Correct Answer - Option 1 : 279.5 Hz
Concept:
Elongation due to axial load is
\(\delta l = \frac{{PL}}{{AE}}\) ------ (1)
Let, k be the stiffness of the material then
P = kδl ------ (2)
Now, substituting equation (2) in equation (1);
\(\therefore k = \frac{{AE}}{L}\)
Frequency (f) \( = \frac{{{\omega _n}}}{{2\pi }}\)
Where, ωn \(= \sqrt {\frac{k}{m}}\)
Calculation:
Given:
d = 100mm; L = 1000mm; W = 5000N; E = 2 ×105 N/mm2 = 2 × 1011 N/m2
\(k = \frac{{AE}}{L}\)
\(k = \frac{{\frac{\pi }{4}{{\left( {0.1} \right)}^2} \times 2 \times {{10}^{11}}}}{1}\)
∴ k = 1570796327 N/m.
Now,
\({\omega _n} = \sqrt {\frac{k}{m}} = \sqrt {\frac{{1570796327}}{{\frac{{5000}}{{9.81}}}}}\)
ωn = 1755.535 rad/s
\({f_n} = \frac{{{\omega _n}}}{{2\pi }} = \frac{{1755.535}}{{2\pi }}\)
∴ fn = 279.402 Hz.