# Young's Modulus: Vibrations in a rod

1. Jun 9, 2013

### AJKing

1. The problem statement, all variables and given/known data

From A.P. French, question 3-10.
I'm having trouble decoding part b.

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A metal rod, 0.5m long, has a rectangular cross section of 2mm2.

(a) This question provided me with information to calculate young's modulus for the metal Y = 5.9*1011 N/m2

(b) The rod is firmly clamped at the bottom, and at the top a force F is applied in the y direction [perpendicular to side a, parallel to side b]. The result is a static deflection, y, given by:

$y = \frac{4L^3}{Yab^3}F$

If the force is removed and a mass m, which is much greater than the mass of the rod, is attached to the top end of the rod, what is the ratio of the frequencies of vibration in the y and x directions (i.e., parallel to edges of length b and a)?

2. Jun 9, 2013

### AJKing

SOLUTION: b/a.

I don't understand why that is.

I don't understand why there is vibration at all - where is this energy coming from?

What is happening here?

3. Jun 9, 2013

### SteamKing

Staff Emeritus
When the rod is clamped at one end and then a force is used to deflect the other end, it takes a certain amount of work or energy to cause this deflection. Once the force is released, and assuming the rod has only undergone an elastic deflection, then the rod will try to recover its undeflected shape. When deflected, the rod acts as a spring, storing the energy which caused it to deflect. When the end of the rod is released, so is the stored energy. Because there is no restraint on the motion of the free end of the rod, after release of the free end, the rod will vibrate about its original undeflected position until the energy stored within the rod dissipates.

Haven't you ever seen a tuning fork?

4. Jun 9, 2013

### AJKing

Ahh - poor interpretation.

What I understood was:

> An applied force reveals these properties. Now consider an idle system of these properties with a mass on top.

What it's clearly saying is:

> An applied force reveals these properties. Now put a mass on top and apply the force again.