Answer:
Natural frequency=21.40 Hz
Time= 0.2936 seconds
Explanation:
Idealizing the question as a cantilever beam with point load of mass M as 20 tons
Lateral stiffness, [tex]k=\frac {3EI}{l^{3}}[/tex] where l is length given as 10 m, E is Young’s modulus given as 30GPa and I is inertia where for a circular cross-section is given by [tex]\frac {\pi d^{4}}{64}[/tex]
k=[tex]\frac {3*(30*10^{9})*(\pi *1.2^{4})}{64*10^{3}}[/tex]= 9160884.178
k= [tex]9.160884178*10^{6}[/tex]
To find the frequency, [tex]w_{n}[/tex], the mass m is given as 20 tons or 20000 Kg
[tex]w_{n}=\sqrt (\frac {k}{m})= \sqrt (\frac {9.160884178*10^{6}}{20000})[/tex]=21.40196741 Hz
Natural frequency=21.40 Hz
Time period,
T=[tex]\frac {2\pi}{w_{n}}=\frac {2\pi}{21.40196741}[/tex]=0.2935798 seconds
T=0.2936 seconds
