Young's Modulus: Vibrations in a rod

[AJKing]

Homework Statement

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

/Do not solve/ please help me find what it is asking.

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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:

[itex]y = \frac{4L^3}{Yab^3}F[/itex]

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)?

 

[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?

 

[SteamKing]

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?

 

[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.

 

[Nickie]

The question is asking for the ratio of the frequencies of vibration in the y and x directions of the rod when a mass m is attached to the top end of the rod. To find this ratio, we need to use the equation for the frequency of vibration, given by:

f = \frac{1}{2\pi}\sqrt{\frac{Y}{\rho}}\frac{w}{L}

where Y is the Young's modulus, \rho is the density of the material, w is the width of the rod (side b), and L is the length of the rod.

In this case, the mass attached to the top end of the rod will change the effective length of the rod, and therefore will affect the frequency of vibration. The ratio of the frequencies can be found by taking the ratio of the frequencies in the y and x directions, given by:

\frac{f_y}{f_x} = \frac{\sqrt{\frac{Y}{\rho}}\frac{w}{L}}{\sqrt{\frac{Y}{\rho}}\frac{L}{w}} = \frac{w^2}{L^2}

Therefore, the ratio of the frequencies of vibration in the y and x directions will be equal to the square of the ratio of the width to the length of the rod. This can also be written as the square of the ratio of the sides of the rectangular cross section, which in this case is given by b^2/a^2. So, the final answer would be:

\frac{f_y}{f_x} = \left(\frac{b}{a}\right)^2