If only you had known about springs and the potential energy stored in them when you were a child, you would have asked your parents to buy you a trampoline with a large spring constant. This would have allowed you to store more energy in the spring and jump higher than all your friends, making you the coolest kid in the neighborhood. As we will see in this article, the potential energy of a spring-mass system is related to the spring's stiffness and the distance that the spring has been stretched or compressed, we will also discuss how we can model an arrangement of multiple springs as a single one.
Overview of Springs
A spring exerts a force when it is stretched or compressed. This force is proportional to the displacement from its relaxed or natural length. The spring force is opposite to the direction of displacement of the object and its magnitude is given by Hooke's Law, in one dimension this is:
$$\boxed{F_s=kx,}$$
where \(k\) is the spring constant that measures the stiffness of the spring in newtons per meter, \(\frac{\mathrm N}{\mathrm m}\), and \(x\) is the displacement in meters, \(\mathrm{m}\), measured from the equilibrium position.
Hooke's Law can be proven by setting up a spring system with hanging masses. Every time that you add a mass, you measure the extension of the spring. If the procedure is repeated, it will be observed that the extension of the spring is proportional to the restoring force, in this case, the weight of the hanging masses, since in physics we consider the spring to have a negligible mass.
A block of mass \(m=1.5\;\mathrm{kg}\) is attached to a horizontal spring of force constant \(k=300\;{\textstyle\frac{\mathrm N}{\mathrm m}}\). After the spring-block system reaches equilibrium it is pulled down \(2.0\ \text{cm}\), then it is released and starts oscillating. Find the equilibrium position before the blocked is pulled down to begin oscillations. What are the minimum and maximum displacements from the spring equilibrium position during the oscillations of the block?
Fig. 1 - Spring-mass system reaches an equilibrium point and is displaced even further. When the mass is released it begins to oscillate due to the spring force.
Solution
Before the block is pulled down to begin oscillating, because of its weight, it stretched the spring by a distance \(d\). Note that when the spring-mass system is in equilibrium, the net force is zero. Therefore, the weight of the block bringing it down, and the force of the spring pulling it up, are equal in magnitude:
$$\begin{align*}F_\text{s}&=w,\\kd&=mg.\end{align*}$$
Now we can find an expression for \(d\):
$$\begin{align*}d&=\frac{mg}k,\\d&=\frac{\left(1.5\;\mathrm{kg}\right)\left(10\;\frac{\mathrm m}{\mathrm s^2}\right)}{300\;\frac{\mathrm N}{\mathrm m}},\\d&=\frac{\left(1.5\;\bcancel{\mathrm{kg}}\right)\left(10\;\bcancel{\frac{\mathrm m}{\mathrm s^2}}\right)}{300\;\frac{\bcancel{kg}\;\bcancel{\frac m{s^2}}}{\mathrm m}},\\d&=0.050\;\mathrm m,\\d&=5.0\;\mathrm{cm}.\end{align*}$$
If the amplitude of the oscillations is \(2.0\;\mathrm{cm}\), it means that the maximum amount of stretch happens at \(5.0\;\mathrm{cm}+2.0\;\mathrm{cm}=7.0\;\mathrm{cm},\) similarly, the minimum is \(5.0\;\mathrm{cm}-2.0\;\mathrm{cm}=3.0\;\mathrm{cm}.\)
A collection of springs can be represented as a single spring with an equivalent spring constant which we represent as \(k_\text{eq}\). The arrangement of these springs may be done in series or in parallel. The way we calculate \(k_\text{eq}\) will vary depending on the type of arrangement we use.
Springs in Series
When the set of springs is arranged in series, the reciprocal of the equivalent spring constant is equal to the sum of the reciprocal of the spring constants, this is:
$$\boxed{\frac1{k_\text{eq series}}=\sum_n\frac1{k_n}}.$$
If the set of springs is arranged in series, the equivalent spring constant will be smaller than the smallest spring constant in the set.
Fig. 2 - Two springs in series.
A set of two springs in series have springs constants of \(1\;{\textstyle\frac{\mathrm N}{\mathrm m}}\) and \(2\;{\textstyle\frac{\mathrm N}{\mathrm m}}\) . What is the value for the equivalent spring constant?
Solution
$$\begin{align*}\frac1{k_\text{eq series}}&=\frac1{1\;\frac{\mathrm N}{\mathrm m}}+\frac1{2\;\frac{\mathrm N}{\mathrm m}},\\\frac1{k_\text{eq series}}&=\frac32{\textstyle\frac{\mathrm m}{\mathrm N},}\\k_\text{eq series}&=\frac23{\textstyle\frac{\mathrm N}{\mathrm m}.}\end{align*}$$
As we indicated previously, when you set up springs in series, \(k_{\text{eq}}\) will be smaller than the smallest spring constant in the setup. In this example the smallest spring constant has a value of \(1\;{\textstyle\frac{\mathrm N}{\mathrm m}}\), while \(k_{\text{eq}}\) is \(\frac23\;\frac{\mathrm N}{\mathrm m}\approx 0.67\;\frac{\mathrm N}{\mathrm m}\).
