Formula for Natural Frequency of Cantilever Beam

  • Thread starter Oscar6330
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  • #1
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I found two different versions of equations to find natural frequency of a cantilever beam. I am not sure which one is right. I would appreciate if someone could make things a bit clear here


F1= k^2*sqrt(E*I/(mpl*L^4))/(2*pi) where k=1.875 for first natural freq and I= b*d^3/12;

OR is it

F1= k^2*sqrt(E*I/(mpl*L^4)) where k=1.875 for first natural freq and I= b*d^3/12;

Basically I am not sure why some equations have /(2*pi) while others do not and which one is correct
 

Answers and Replies

  • #2
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Angular frequency , [itex]\omega[/itex] radians /second = 2[itex]\pi[/itex]f cycles per second ?
 
  • #3
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Angular frequency , [itex]\omega[/itex] radians /second = 2[itex]\pi[/itex]f cycles per second ?

Thanks. So the first equation gives freq in Hz while other one gives answer in radians per second?
 
  • #4
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Yes the solution to the governing differential equation for say a cantilever with mass m at the end is

[tex]m\ddot x + kx = 0[/tex]

which has solution

[tex]\omega = \sqrt {\frac{k}{m}} [/tex]

where omega is in rads/second
 
Last edited:
  • #5
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Yes the solution to the governing differential equation for say a cantilever with mass m at the end is

[tex]m\ddot x + kx = 0[/tex]

which has solution

[tex]\omega = \sqrt {\frac{k}{m}} [/tex]

where omega is in rads/second

thanks for your kind reply.

So

F1= k^2*sqrt(E*I/(mpl*L^4))/(2*pi)


w1=k^2*sqrt(E*I/(mpl*L^4))
 

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