The simple harmonic motion of a mass point m is obtained by the force F=−kx where k is a positive constant, and was already discussed in (2.1.4.18-24). In the present case the initial conditions are
, where x is the coordinate and v the velocity.
Obtain the expression of energy conservation.
What is the maximal possible kinetic energy Kmax ?
Obtain the mean value of the kinetic energy with respect to x in terms of the maximal kinetic energy! Use the interval
.
Use the already known solution of the motion:
in order to obtain the mean value of the kinetic energy with respect to time, integrated over one time period!
Reminder
(2.2.2.10)
The expression (2.2.210) is called the mean value of T with respect to x.
... the second law of Newton for the motion of a mass point m in one dimension with a force F dependent on the coordinate x only is
(2.2.2.14)
where v is the velocity and t - the time.
... we'll call U the potential energy.
(2.2.2.16)
One can obtain the total energy from the initial conditions ... as
(2.2.2.18)
which represents a very useful relation between v² and the coordinate x.
Parts 1-2
Solution of question 1
The potential energy for F=−kx is
For the given initial conditions the expression of energy conservation is
Part 3
Solution of question 2
It follows from part 2 that the maximal kinetic energy is obtained at x = 0 , therefore
Parts 4-5
Solution of question 3
From parts 2 the mean kinetic energy can be expressed as
and according to (2.2.2.10)
where Kmax from part 3 was substituted.
Parts 6-8
Solution of question 4
The kinetic energy as a function of time is
One period of time corresponds to
therefore the mean kinetic energy with respect to time becomes
Score
By parts:
Parts 1,2,3,4,6,7 are each worth one point.
Parts 5,8 are each worth 2 points.
By questions:
Questions 1 is worth 2 points.
Questions 2 is worth 1 point.
Question 3 is worth 3 points.
Question 4 is worth 4 points.