Springs in Parallel
When the set of springs is arranged in parallel, the equivalent spring constant will be equal to the sum of the spring constants:
$$\boxed{k_\text{eq parallel}=\sum_nk_n}.$$
In this case, the equivalent spring constant will be greater than every individual spring constant in the set of springs involved.
Fig. 3 - Two springs in parallel.
Spring Potential Energy Units
Potential energy is the energy stored in an object because of its position relative to other objects in the system.
The unit for potential energy is joules, \(\mathrm J\), or newton meters, \(\mathrm N\;\mathrm m\). It is important to notice that potential energy is a scalar quantity, meaning that it has a magnitude, but not a direction.
Spring Potential Energy Equation
Potential energy is deeply related to conservative forces.
The work done by a conservative force is path independent and only depends on the initial and final configurations of the system.
This means that it does not matter the direction or trajectory that the objects of the system followed when they were being moved around. The work only depends on the initial and final positions of these objects. Because of this important property, we can define the potential energy of any system made by two or more objects that interact via conservative forces.
Since the force exerted by a spring is conservative, we can find an expression for the potential energy in a spring-mass system by calculating the work done over the spring-mass system when displacing the mass:
$$\Delta U=W.$$
In the above equation we are using the notation \(\Delta U=U_f-U_i\).
The idea is that this work is done against the conservative force, thus storing energy in the system. Alternatively, we can calculate the potential energy of the system by calculating the negative of the work done by the conservative force \( \Delta U = - W_\text{conservative}, \) which is equivalent.
The expression of the potential energy of a spring-mass system can be simplified if we chose the equilibrium point as our point of reference so that \( U_i = 0. \) Then we are left with the following equation
$$U=W.$$
In the case of a system with multiple objects, the total potential energy of the system will be the sum of the potential energy of every pair of objects inside the system.
As we will see in more detail in the next section, the expression for the potential energy of a spring is
$$\boxed{U=\frac12kx^2}$$
As an example to use this equation, let's explore the situation we discussed at the beginning of this article: a trampoline with multiple springs.
A trampoline with a set of \(15\) springs in parallel have springs constants of \(4.50\times10^3\,{\textstyle\frac{\mathrm N}{\mathrm m}}\). What is the value for the equivalent spring constant? What is the potential energy of the system due to the springs if they get stretched by \(0.10\ \text{m}\) after landing from a jump?
Solution
Remember that to find the equivalent constant for a set of springs in parallel we sum all the individual spring constants. Here all the spring constants in the set have the same value so it's easier to just multiply this value by \( 15 \),
\begin{aligned}k_\text{eq parallel}&=15\times4.50\times10^3\;{\textstyle\frac{\mathrm N}{\mathrm m}}\\k_\text{eq parallel}&=6.75\times 10^4\textstyle\frac{\mathrm N}{\mathrm m}\end{aligned}
Now we can find the potential energy of the system, using the equivalent spring constant.
\begin{aligned}U&=\frac12k_{\text{eq}}x^2,\\[6pt]U&=\frac12\left(6.75\times 10^4\textstyle\frac{\mathrm N}{\mathrm m}\right)\left(0.10\ \text m\right)^2,\\[6pt] U&=338\,\mathrm{J}. \end{aligned}
Spring Potential Energy Derivation
Let's find the expression of the potential energy stored in a spring, by calculating the work done over the spring-mass system when moving the mass from its equilibrium position \(x_{\text{i}}=0\) to a position \(x_{\text{f}} = x.\) Since the force we need to apply is constantly changing as it depends on the position we need to use an integral. Note that the force we apply \(F_a\) over the system must be equal in magnitude to the force of the spring and opposite to it so that the mass is moved. This means that we need to apply a force \(F_a = kx\) in the direction of the displacement we want to cause:
$$\begin{align*}\Delta U&=W\\[8pt]\Delta U&=\int_{x_{\text{i}}}^{x_{\text{f}}}{\vec F}_{\mathrm a}\cdot\mathrm{d}\vec{x}\\[8pt]\Delta U&=\int_{x_{\text{i}}}^{x_{\text{f}}}\left|{\vec F}_{\mathrm a}\right|\left|\mathrm{d}\vec{x}\right|\cos\left(0^\circ\right)\\[8pt]\Delta U&=\int_{x_{\text{i}}}^{x_{\text{f}}}kx\mathrm dx\\[8pt]\Delta U&=\frac12kx_{\mathrm f}^2-\frac12kx_{\mathrm i}^2.\end{align*}$$
However, since \(x_{\text{i}}=0\) is the equilibrium point, recall that we can choose it to be our reference point to measure the potential energy, so that \(U_{\text{i}}=0,\) leaving us with the simpler formula:
$$U = \frac12kx^2,$$
where \( x \) is the distance from the equilibrium position. There is an easier way to arrive at this expression without the use of calculus. We can plot the spring force as a function of position and determine the area under the curve.
