Torsional Oscillator Calculator
Our rotational & periodic motion calculator computes torsional oscillator accurately. Enter measurements for results with formulas and error analysis.
Reviewed by Manoj Kumar, Mathematics Educator
Formula
omega_n = sqrt(kappa / I), zeta = c / (2 sqrt(kappa I))
Where omega_n is the natural frequency in rad/s, kappa is the torsional stiffness (Nm/rad), I is the mass moment of inertia (kg m^2), zeta is the damping ratio, and c is the damping coefficient (Nm s/rad). The damped frequency is omega_d = omega_n x sqrt(1 - zeta^2) for underdamped systems.
Worked Examples
Example 1: Engine Crankshaft Torsional Vibration
Problem:A crankshaft has torsional stiffness of 200 Nm/rad, moment of inertia 0.8 kg m^2, and damping coefficient 2.0 Nm s/rad. Find the natural frequency, damping ratio, and damped frequency.
Solution:Natural frequency: omega_n = sqrt(200/0.8) = sqrt(250) = 15.81 rad/s = 2.517 Hz\nPeriod = 1/2.517 = 0.3973 s\nCritical damping = 2 x sqrt(200 x 0.8) = 2 x 12.649 = 25.298 Nm s/rad\nDamping ratio = 2.0 / 25.298 = 0.0791 (underdamped)\nDamped frequency = 15.81 x sqrt(1 - 0.0791^2) = 15.81 x 0.9969 = 15.76 rad/s = 2.508 Hz
Result:Natural: 15.81 rad/s (2.517 Hz) | Damping ratio: 0.079 | Damped: 15.76 rad/s
Example 2: Torsion Pendulum for Physics Lab
Problem:A torsion pendulum has a wire with stiffness 0.5 Nm/rad and a disk with I = 0.01 kg m^2. Damping is negligible (c = 0.001). Initial displacement is 30 degrees.
Solution:Natural frequency: omega_n = sqrt(0.5/0.01) = sqrt(50) = 7.071 rad/s = 1.125 Hz\nPeriod = 0.889 s\nDamping ratio = 0.001 / (2 x sqrt(0.5 x 0.01)) = 0.001 / 0.1414 = 0.00707\nMax energy = 0.5 x 0.5 x (30 x pi/180)^2 = 0.5 x 0.5 x 0.2741 = 0.0685 J\nMax angular velocity = 0.5236 x 7.071 = 3.703 rad/s
Result:Period: 0.889 s | Damping ratio: 0.007 (underdamped) | Max energy: 0.069 J
Frequently Asked Questions
What is a torsional oscillator and where is it used?
A torsional oscillator is a mechanical system that undergoes rotational vibrations about its axis. It consists of a rotating mass (disk, flywheel, or rotor) connected to a restoring element such as a shaft or torsion spring that resists angular displacement. When the mass is displaced from its equilibrium angle and released, it oscillates back and forth around the rest position. Torsional oscillators appear in many engineering contexts including crankshafts in internal combustion engines, drive shafts in vehicles, turbine rotors, quartz crystal oscillators in watches and electronics, and torsion pendulums used in precision measurements. Understanding torsional oscillations is critical for preventing resonance failures in rotating machinery.
How is the natural frequency of a torsional oscillator calculated?
The natural frequency of a torsional oscillator is determined by the ratio of torsional stiffness to the moment of inertia. The formula is omega_n equals the square root of kappa divided by I, where kappa is the torsional stiffness in Newton-meters per radian and I is the mass moment of inertia in kilogram-meters squared. The result is in radians per second. To convert to Hertz (cycles per second), divide by 2 times pi. For example, a system with kappa = 50 Nm/rad and I = 0.5 kg m squared has omega_n = sqrt(50/0.5) = 10 rad/s, which equals 1.592 Hz with a period of 0.628 seconds. Higher stiffness increases frequency while greater inertia decreases it.
How does shaft geometry affect torsional stiffness?
Torsional stiffness of a solid circular shaft depends on its material shear modulus, diameter, and length according to the formula kappa = G times J divided by L, where G is the shear modulus (79.3 GPa for steel), J is the polar moment of inertia (pi times d to the fourth power divided by 32), and L is the shaft length. Stiffness increases with the fourth power of diameter, meaning a small increase in diameter dramatically raises stiffness. Doubling the shaft diameter increases stiffness by a factor of 16. Conversely, doubling the length halves the stiffness. For hollow shafts, the polar moment of inertia is pi times (outer diameter to the fourth minus inner diameter to the fourth) divided by 32. This relationship makes shaft sizing a critical engineering design decision.
References
Reviewed by Manoj Kumar, Mathematics Educator · Editorial policy