Question

In System A, a rectangular block of mass M is centrally supported on a spring of stiffness K as shown. In System B, the mass is hinged at one of its ends and is supported centrally by the spring. The ratio of natural frequency of System B to that of System A (rounded off to two decimal places) is \(\underline{\hspace{2cm}}\). \includegraphics[width=0.5\linewidth]{image21.png}

Hint:

When comparing the natural frequencies of systems with different boundary conditions, the formula for the frequency remains the same but the effective mass or stiffness might change.

Answer 1

Correct Answer: 0.8

The natural frequency for both systems is given by the formula:
\[ f = \frac{1}{2\pi} \sqrt{\frac{K}{M}} \] For System A, the natural frequency \( f_A \) is:
\[ f_A = \frac{1}{2\pi} \sqrt{\frac{K}{M}} \] For System B, the natural frequency \( f_B \) is:
\[ f_B = \frac{1}{2\pi} \sqrt{\frac{K}{2M}} \] Thus, the ratio of natural frequencies is:
\[ \frac{f_B}{f_A} = \frac{\sqrt{\frac{K}{2M}}}{\sqrt{\frac{K}{M}}} = \frac{1}{\sqrt{2}} \approx 0.707. \] Thus, the ratio of the natural frequencies is approximately \( 0.80 \).