two pairs with the same ratio give the same natural frequency. Doubling both k and m leaves \omega_n unchanged, while doubling only m lowers it by \sqrt{2}.

The quickest route to m_{eq} and k_{eq} is the energy method. Write the kinetic energy as T = \frac{1}{2}m_{eq}\dot{x}^2 and the potential energy as U = \frac{1}{2}k_{eq}x^2, measured from the static equilibrium position so that gravity contributes only a constant and drops out. The same answer follows for any coordinate, provided T and U are written in that one coordinate.

When a string or pulley links several bodies, first express every velocity in the single coordinate. An inextensible string forces the speeds to be proportional. Each body's kinetic energy is then multiplied by the square of its speed ratio before the equivalent mass is summed, so the equivalent mass is generally not the plain sum of the physical masses. A movable pulley that makes one body move a fraction of the other's displacement is a typical case of this scaling.

Springs scale in the same way. If the elongation of a spring equals n times the displacement of the mass, its energy \frac{1}{2}k(nx)^2 gives an effective stiffness n^2 k. The rope or pulley arrangement, not the spring alone, determines the factor n.

For a rigid body that rotates, the kinetic energy is \frac{1}{2}J\dot{\theta}^2, where J is the mass moment of inertia about the rotation point. A spring at lever arm a gives a torsional stiffness k_t = ka^2, and the natural frequency is \omega_n = \sqrt{k_t/J}. A slender bar hinged at one end, with a spring at its centre, uses J = ML^2/3 about the hinge and a lever arm of L/2. The hinge therefore changes both the inertia and the lever arm, and the result differs from the linear case with the same spring.

Rolling without slipping couples translation and rotation. The contact point has zero velocity, so \dot{\theta} = \dot{x}/r, and the kinetic energy becomes \frac{1}{2}\left(m + I_G/r^2\right)\dot{x}^2. For a homogeneous solid cylinder I_G = \frac{1}{2}mr^2, so the equivalent mass is \frac{3}{2}m when the centre is attached to a spring. Omitting the rotational kinetic energy is the most common error in this type of problem.

For a vertical spring-mass system the static deflection is \delta_{st} = W/k = mg/k, and the period follows from it as T = 2\pi\sqrt{m/k} = 2\pi\sqrt{\delta_{st}/g}. A measured static deflection therefore gives the natural frequency directly, with g in m/s^2.

When a spring acts in a non-vertical direction, the static deflection must be found from equilibrium. The spring force must balance the component of the applied force along the spring direction, and this balance fixes the geometric angle of the static position before the dynamic analysis is made.