# 15.2 Energy in simple harmonic motion

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By the end of this section, you will be able to:
• Describe the energy conservation of the system of a mass and a spring
• Explain the concepts of stable and unstable equilibrium points

To produce a deformation in an object, we must do work. That is, whether you pluck a guitar string or compress a car’s shock absorber, a force must be exerted through a distance. If the only result is deformation, and no work goes into thermal, sound, or kinetic energy, then all the work is initially stored in the deformed object as some form of potential energy.

Consider the example of a block attached to a spring on a frictionless table, oscillating in SHM. The force of the spring is a conservative force (which you studied in the chapter on potential energy and conservation of energy), and we can define a potential energy for it. This potential energy is the energy stored in the spring when the spring is extended or compressed. In this case, the block oscillates in one dimension with the force of the spring acting parallel to the motion:

$W=\underset{{x}_{i}}{\overset{{x}_{f}}{\int }}{F}_{x}dx=\underset{{x}_{i}}{\overset{{x}_{f}}{\int }}\text{−}kxdx={\left[-\frac{1}{2}k{x}^{2}\right]}_{{x}_{i}}^{{x}_{f}}=\text{−}\left[\frac{1}{2}k{x}_{f}^{2}-\frac{1}{2}k{x}_{i}^{2}\right]=\text{−}\left[{U}_{f}-{U}_{i}\right]=\text{−}\text{Δ}U.$

When considering the energy stored in a spring, the equilibrium position, marked as ${x}_{i}=0.00\phantom{\rule{0.2em}{0ex}}\text{m,}$ is the position at which the energy stored in the spring is equal to zero. When the spring is stretched or compressed a distance x , the potential energy stored in the spring is

$U=\frac{1}{2}k{x}^{2}.$

## Energy and the simple harmonic oscillator

To study the energy of a simple harmonic oscillator, we need to consider all the forms of energy. Consider the example of a block attached to a spring, placed on a frictionless surface, oscillating in SHM. The potential energy stored in the deformation of the spring is

$U=\frac{1}{2}k{x}^{2}.$

In a simple harmonic oscillator    , the energy oscillates between kinetic energy of the mass $K=\frac{1}{2}m{v}^{2}$ and potential energy $U=\frac{1}{2}k{x}^{2}$ stored in the spring. In the SHM of the mass and spring system, there are no dissipative forces, so the total energy is the sum of the potential energy and kinetic energy. In this section, we consider the conservation of energy of the system. The concepts examined are valid for all simple harmonic oscillators, including those where the gravitational force plays a role.

Consider [link] , which shows an oscillating block attached to a spring. In the case of undamped SHM, the energy oscillates back and forth between kinetic and potential, going completely from one form of energy to the other as the system oscillates. So for the simple example of an object on a frictionless surface attached to a spring, the motion starts with all of the energy stored in the spring as elastic potential energy    . As the object starts to move, the elastic potential energy is converted into kinetic energy, becoming entirely kinetic energy at the equilibrium position. The energy is then converted back into elastic potential energy by the spring as it is stretched or compressed. The velocity becomes zero when the kinetic energy is completely converted, and this cycle then repeats. Understanding the conservation of energy in these cycles will provide extra insight here and in later applications of SHM, such as alternating circuits.

A spring with 50g mass suspended from it,has its length extended by 7.8cm 1.1 determine the spring constant? 1.2 it is observed that the length of the spring decreases by 4.7cm,from its original length, when a toy is place on top of it. what is the mass of the toy?
solution mass = 50g= 0.05kg force= 50 x 10= 500N extension= 7.8cm = 0.078m using the formula Force= Ke K = force/extension 500/.078 = 6410.25N/m
Sampson
1.2 Decrease in length= -4.7cm =-0.047m mass=? acceleration due to gravity= 10 force = K x e force= mass x acceleration m x a = K x e mass = K x e/acceleration = 6410.25 x 0.047/10 = 30.13kg
Sampson
1.1 6.28Nm-¹
Anita
1.2 0.03kg or 30g
Anita
I used g=9.8ms-²
Anita
you should explain how yoy got the answer Anita
Grant
ok
Anita
with the fomular F=mg I got the value for force because now the force acting on the spring is the weight of the object and also you have to convert from grams to kilograms and cm to meter
Anita
so the spring constant K=F/e where F is force and e is extension
Anita
In this first example why didn't we use P=P° + ¶hg where ¶ is density
Density = force applied x area p=fA =p = mga, then a=h therefore substitute =p =mgh
Hlehle
Hlehle
sorry I had a little typo in my question
Anita
Density = m/v (mass/volume) simple as that
Augustine
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