Spring Forces: Definition, Formula, and Examples
A spring has negligible mass and exerts a force, when stretched or compressed, that is proportional to the displacement from its relaxed length. When you grab an object attached to a spring, pull it a distance from its equilibrium position, and release it, the restoring force will pull the object back to equilibrium. For a spring-mass system on a horizontal table, the only force acting on the mass in the direction of displacement is the restoring force exerted by the spring. Using Newton's Second Law, we can set up an equation for the motion of the object. The direction of the restoring force will always be opposite and antiparallel to the displacement of the object. The restoring force acting on the spring-mass system depends on the spring constant and the object's displacement from the equilibrium position.
Fig. 1 - Representation of a spring-mass system, where the mass oscillates about an equilibrium position.
$$\vec{F_{\text{net}}}=m\vec a$$
Along the direction of displacement \(\widehat x\):
$$-kx=m\frac{\operatorname d^2x}{\operatorname dt^2}$$
$$\frac{\operatorname d^2x}{\operatorname dt^2}=-\frac km x$$
Where \(m\) is the mass of the object at the end of the spring in kilograms \((\mathrm{kg})\), \(a_x\) is the acceleration of the object on the \(\text{x-axis}\) in meters per second squared \((\frac{\mathrm m}{\mathrm s^2})\), \(k\) 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 \((\mathrm m)\).
This relationship is also known as Hooke's Law, and 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.
The above expression looks a lot like the differential equation for simple harmonic motion, so the spring-mass system is a harmonic oscillator, where its angular frequency can be expressed in the below equation.
$$\omega^2=\frac km$$
$$\omega=\sqrt{\frac km}$$
A \(12\;\mathrm{cm}\) spring has a spring constant of \(400\;{\textstyle\frac{\mathrm N}{\mathrm m}}\). How much force is required to stretch the spring to a length of \(14\;\mathrm{cm}\)?
The displacement has a magnitude of
$$x=14\;\mathrm{cm}\;-\;12\;\mathrm{cm}=2\;\mathrm{cm}=0.02\;\mathrm m$$
The spring force has a magnitude of
$$F_s=kx=(400\;{\textstyle\frac{\mathrm N}{\mathrm m}})(0.02\;\mathrm m)=8\;\mathrm N$$
A spring-mass system is said to be in equilibrium if there is no net force acting on the object. This can happen when the magnitude and direction of the forces acting on the object are perfectly balanced, or simply because no forces are acting on the object. Not all forces try to restore the object back to equilibrium, but forces that do so are called restoring forces, and the spring force is one of them.
A restoring force is a force acting against the displacement to try and bring the system back to equilibrium. This type of force is responsible for generating oscillations and is necessary for an object to be in simple harmonic motion. Furthermore, the restoring force is what causes the change in acceleration of an object in simple harmonic motion. As the displacement increases, the stored elastic energy increases and the restoring force increases.
In the diagram below, we see a complete cycle that begins when the mass is released from point \(\text{A}\). The spring forces cause the mass to pass through the equilibrium position all the way up to \(\text{-A}\), just to pass again through the equilibrium position and reach point \(\text{A}\) to complete an entire cycle.
Fig. 2 - Complete oscillation cycle of a spring-mass system.
Combination of Springs
A collection of springs may act as a single spring, with an equivalent spring constant which we will call \(k_{\text{eq}}\). The springs may be arranged in series or in parallel. The expressions for \(k_{\text{eq}}\) will vary depending on the type of type of arrangement. In series, the inverse of the equivalent spring constant will be equal to the sum of the inverse of the individual spring constants. It is important to note that in an arrangement in series, the equivalent spring constant will be smaller than the smallest individual spring constant in the set.
$$\frac1{k_{eq\;series}}=\sum_n\frac1{k_n}$$
Fig. 3 - Two springs in series.
A set of 2 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?
$$\frac1{k_{eq\;series}}=\frac1{1\frac{\mathrm N}{\mathrm m}}+\frac1{2\frac{\mathrm N}{\mathrm m}}$$
$$\frac1{k_{eq\;series}}=\frac32{\textstyle\frac{\mathrm m}{\mathrm N}}$$
$$k_{eq\;series}=\frac23{\textstyle\frac{\mathrm N}{\mathrm m}}$$
In parallel, the equivalent spring constant will be equal to the sum of the individual spring constants.
$$k_{eq\;parallel}=\sum_nk_n$$
Fig. 4 - Two springs in parallel.
