## Spring-Mass System Meaning

The simplest case of a spring-mass system consists of a mass attached to the end of a spring. There can be other arrangements with more masses and more springs attached in different ways, but we are just going to consider this simple case. An example of a spring-mass system on a horizontal surface is shown below.

A spring is attached to a fixed point on one side and to a mass $m$ on the other side. In the diagram, $k$ represents the spring constant and is a measure of the stiffness of the sprint. A larger spring constant means a stiffer spring. The more stiff a string is, the more force is exerted by the string when it is displaced by a certain amount. The spring constant is measured in newtons per meter, $\mathrm{N}/\mathrm{m}$. The spring-mass system above is shown to be in its **relaxed state**, meaning that the string is at its natural length - it is not stretched or compressed - and so there is no force on the mass. We say that the mass is at its **equilibrium position**.

## Spring-Mass System Equation

Now we can consider the forces acting on the mass when it is moved from its equilibrium position. If we move the mass to a displacement of$x$meters to the right, the spring will be stretched past its natural length, as shown below.

If the mass is pulled to the right, the string will be stretched and exert a force on the mass directed back towards the left, ux1.eiu.

This causes the spring to exert a force on the mass as the spring is trying to return to its natural length, hence the mass accelerates back towards the equilibrium position when released.

On the other hand, if the mass is moved to the left, the spring will be compressed.

If the mass is pushed to the left, the string will be stretched and exert a force on the mass directed back towards the right, ux1.eiu.

The spring will attempt to return to its natural length and will exert a force on the mass to the right which will cause it to move back towards its equilibrium position.

From the two situations above, you can see that when the mass is displaced, the spring will always exert a force to bring the mass back to its equilibrium position - this is called a **restoring force**. This force is given by a simple relation:

$F=-kx$.

$F$ is the force acting on the mass in Newtons,$\mathrm{N}$$k$ is the spring constant in newtons-per-meter,$\mathrm{N}/\mathrm{m}$

$x$ is the displacement of the mass from its equilibrium position in meters,$\mathrm{m}$.

This equation is Hooke's law. It was named after the English scientist Robert Hooke, who did work on a variety of subjects in the 17^{th} century including on elasticity.

We must be careful to say 'displacement' and not 'distance' as the displacement can be both positive and negative, which is important to find the right direction of the force.

Hooke's law clearly shows that the force acting on the mass increases as the distance of the mass from the equilibrium position increases. However, the minus sign in the equation shows that the force is always directed in the *opposite *direction to the displacement - when the mass is on the right of the equilibrium, the force will be directed towards the left and vice versa. The force is always directed towards the equilibrium position.

Another example of a spring-mass system that you will often find in problems is a vertically oriented spring attached to a ceiling at its top end and a mass at its bottom end.

Hooke's law can be applied to this situation in exactly the same way as for the horizontal system, but this time the equilibrium position has shifted due to gravity. The point where there is no force acting on the mass occurs when the upwards force due to the stretched spring is equal to the downwards force due to the mass. The mass of the spring is assumed to be zero for an ideal mass-spring system. The weight of the mass is equal to:

$W=mg$,

where $g$ is the gravitational constant on the surface of the Earth in meters-per-second-squared,$\mathrm{m}/{\mathrm{s}}^{2}$. This can be equated to the spring force to find the equilibrium position ${x}_{0}$.

$-k{x}_{0}=mg\phantom{\rule{0ex}{0ex}}{x}_{0}=\frac{-mg}{k}$

The minus sign shows that the equilibrium position is below the position of the mass when the spring is at its natural length, which is as expected.

## Spring-Mass System Simple Harmonic Motion

If the mass on a spring-mass system is displaced from its equilibrium position and released, it will exhibit **simple harmonic motion**. An object performing simple harmonic motion (SHM) moves back and forth between points of maximum displacement from the equilibrium position on either side. The mass will be accelerated towards its equilibrium position when it is displaced. When it reaches this position, it will have a velocity and so carry on to the other side, at which point it will experience a restoring force in the opposite direction and the process will be repeated. This motion will continue indefinitely under ideal conditions (ignoring any frictional forces present).

**Simple Harmonic Motion (SHM)** is defined as an oscillation in which the acceleration of an object is inversely proportional to its displacement from the equilibrium position and this acceleration must always be directed towards the equilibrium.

