Question

Learning Goal: To learn to apply the law ofconservation of energy to the analysis of harmonic oscillators.

Systems in simple harmonic motion, or harmonicoscillators, obey the law of conservation of energy just likeall other systems do. Using energyconsiderations, one can analyzemany aspects of motion of the oscillator. Such an analysis can besimplified if one assumes that mechanical energy is notdissipated.In other words,

,

where is the total mechanical energy of the system, is the kinetic energy, and is the potential energy.As you know, a common example of aharmonic oscillator is a mass attached to a spring. In thisproblem, we will consider a horizontally movingblockattached to a spring. Note that, since the gravitational potentialenergy is not changing in this case, it can be excluded from thecalculations.

For such a system, the potential energy is stored in the springand is given by

$U=frac{1}{2}kx^2$ ,

where is the force constant of the spring andis the distance from the equilibrium position.

The kinetic energy of the system is, as always,

$K=frac{1}{2}mv^2$ ,

where is the mass of the block and is the speed of the block.

We will also assume that there are no resistive forces; that is,.

Consider a harmonic oscillator at four different moments,labeled A, B, C, and D, as shown in the figure. Assume that theforce constant ,the mass of the block, ,and the amplitude of vibrations, ,are given. Answer the following questions.

When the block is displaced a distancefrom equilibrium, thespring is stretched (or compressed) the most,and the block is momentarily at rest. Therefore, the maximumpotential energy is .At that moment, of course, . Recall that $E=K+U$ . Therefore,

$E=frac{1}{2}kA^2$ .

In general, the mechanical energy of a harmonic oscillatorequals its potential energy at the maximum or minimumdisplacement.

When the block is at the equilibrium position,the spring is not stretched (or compressed) at all. At that moment,of course, . Meanwhile, the block is at its maximumspeed( $v_max$ ). The maximum kinetic energy can then bewritten as. Recall that $E=K+U$ and that $U=0$ at the equilibrium position. Therefore,

.

Recalling what we found out before,

$E=frac{1}{2}kA^2$ ,

we can now conclude that

,

or

.

Question:PartGFind the kinetic energy of the block at the momentlabeled B.Express your answer in terms ofand .

Answer 1

Complete energy can be written like:k.A²/2(AtpositonofA)which is equal to the position at the B also(Energy Conservation)AtAposition x=A At B position x=A/2k.A²/2=k.A²/8+E_kE_k=EnergykineticWeknowthat Potential Energyofspring= kx²/2soE_k=3k.A²/8