A set of 2 springs in parallel 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?
$$k_{eq\;parallel}=1\;{\textstyle\frac{\mathrm N}{\mathrm m}}+\;2{\textstyle\frac{\mathrm N}{\mathrm m}}=3\;{\textstyle\frac{\mathrm N}{\mathrm m}}$$
Force vs. Displacement Graph
We can plot the spring force as a function of position and determine the area under the curve. Performing this calculation will provide us with the work done on the system by the spring force and the difference in potential energy stored in the spring due to its displacement. Because in this case, the work done by the spring force depends only on initial and final positions, and not on the path between them, we can derive the change in potential energy from this force. These types of forces are called conservative forces.
Using calculus, we can determine the change in potential energy.
$$\begin{array}{rcl}\triangle U&=&-\int_i^f{\overset\rightharpoonup F}_{cons}\cdot\overset\rightharpoonup{dx},\\\triangle U&=&-\int_i^f\left|{\overset\rightharpoonup F}_{\mathrm{cons}}\right|\left|\overset\rightharpoonup{dx}\right|\cos\left(180^\circ\right),\\\triangle U&=&-\int_i^f\left(kx\right)\left(\mathrm dx\right)\cos\left(180^\circ\right),\\\triangle U&=&\frac12kx_{\mathrm f}^2-\frac12kx_{\mathrm i}^2.\end{array}$$
Fig. 5 - Force vs Displacement graph, the spring constant is the slope and the potential energy is the area below the curve.
Spring Force - Key takeaways
- A spring has negligible mass and exerts a force, when stretched or compressed, that is proportional to the displacement from its relaxed length. When you grab an object attached to a spring, pull it a distance from its equilibrium position, and release it, the restoring force will pull the object back to equilibrium.
- The magnitude of spring force is described by Hooke's Law, \(kx=m\frac{\operatorname d^2x}{\operatorname dt^2}\).
- The direction of the restoring force will always be opposite and antiparallel to the displacement of the object.
- A collection of springs may act as a single spring, with an equivalent spring constant, which we will call \(k_eq\).
- In series, the inverse of the equivalent spring constant will be equal to the sum of the inverse of the individual spring constants, \(\frac1{k_{eq\;series}}=\sum_n\frac1{k_n}\).
- In parallel, the equivalent spring constant will be equal to the sum of the individual spring constants \(k_{eq\;parallel}=\sum_nk_n\).
References
- Fig. 1 - Representation of a spring-mass system, where the mass oscillates about an equilibrium position, StudySmarter Originals
- Fig. 2 - Complete oscillation cycle of a spring-mass system, StudySmarter Originals
- Fig. 3 - Two springs in series, StudySmarter Originals
- Fig. 4 - Two springs in parallel, StudySmarter Originals
- Fig. 5 - Force vs Displacement graph, the spring constant is the slope and the potential energy is the area below the curve, StudySmarter Originals
How we ensure our content is accurate and trustworthy?
At StudySmarter, we have created a learning platform that serves millions of students. Meet
the people who work hard to deliver fact based content as well as making sure it is verified.
Content Creation Process:
Lily Hulatt is a Digital Content Specialist with over three years of experience in content strategy and curriculum design. She gained her PhD in English Literature from Durham University in 2022, taught in Durham University’s English Studies Department, and has contributed to a number of publications. Lily specialises in English Literature, English Language, History, and Philosophy.
Get to know Lily
Content Quality Monitored by:
Gabriel Freitas is an AI Engineer with a solid experience in software development, machine learning algorithms, and generative AI, including large language models’ (LLMs) applications. Graduated in Electrical Engineering at the University of São Paulo, he is currently pursuing an MSc in Computer Engineering at the University of Campinas, specializing in machine learning topics. Gabriel has a strong background in software engineering and has worked on projects involving computer vision, embedded AI, and LLM applications.
Get to know Gabriel