弹性势能

Okay, so now I'm gonna present an argument for calculating kinetic energy in a spring-block system when the spring itself is massive.

Let's consider the mass of spring to be $\Large m$. So let's begin.

So let us take a small element of mass $dm$ of the spring, so total energy can be calculated by adding all the length elements' kinetic energy, and requires the following integral:

$$E = \int \dfrac{1}{2}u^{2} \ dm$$

Since the spring is uniform,

$$dm = \left(\dfrac{dy}{L}\right) \cdot m,$$

so $E = \displaystyle\int_{0}^{L} \dfrac{1}{2}u^{2} \cdot \left(\dfrac{dy}{L}\right) \cdot m$.

The velocity of each mass element of the spring is directly proportional to its length, i.e.

$u =\dfrac {vy}{L}$

$E = \dfrac{1}{2} \dfrac{m}{L} \cdot \displaystyle\int_{0}^{L} \left(\dfrac{vy}{L}\right)^{2} \ dy$

After some integration, this gives

$$E= \dfrac{1}{2} \dfrac{m}{3} v^{2}.$$

Comparing to the expected original kinetic energy formula:

$$\text{K.E.} = \dfrac{1}{2}mv^{2}$$

we can conclude that effective mass of spring in this case is $\dfrac{m}{3}$.

The effective mass of the spring in a spring-mass system when using an ideal spring of uniform linear density is $\dfrac{1}{3}$ of the mass of the spring.

And we're done!

Suppose that we have a spring of spring constant $k$, and we stretch it from its relaxed length by a displacement $x$.

We would like to find the potential energy in this configuration.

To do so, we will release the spring and let it go to equilibrium, at $x=0$.

We use Hooke's Law!

By Hooke's Law, $$F(x)=-kx$$ where $F(x)$ is the spring force for a certain value of $x$.

Calculating the work done by the spring force (and using a dummy variable to avoid confusion), $$W=\int_{x}^{0}F(t)dt$$ $$W=\int_{x}^{0}-ktdt$$ $$W=\left[-\frac{1}{2}kt^2\right]_{x}^{0}$$ $$W=\frac{1}{2}kx^2$$ We may verify that we have not made sign errors by noticing that the force and the motion are in the same direction, so the work done is positive.

By the Work-Energy Theorem, this work is equal to the increase in kinetic energy. But since energy is conserved, there is an equal drop in potential energy.

Thus, we have the following: $$U_0-U_x=-\frac{1}{2}kx^2$$ But at displacement $x=0$, the spring is relaxed! Thus, the potential energy is zero: $$U_x=\frac{1}{2}kx^2$$ The potential energy in a spring is equal to one half times the spring constant times the displacement squared!

Of course, in the general case of an elastic material, the derivation and corresponding equations are much more complicated. However, the use of Hooke's Law is the central component in any case.