Spring Mass System

Springs are very basic objects with a simple function. We all have intuitive ideas of how springs work, whether that be from playing with toys with springs in them when we were young or from jumping on a trampoline. However, springs are probably more important than you realize and can be found everywhere. They play a key role in cars, electrical circuits and clocks, as well as even being useful for many scientific models! Springs and elasticity have been studied greatly for a long time and the function of systems involving springs can be explained by simple equations. Read on to learn more about them.

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 simple horizontal spring-mass system consists of a spring attached to a wall on one side and a mass on the other, ux1.eiu.

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.

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.

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:

$F=-kx$.

$k$ 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.

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.

A mass hanging from a vertical spring can be treated in the same way as a horizontal spring-mass system, khanacademy.

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:

$W=mg$,

$$\begin{gathered} -k{x}_0=mg \\ {x}_0=\frac{-mg}{k} \end{gathered}$$

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.

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

$\omega =\frac{2\pi }{T}$.

$T$ 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.

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

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

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.

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,$\mathrm{J}$,

$x$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.

$$\begin{gathered} \frac{1}{2}m{v}^2=\frac{1}{2}k{A}^2 \\ m{v}^2=k{A}^2 \\ {v}^2=\frac{k{A}^2}{m} \\ v=\sqrt{\frac{k}{m}}\times A=\omega A \end{gathered}$$

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.

Shock absorbers are placed just above each wheel on a car to prevent damage from bumps, cardealpage.

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.

Oxygen exists as a diatomic molecule and can be approximated as a spring-mass system, bartlby.