# Elasticity of spring and conservation of energy [closed]

I have a problem of one of my students mechanics homework that I am unsure on and could do with resolving by tomorrow is possible! A spring of natural length $$0.5m$$ is attached to the ceiling and a particle of mass $$0.3kg$$ is attached to the other end. The modulus of elasticity of the spring is $$6N$$. The particle is raised so that it is $$0.4m$$ from the ceiling (less than the natural length of the spring) and dropped. The question wants to know the distance between the particle and the ceiling at the point which the particle first comes to rest. It also insists that we use the Conservation of Energy. Now clearly the particle has an initial and a final KE of $$0$$, and so its energy loss is equal to its change in GPE. I am assuming that this loss in mechanical energy is equal to the work done by the tension in the spring. However this tension is constantly changing and so I'm not sure how I would calculate this. I know we can use $$T=\frac{\lambda x}{L}$$ to calculate the tension at any given point, but this is not sufficient.

Could someone please shine some light on this for me, as it is not covered in any textbook I possess and I can't seem to resolve it with a google search.

• Why are you using the $T=\lambda x/L$ formula? Isn't the spring ideal? Also, your dimensions for the modulus of elasticity seem wrong, shouldn't it be force per unit area? You also need to convert this modulus to the spring constant $k$, for which (I'm guessing) you also need the cross-sectional area. Have you been provided with that? Jan 16, 2021 at 13:59
• HI Philip, thanks for your response. There is no other information given in the question. Why should $\lambda$ be force per area? $x$ and $L$ are both lengths and $T$ is force. Jan 16, 2021 at 14:06
• What are you calling $\lambda$? The modulus of elasticity? As far as I know, the dimensions of such a quantity should be $N m^{-2}$. But maybe I'm wrong. Anyway, if this is an ideal spring, all you need to do is find the spring constant $k$ for this problem, in which case the energy stored in a spring extended by some $x$ from equilibrium is just $1/2 k x^2$. Jan 16, 2021 at 14:15
• Yes $\lambda$ is the modulus of elasticity and yes, let's assume it's ideal. That's great. So how do I find the spring constant $k$? Jan 16, 2021 at 14:20
• A modulus of elasticity cannot be expressed in $\mathrm{N}$. Once you've figured out the spring constant this is a problem of Simple Harmonic Oscillation.
– Gert
Jan 16, 2021 at 14:27

The elastic potential energy of a spring is

$$\mathrm{EPE} = \frac{1}{2}kx^2$$

where $$k$$ is the spring stiffness (and I believe is related to your $$\lambda$$ by $$\lambda = kL$$) and $$x$$ is the spring extension.

Let $$y_1$$ be the initial height of the mass, and $$y_2$$ be the final height. Since we know that the mass has no kinetic energy at either of these positions, we can express the conservation of energy as follows:

$$\mathrm{GPE}_1 + \mathrm{EPE}_1 = \mathrm{GPE}_2 + \mathrm{EPE}_2$$

The spring extensions $$x_1$$ and $$x_2$$ can be expressed in terms of $$y_1$$ and $$y_2$$.

Since the initial height $$y_1$$ is know, then the initial extension $$x_1$$ can be determined directly, and so the left hand side of the above energy balance can be evaluated directly.

However, the right hand side will be expressed in terms of the unknown $$y_2$$, and so you obtain an equation in which can you solve for $$y_2$$.

Yes, the tension in the spring is changing continuously. At any short duration of time the increment in the stored energy in the spring can be written as: $$$$\Delta{W} = Tdx = {\frac{\lambda x}{L}}dx$$$$ Now, if the limit is from 0 to x. then, $$$$W= \int_{0}^{x}\Delta{W} = \int_{0}^{x} {\frac{\lambda x}{L}}dx = {\frac{\lambda x^2}{2L}}$$$$ $$$$W= \frac{1}{2}kx^2 ~~\text{where}~~k=\frac{\lambda}{L}$$$$ Remember this form of the total stored energy is only valid if your choice of coordinate system is such that $$at~ x=0, T=0$$. Now let's use this understanding to solve the problem. Below is a cartoon picture of the problem:

Assume that the spring reaches maximum displacement of $$h_2$$ in the other direction from it's relaxed position at $$x=0$$.The balance of total energy at the beging and at the time of maximum displacement becomes: \begin{align} & \frac{1}{2}kh_1^2 + mgh_1 = \frac{1}{2}k(-h_2)^2 + mg(-h_2) \\ => & kh_2^2 -2mgh_2 -(kh_1^1 + 2mgh_1) = 0 \\ & \text{This is quadratic equation of } h_2 \\ => & h_2 = \frac{2mg \pm \sqrt{(2mg)^2+4k(kh_1^2+2mgh_1)}}{2k} \\ & ~~~= \frac{mg \pm \sqrt{(mg)^2+k(kh_1^2+2mgh_1)}}{k} \\ \end{align}

For your proble $$m=0.3Kg,~ g=10m/s^2,~ k=\frac{\lambda}{L}=6/0.5=12N/m,~ h_1=0.1m$$. Clearly, it has two solutions. The solution with negatize sign will give you back the configuration that you already know i.e $$h_2=-0.1m$$. The negative sign only tells that the solution is in the opposite to the assued direction of $$h_2$$.

The solution corresponding to positive sign becomes $$h_2=0.6m$$. That is the actual solution you are looking for.

Another way to solve this problem is by using the equation of motion of a spring-mass system. \begin{align} & m\ddot{y} = -mg -ky \\ & m\ddot{y} +ky +mg=0 \\ & \text{solve this equation with intial condition of:} \\ & \dot{y}(t=0)=0 ~and~ y(t=0)=h_1 \end{align} Get the extreme value of displacements at time corresponding to the position where velocity is zero i.e, $$\dot{y}=0$$.