Natural Stretching of a metal rod Let’s say a metal rod of Young’s modulus $Y$ and cross section $A$ with length $l$ was hung from the top of a ceiling straight down. By the theory of Young’s modulus, the metal rod should lengthen by some length $\delta l$ just by the gravitational force, how much would the rod be stretched by?
Also, if a mass $m$ was attached to the bottom of the rod, would the rod display vertical oscillations like a spring attempting to bring the mass back to equilibrium position? Would the same be true if there was no mass at all attached to the bottom of the rod?
 A: I will attempt to answer your questions in order.

Let’s say a metal rod of Young’s modulus Y and cross section A with length l was hung from the top of a ceiling straight down. By the theory of Young’s modulus, the metal rod should lengthen by some length δl just by the gravitational force, how much would the rod be stretched by?

1. The rod will naturally stretch by a length $\Delta l$ as you have said.
We can take the average of the sums of all the stresses on the rod and treat this as the overall stress given to the rod. This is because the top of the rod is under the most stress while the bottom is under almost none. And since the stress doesn't increase exponentially, but rather linearly, the effective stress upon the cylinder at the center will be treated as the overall stress given upon the rod. Assuming the metal rod has a radius $r$ and is uniform throughout, then the stress at the center will be given by (the force is given by the amount of weight the metal rod has to support from halfway)
$$\sigma=\frac{F}{A}=\frac{\pi r^2\rho g\frac{l}{2}}{\pi r^2}=\frac{1}{2}\rho gl.$$
We know that the stress-strain relationship gives us
$$\epsilon = \dfrac{\Delta l}{l} = \dfrac{1}{Y} \sigma.$$
Since we know $\sigma$, replacing our expression for sigma into the relation gives us
$$\frac{\Delta l}{l} = \dfrac{1}{2Y}\rho gl$$
$$\Delta l = \frac{1}{2Y}\rho gl^2$$

Also, if a mass m was attached to the bottom of the rod, would the rod display vertical oscillations like a spring attempting to bring the mass back to equilibrium position?

2. Yes, to be more specific, we can carry out the calculations for the frequency of the oscillations. The tension in the wire would defined as
$$T=PA=AY\frac{\Delta L}{L}$$
When the mass is distance $\Delta L$ down it will recieve a restoring force of $k\Delta{L}$. This means that the rod can be modeled as a spring with spring constant
$$k=\frac{T}{\Delta L}=Y\frac{A}{L}$$
The frequency of oscillations is then
$$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}=\frac{1}{2\pi}\sqrt{\frac{YA}{ml}}$$
The same will be true for no mass at all, but it will be rather easy now to do yourself with the calculations I have given.
A: If we take the coordinate $x$ pointing downward with its origin at the top of the rod, the tensile normal force carried at some location $x$ along the rod is $N(x) = \rho g A (l-x)$. The stress is just this divided by the cross-sectional area of the bar, and the strain is the stress divided by Young's modulus.  The strain is also related to the displacement of the bar by $\epsilon = du/dx$. You can then integrate the strain over the length of the bar to get a total stretch of the bar as $\rho g l^2 / 2 Y$.
$\delta = \int^{l}_0 \rho g (l-x)/Y dx=\rho g l^2 / 2Y$
The answers to your oscillation questions are yes and yes.  However, it is not exactly like a simple spring-mass system.  For example, for a uniform bar stretched at one end and then released, the displacement oscillations take a saw-tooth form. For the attached mass, it will approach the simple spring-mass system when the mass of the attached mass is much greater than the mass of the rod.
