Question

Relation between angular frequency and young's modulus

Answer 1

Answer:

The frequency is a function of the dimensions of the bar and its Young's modulus.

You need to know what mode of oscillation you are exciting in your bar - there is a hug difference between the flexural and longitudinal modes.

If the rod is bending, you can find the formulas here. The derivation goes on and on... but you should be able to use the formula on the first page

$f = \frac{1}{2\pi } [\frac{22.373}{L^{2} } ] \sqrt\frac{EI}{rho}$

In this formula, I is the second moment of area of the rod - see the wiki article for an explanation and to find the appropriate value for the shape of your rod.

If you have a higher mode, you can find the position of two fixed nodes and use the fixed-fixed equation instead.

And if you have longitudinal vibration, the answer is much simpler - you just have to look at the transit time of the sound wave from end to end. One round trip corresponds to the fundamental frequency, so

$f = \frac{v}{2L} = \sqrt\frac{E}{4L^{2}rho }$

$E = Rho (2 L f)^{2}$

Explanation: rho here is the symbol ρ