Consider a uniform shaft of length L fixed at its upper end and carrying a disc of the moment of inertia I at its lower end. The disc is twisted about the vertical axis and released. 'fa' is the natural frequency of the system when the shaft is assumed as massless, and 'fb' is the natural frequency of the system when the shaft is considered of the same moment of inertia as that of the disc. Find the ratio fa/fb.

Concept:-

Free torsional vibration:-

The natural frequency for torsional vibration, without considering the mass moment of inertia of the shaft, is given as, 

⇒ \(f = \sqrt{\frac{k_T}{q}}\;~\).....Eqn(1)

The natural frequency for torsional vibration, considering the mass moment of inertia of the shaft, is given as, 

⇒ \(f = \sqrt{\frac{k_T}{q+\frac{q_S}{3}}}\;~\)......Eqn(2)

Where, 

f = Natural frequency in rad/s

k_T = Torsional stiffness

q = Mass moment inertia of the disc attached at the end of the shaft

q_S = Mass moment inertia of the shaft

Calculation:

Given:

f_a = Natural frequency of the system when the shaft is assumed as massless

f_b = Natural frequency of the system when the shaft is considered of the same moment of inertia as that of the disc.

q_S = q

By using equation 1, we get

 \(fa = \sqrt{\frac{k_T}{q}}\;~\)

By using equation 2, we get

 \(fb = \sqrt{\frac{k_T}{q+\frac{q_S}{3}}}\;~\)

Then applying the given condition, i.e., q = q_S. 

 \(fb = \sqrt{\frac{k_T}{q+\frac{q}{3}}}=\sqrt{\frac{3k_T}{4q}}\;~\)

Then the ratio fa/fb we get as,

 \(\frac{fa}{fb}= \sqrt{\frac{4}{3}}\;~\)