A uniform rigid prismatic bar of total mass \( m \) is suspended from a ceiling by two identical springs as shown in the figure.
Let \( \omega_1 \) and \( \omega_2 \) be the natural frequencies of mode I and mode II respectively (with \( \omega_1<\omega_2 \)).
The value of \( \omega_2/\omega_1 \) is _________ (rounded off to one decimal place).
Show Hint
In multi-spring systems, the natural frequencies for each mode depend on the spring constants and the mass. The ratio \( \omega_2/\omega_1 \) can be derived using the formulas for each mode's frequency.
For the system of two springs with identical spring constants and a mass \( m \), the natural frequencies for modes I and II can be derived from the equations for a two-degree-of-freedom system.
For mode I, the natural frequency \( \omega_1 \) is given by:
\[
\omega_1 = \sqrt{\frac{k}{m}},
\]
where \( k \) is the spring constant, and \( m \) is the total mass.
For mode II, the natural frequency \( \omega_2 \) is given by:
\[
\omega_2 = \sqrt{\frac{2k}{m}}.
\]
Thus, the ratio of \( \omega_2/\omega_1 \) is:
\[
\frac{\omega_2}{\omega_1} = \frac{\sqrt{\frac{2k}{m}}}{\sqrt{\frac{k}{m}}} = \sqrt{2} \approx 1.414.
\]
Rounded to one decimal place, the ratio is:
\[
\boxed{1.4}.
\]
