Answer:
The natural angular frequency of the rod is 53.56 rad/sec
Explanation:
Since the beam is free at one end and fixed at the other hence the beam is a cantilevered beam as shown in the attached figure
We know that when a unit force is placed at the end of a cantilever the displacement of the free end is given by
[tex]\Delta x=\frac{PL^3}{3EI}[/tex]
Hence we can write
[tex]P=\frac{3EI\cdot \Delta x}{L^3}[/tex]
Comparing with the standard spring equation [tex]F=kx[/tex] we find the cantilever analogous to spring with [tex]k=\frac{3EI}{L^3}[/tex]
Now the angular frequency of a spring is given by
[tex]\omega =\sqrt{\frac{k}{m}}[/tex]
where
'm' is the mass of the load
Thus applying values we get
[tex]\omega _{beam}=\sqrt{\frac{\frac{3EI}{L^{3}}}{Area\times density}}[/tex]
[tex]\omega _{beam}=\sqrt{\frac{\frac{3\times 20.5\times 10^{10}\times \frac{0.1\times 0.3^3}{12}}{5.9^{3}}}{0.3\times 0.1 \times 7830}}=53.56rad/sec[/tex]