From the definition of simple harmonic motion, the acceleration of an object undergoing SHM is proportional to its displacement and is given by:

$a=-{\omega}^{2}x$,where $a$ is the acceleration in$\mathrm{m}/{\mathrm{s}}^{2}$,$\omega $is the angular frequency of the motion is$\mathrm{rad}/\mathrm{s}$and$x$is the displacement of the object from its equilibrium position in$\mathrm{m}$. The angular frequency$\omega $is defined to be equal to:$\omega =\frac{2\pi}{T}$.

$T$ is the time period of the motion, which is equal to the time for the mass to move from one position of maximum displacement and back again. For the spring-mass system, we can find the acceleration of the mass from the equation for the restoring force acting on the mass:

$a=\frac{F}{m}=-\frac{k}{m}x$.

We can compare this to the equation for SHM stated above, from which we can see that the angular frequency for a mass-spring system is:

$\omega =\sqrt{\frac{k}{m}}$.

We can then use the equation for angular frequency to find the time period in$\mathrm{s}$of the simple harmonic motion of a spring-mass system.

$\frac{2\pi}{T}=\sqrt{\frac{k}{m}}$

$T=\frac{2\pi}{\sqrt{\frac{k}{m}}}=2\pi \sqrt{\frac{k}{m}}$

This does not depend on the initial displacement of the system - known as the**amplitude**of the oscillation. A spring-mass system will always oscillate with the same time period no matter how far the mass is originally displaced (as long as frictional forces are ignored).

## Spring-Mass System Examples

The equations stated above can be utilized in many different practice problems - you just need to identify which equation is the correct one to use.

A$2\mathrm{kg}$mass is hung from a ceiling from a spring of natural length$2\mathrm{m}$and spring constant$k=100\mathrm{N}/\mathrm{m}$. After initially being supported so that the spring is at its natural length, the mass is released and undergoes simple harmonic motion about an equilibrium position.

What is the distance of this equilibrium position from the ceiling? What is the amplitude of the oscillation? What is the angular frequency of the oscillation? The spring can be assumed to be massless.

At the equilibrium position, the weight of the mass has to match the upwards force due to the extension of the spring. We considered this situation earlier on in the article and found an equation for the equilibrium position, ${x}_{0}$. We can plug the given values in to that equation to find the answer:

${x}_{0}=\frac{-mg}{k}=-0.196\mathrm{m}$.

This position is below the unstretched length of the string, so we must add its magnitude to the unstretched length of the string to find the distance of the equilibrium position from the ceiling, $x$:

$x=(2+0.196)\mathrm{m}=2.20\mathrm{m}$

We have already found the amplitude$A$of the oscillation; it is equal to the extra distance between the unstretched string and the equilibrium point:

$A=0.196m$

To find the angular frequency$\omega $of the simple harmonic motion of this system, we use the equation in terms of the mass and spring constant that was stated above:

$\omega =\sqrt{\frac{k}{m}}=7.07\mathrm{rad}/\mathrm{s}$

The same spring is used as in the question above but this time it is attached to a$5\mathrm{kg}$mass at one end and a wall at the other. The mass rests on the floor. The mass is pulled to a distance of$2.5\mathrm{m}$from the wall and released, causing it to undergo simple harmonic motion.

What is the time period of this motion? What is the closest distance to the wall that the mass will reach during the motion? Assume that the spring is massless and the floor is frictionless.

The time period of the oscillation of the mass can be found from the equation stated above as we are given values for the mass and the spring constant in the question.

$T=2\pi \sqrt{\frac{m}{k}}=1.4\mathrm{s}$

The mass will be closest to the wall when its displacement is equal to the negative of the amplitude of the oscillation. The amplitude of the oscillation is given by the initial position minus the length of the spring, as in a horizontal position the equilibrium point will be the unstretched spring.

$A=(2.5-2)\mathrm{m}=0.5\mathrm{m}$

The shortest distance from the mass to the wall $S$ will be the amplitude of the oscillation subtracted from the natural length of the spring.

$S=(2-0.5)\mathrm{m}=1.5\mathrm{m}$

## Spring Mass System Energy

The work done by a force moving an object is given by:

$W=Fx$

where$W$is the work done measured in Joules,$\mathrm{J}$,

$F$is the applied force measured in Newtons,$\mathrm{N}$,$x$is the distance moved in the direction of the force measured in metres,$\mathrm{m}$.

The equation for the force applied by a spring is:

$F=-kx$.

We want to use this in the equation to find the work done by a spring to move a mass from its equilibrium position to its position of maximum displacement. However, the force changes as the length of the spring changes. It can be seen from the above equation that the force is directly proportional to the distance that the spring is stretched. As the relationship between these two quantities is linear, we can simply suppose that a constant force acts which is equal to the average of the initial and final forces.

