# Natural Frequency of Spring

1. May 4, 2009

### Sami Lakka

Hi

Natural frequency of spring is fn = 1/(2*pi) * sqrt(k/m), where k = spring constant and m is the mass.

Ok, so I have a spiral spring which means that spring is coiled couple of times. Its spring constant is (according to Hartog's Mech. Vibrations), k = E*I/l, E=Elastic Modulus, I = Inertia of cross section and l = length. All good at this point.

Now I straighten the coil so I get an cantilevered beam. Its natural frequency is (according Hartog):
f = 1/(2*pi) * sqrt((3*E*I)/(l^3*m))

So assume we have a beam with 1 mm width and 10 mm height. Lets plug couple of values:
E=22*10^9 Pa, I = b^3*h*1/12, b=1*10^-3 m, h= 10*10^-3 m, l = 40*10^-3 m
m = 0.01 kg (just some value)
f (spiral spring) = 1 Hz
f (cantilevered beam) = 46.6 Hz (??)

Why is the natural frequency of the cantilever beam so much higher than the spiral spring? Am I using wrong equations?

2. May 4, 2009

### minger

For a standard helical spring, Shigley states the stiffness is:
$$k = \frac{d^4 G}{8D^3 N}$$
Where d is the wire diameter, G is the shear modulus, D is the spring diameter, and N is the number of coils. Your formula doesn't take into consideration the number of coils. May I ask where you got it from.