What Is the Natural Angular Frequency of a Cantilevered Body?

[zorro]

Homework Statement

A body A weighing 22N is positioned on the end of a slender horizontal cantilever fixed at one end (mass of the plank is negligible). A force (vertically downwards) on A causes a deflection of 12.5mm. The natural angular frequency in rad/s of the body A is (nearly) ?

1)2
2)6
3)24
4)1.2

The Attempt at a Solution

F=kx
k=22/0.0125

w=(k/m)^1/2=28rad/s, not in the choices. Did I make any mistake?

 

[rock.freak667]

Usually your method would work for a spring, but I think a cantilever is a bit different.

If you look in this thread, you will find a different formula

https://www.physicsforums.com/showthread.php?t=275927

You can get the value for EI/L³ using

deflection = PL³/3EI where P = load and L = length of beam

 

[zorro]

What are W,E and I?

 

[rock.freak667]

What are W,E and I?

Actually now that I see that formula, it is essentially the same as what you did.

The most I can say is 24 rad/s would be the best answer given that you need to choose from those 4.

 

[rwisz]

Your calculation for the spring constant (k) is correct, but you have made a mistake in calculating the angular frequency (w). The correct formula for the natural angular frequency of a cantilever is given by w = (k/m)^0.5, where m is the mass of the body (not the plank). Since the mass of the body A is not given, we cannot calculate the exact angular frequency. However, we can use the given information to approximate the answer.

Assuming the mass of the body A is negligible compared to the mass of the plank, we can use the given weight of 22N to estimate the mass of the body as m = F/g = 22/9.8 = 2.24kg. Plugging this into the formula, we get w = (k/m)^0.5 = (22/0.0125/2.24)^0.5 = 11.1 rad/s. This is closest to option 2) 6 rad/s, which would be the correct answer for this approximation. So, the approximate natural angular frequency of the body A in this scenario is 6 rad/s.