${F}_{average}=\frac{{F}_{final}-{F}_{initial}}{2}=\frac{kA-0}{2}=\frac{kA}{2}$

In the above expression,$A$represents the amplitude of a spring mass oscillation. The magnitude of the force has been taken as we are only considering distances and not displacements. We can now use this expression for the average force to find the work done in moving a mass from its equilibrium position to its initial position.

$W={F}_{average}x=\frac{kA}{2}\times A=\frac{1}{2}k{A}^{2}$

The work done in stretching a spring is converted into the elastic potential energy stored in the spring, so this expression above gives the maximum potential energy of a spring mass system.

There are presumed to be no frictional forces in the spring-mass systems that we are considering, this means that no energy is lost in the motion. The energy is shared between the potential and the kinetic energies of the system.

${E}_{total}=PE+KE$

When the mass is at maximum displacement, the spring is the most stretched so the potential energy is maximum. At this point, the speed is zero so its kinetic energy is zero. When the mass passes through its equilibrium position, its speed is a maximum and hence so is its kinetic energy. The spring is not stretched at this point so the potential energy is zero. Along with the equation above, this leads us to conclude that the maximum kinetic energy of the mass of a mass-spring system is equal to the maximum potential energy stored in the spring, which are both also equal to the total energy.

The formula for kinetic energy is

$KE=\frac{1}{2}m{v}^{2}$.

We can equate the maximum kinetic energy - when the mass is moving at its maximum speed${v}_{0}$through the equilibrium position - to the maximum potential energy to find a value for the maximum speed.

$\frac{1}{2}m{v}^{2}=\frac{1}{2}k{A}^{2}\phantom{\rule{0ex}{0ex}}m{v}^{2}=k{A}^{2}\phantom{\rule{0ex}{0ex}}{v}^{2}=\frac{k{A}^{2}}{m}\phantom{\rule{0ex}{0ex}}v=\sqrt{\frac{k}{m}}\times A=\omega A\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

## Spring-Mass System Application

Spring-mass systems can be seen very often in everyday life and they are also used in scientific models.

### Car Shock Absorbers

A spring-mass system is used to great effect in cars in the form of shock absorbers, which are placed above the wheel. They are designed for preventing the car from being damaged when it passes over bumps and other obstacles. As a car goes over a bump, the spring above the shock absorber will compress and these springs must be designed to oscillate at the right amplitude and frequency to make the ride of the car comfortable.

### Diatomic Molecules

Diatomic molecules consist of two atoms that are chemically bonded together. These molecules can oscillate and (for small vibrations) the bond can be approximated as a spring with the two atoms as masses on the ends of the spring. The oscillations obey Hooke's law for small amplitudes. This approximation is used to simplify models for scientific research.

## Spring Mass System - Key takeaways

- The simplest case of a spring-mass system consists of a mass attached to the end of a spring.
- Hooke's law states that the force generated by a spring is proportional to its extension or compression from the natural length.
- A spring will always exert a force to bring itself back to its natural length.
- Hooke's law applies in the same way to horizontal and vertical spring-mass systems.
- For a vertical spring-mass system, the equilibrium position is displaced downwards due to the gravitational force on the mass.
- Springs are presumed to be massless in ideal spring-mass systems.
- There are presumed to be no frictional forces in ideal spring-mass systems.
- The oscillations of mass on springs can be described by the equations of simple harmonic motion.
- Simple harmonic motion is defined as an oscillation in which the acceleration of an object is directly proportional to its displacement from its equilibrium position and this acceleration must always be directed towards the equilibrium.
- The period of oscillation of a spring-mass system does not depend on the amplitude of oscillation.

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##### Frequently Asked Questions about Spring Mass System

What is the formula for spring mass system?

The formula to find the time period T of a spring-mass system is:

T=2π sqrt(m/k)

What is an ideal mass spring system?

In an ideal mass system, the spring is massless and there are no frictional forces present.

Where are mass spring systems used?

Spring-mass systems have many applications such as:

- Shock absorbers in cars.
- Scientific models.
- Trampolines.
- Pogo sticks.

What is an example of a spring mass system?

An example of a spring mass system is a mass hung vertically from a spring that is attached to a fixed point on a ceiling.

How do you find the mass of a spring?

The mass of a spring can be found by placing it on a weight measuring scale, just like any other object.

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