Fig. 4 - We can determine the spring's potential energy by calculating the area below the curve \(F_s(x)\).
From the above figure, we see that the area under the curve is a triangle. And, since the work equals the area under a force vs position graph, we can determine the expression of the spring's potential energy by finding this area.
\begin{aligned}U&=W\\[6pt]U&=\text{area under }F(x)\\[6pt]U&=\frac12\left(\text{ triangle's base}\right)\left(\text{triangle's height}\right)\\[6pt]U&=\frac12\left(x\right)\left(kx\right)\\[6pt]U&=\frac12kx^2.\end{aligned}
As you can see, we arrived at the same result. Where \(k\) is the spring constant that measures the stiffness of the spring in newtons per meter, \(\frac{\mathrm N}{\mathrm m}\), and \(x\) is the mass position in meters, \(\mathrm m,\) measured from the point of equilibrium.
Spring Potential Energy Graph
By plotting the potential energy as a function of position, we can learn about different physical properties of our system. The points where the slope is zero are considered equilibrium points. We can know that the slope of \( U(x) \) represents the force, since for a conservative force
$$F = -\frac{\mathrm{d}U}{\mathrm{d}x}$$
This implies that the points where the slope is zero identify locations where the net force on the system is zero. These can either be local maximums or minimums of \( U(x). \)
Local maximums are locations of unstable equilibrium because the force would tend to move our system away from the equilibrium point at the slightest change in position. On the other hand, local minimums indicate locations of stable equilibrium because at a small displacement of the systems the force would act against the direction of displacement, moving the object back to the equilibrium position.
Below we can see a graph of the potential energy as a function of position for a spring-mass system. Notice that it is a parabolic function. This is because the potential energy depends on the square of the position. Take a look at the point \(x_1\) located in the graph. Is it a stable or unstable equilibrium point?
Potential energy as a function of position and equilibrium point for a spring-mass system.
Solution
Point \(x_1\) is a location of stable equilibrium as it is a local minimum. We can see that this makes sense with our previous analysis. The force at \( x_1 \) is zero as the slope of the function is zero there. If we move the left of \( x_1 \) the slope is negative, this means that the force \( f = - \frac{\mathrm{d}U}{\mathrm{d}x}, \) points to the positive direction, tending to move the mass towards the equilibrium point. Finally, at any position to the right of \( x_1 \) the slope becomes positive, therefore the force is negative, pointing to the left and, once more, tending to move the mass back, towards the equilibrium point.
Fig. 6 - Visualization of the relation between the force and potential energy. We see that when the net force is zero, the slope of the potential energy as a function of the position is also zero. This represents the equilibrium position. Whenever the mass is out of the equilibrium position the spring force will act to restore the mass into its equilibrium position.
Spring Potential Energy - Key takeaways
- A spring in consider to have negligible mass and it exerts a force, when stretched or compressed, which is proportional to the displacement from its relaxed length. This force is opposite in the direction of displacement of the object. The magnitude of the force exerted by the spring is given by Hooke's Law, $$F_s=k x.$$
We can model a collection of springs as a single spring, with an equivalent spring constant which we will call \(k_\text{eq}\).
For spring that are arranged in series, the inverse of the equivalent spring constant will be equal to the sum of the inverse of the individual spring constants $$\frac1{k_\text{eq series}}=\sum_n\frac1{k_n}.$$
For springs that are arranged in parallel, the equivalent spring constant will be equal to the sum of the individual spring constants, $$k_\text{eq parallel}=\sum_nk_n.$$
Potential energy is the energy stored in an object because of its position relative to other objects in the system.
The work done by a conservative force does not depend on the direction or path that the object comprising the system followed. It only depends on their initial and final positions.
The force exerted by the spring is a conservative force. This allows us to define the change in the potential energy in a spring-mass system as the amount of work done over the system when moving the mass, \(\Delta U=W\).
The expression of the potential energy for a spring-mass system is $$U=\frac12kx^2.$$
In the case of a system with more than three objects, the total potential energy of the system would be the sum of the potential energy of every pair of objects inside the system.
If we examine the energy of the system in a potential energy vs position graph, points where the slope is zero are considered equilibrium points. The locations with local maximums are locations of unstable equilibrium, while local minimums indicate locations of stable equilibrium.
References
- Fig. 1 - Vertical spring-mass system, StudySmarter Originals
- Fig. 2 - Two springs in series, StudySmarter Originals
- Fig. 3 - Two springs in parallel, StudySmarter Originals
- Fig. 4 - Spring force as a function of position, StudySmarter Originals
- Fig. 5 - Spring potential energy as a function of position, StudySmarter Originals
- Fig. 6 - Relation between the force and potential energy of a spring, StudySmarter Originals
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