This problem asks us to find the ratio of natural frequencies for a torsional pendulum system under two different assumptions about the shaft's inertia. A torsional pendulum consists of a uniform shaft fixed at one end and carrying a disc at the other end. The natural frequency is a crucial parameter in understanding the vibration behavior of such a mechanical system.

Understanding Torsional Pendulum Natural Frequency

The natural frequency \((f)\) of a torsional pendulum is determined by the torsional stiffness \((K)\) of the shaft and the equivalent moment of inertia \((I_{eq})\) of the rotating components. The formula for angular natural frequency \(( \omega )\) is:

\( \omega = \sqrt{\frac{K}{I_{eq}}} \)

And the natural frequency in Hz is related to the angular frequency by:

\( f = \frac{\omega}{2\pi} \)

The torsional stiffness \(K\) for a uniform circular shaft is given by:

\( K = \frac{GJ}{L} \)

where:

• \(G\) is the shear modulus of the shaft material.
• \(J\) is the polar moment of inertia of the shaft's cross-section.
• \(L\) is the length of the shaft.

Since the shaft properties (G, J, L) remain constant in both scenarios, the torsional stiffness \(K\) will be the same for both 'fa' and 'fb' calculations.

Calculating Natural Frequency 'fa' (Massless Shaft)

In this scenario, the uniform shaft is assumed to be massless. This means its own moment of inertia is considered negligible and does not contribute to the system's total inertia. The entire moment of inertia for the system comes only from the disc, which is given as I.

Therefore, the equivalent moment of inertia for this case, \(I_{eq,a}\), is:

\( I_{eq,a} = I \)

The angular natural frequency for scenario 'a' \(( \omega_a )\), considering the massless shaft, is:

\( \omega_a = \sqrt{\frac{K}{I_{eq,a}}} = \sqrt{\frac{K}{I}} \)

And the natural frequency 'fa' is:

\( f_a = \frac{1}{2\pi}\sqrt{\frac{K}{I}} \)   (Equation 1)

Calculating Natural Frequency 'fb' (Shaft with Moment of Inertia)

For the second scenario, the uniform shaft is considered to have a moment of inertia. The problem states that the shaft has "the same moment of inertia as that of the disc". This means the total moment of inertia of the shaft, if treated as a rigid body rotating about its axis, is also I.

However, for a uniform shaft acting as part of a torsional pendulum with a disc at its free end, the effective moment of inertia of the shaft that contributes to the system's inertia is not its total moment of inertia \((I_{shaft})\) but rather one-third of its total moment of inertia. This is due to the distributed nature of the shaft's mass and how it contributes to the overall equivalent inertia of the torsional system.

So, the effective moment of inertia of the shaft \((I_{shaft,effective})\) is:

\( I_{shaft,effective} = \frac{1}{3}I_{shaft} \)

Given that \(I_{shaft} = I\), we have:

\( I_{shaft,effective} = \frac{1}{3}I \)

The equivalent moment of inertia for this case, \(I_{eq,b}\), includes both the moment of inertia of the disc and the effective moment of inertia of the shaft:

\( I_{eq,b} = I_{disc} + I_{shaft,effective} \)

\( I_{eq,b} = I + \frac{1}{3}I = \frac{4}{3}I \)

The angular natural frequency for scenario 'b' \(( \omega_b )\), considering the shaft's moment of inertia, is:

\( \omega_b = \sqrt{\frac{K}{I_{eq,b}}} = \sqrt{\frac{K}{\frac{4}{3}I}} = \sqrt{\frac{3K}{4I}} \)

And the natural frequency 'fb' is:

\( f_b = \frac{1}{2\pi}\sqrt{\frac{3K}{4I}} \)   (Equation 2)

Finding the Ratio fa/fb

Now, we need to find the ratio of the natural frequency 'fa' to the natural frequency 'fb'. We will divide Equation 1 by Equation 2:

\( \frac{f_a}{f_b} = \frac{\frac{1}{2\pi}\sqrt{\frac{K}{I}}}{\frac{1}{2\pi}\sqrt{\frac{3K}{4I}}} \)

We can cancel out the common terms \(\frac{1}{2\pi}\) from both the numerator and the denominator:

\( \frac{f_a}{f_b} = \frac{\sqrt{\frac{K}{I}}}{\sqrt{\frac{3K}{4I}}} \)

Combine the terms under a single square root:

\( \frac{f_a}{f_b} = \sqrt{\frac{\frac{K}{I}}{\frac{3K}{4I}}} \)

To simplify the complex fraction inside the square root, multiply the numerator by the reciprocal of the denominator:

\( \frac{f_a}{f_b} = \sqrt{\frac{K}{I} \times \frac{4I}{3K}} \)

Cancel out the common terms \(K\) and \(I\):

\( \frac{f_a}{f_b} = \sqrt{\frac{4}{3}} \)

Conclusion on Natural Frequency Ratio

The ratio of the natural frequency 'fa' (when the shaft is massless) to 'fb' (when the shaft has the same moment of inertia as the disc) for the torsional pendulum system is \( \sqrt{\frac{4}{3}} \). This result highlights how considering the moment of inertia of the shaft affects the overall natural frequency of the system, reducing it due to increased effective inertia.