Math Animated™
Mathematical Introduction for Physics and Engineering
by Samuel Dagan (Copyright © 2007)

Chapter 2: Integration; Section 2: Definite Integrals; page 2

Applications, Exercise 4

Question

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 $x_0\ne 0\text{and}{v}_0=0$, where x is the coordinate and v the velocity.

1. Obtain the expression of energy conservation.
2. What is the maximal possible kinetic energy  K_max ?
3. Obtain the mean value of the kinetic energy with respect to  x  in terms of the maximal kinetic energy! Use the interval $\left(0,x_0\right)$ .
4. Use the already known solution of the motion: $$\begin{array}{l}x=x_0\cos \left(\omega t\right)\\ v=-\omega x_0\sin \left(\omega t\right)\\ \omega =\sqrt{\frac{k}{m}}\end{array}\}$$ in order to obtain the mean value of the kinetic energy with respect to time, integrated over one time period!

Reminder

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

where v is the velocity and t - the time.

... we'll call U the potential energy.

One can obtain the total energy from the initial conditions ... as

which represents a very useful relation between v² and the coordinate x.

Parts 1-2

Solution of question 1

1. The potential energy for F=−kx is
   
   $$U\left(x\right)=k\int x\text{d}x=\frac{k{x}^2}{2}$$


2. For the given initial conditions the expression of energy conservation is
   
   $$\frac{m{v}^2}{2}+\frac{k{x}^2}{2}=\frac{k{x}_0^2}{2}$$

Part 3

Solution of question 2

3. It follows from part 2 that the maximal kinetic energy is obtained at  x = 0 , therefore
   
   $$K_{\max }=\frac{k{x}_0^2}{2}$$

Parts 4-5

Solution of question 3

4. From parts 2 the mean kinetic energy can be expressed as
   
   $$\langle K\rangle =\frac{k}{2}\left(x_0^2-\langle x^2\rangle \right)$$


5. and according to (2.2.2.10)
   
   $$\begin{array}{l}{\langle K\rangle }_x=\frac{k}{2}\left(x_0^2-\frac{{\displaystyle \underset{0}{\overset{x_0}{\int }}{x}^2\text{d}x}}{{\displaystyle \underset{0}{\overset{x_0}{\int }}\text{d}x}}\right)=\\ =\frac{k}{2}\left(x_0^2-\frac{x_0^3}{3x_0}\right)=\frac{k{x}_0^2}{2}\frac{2}{3}=\frac{2}{3}{K}_{\max }\end{array}$$
   where  K_max  from part 3 was substituted.

Parts 6-8

Solution of question 4

6. The kinetic energy as a function of time is
   
   $$K=\frac{m{v}^2}{2}=\frac{m{\omega }^2{x}_0^2{\sin }^2\left(\omega t\right)}{2}=\frac{k{x}_0^2{\sin }^2\left(\omega t\right)}{2}$$


7. One period of time corresponds to
   
   $$\omega t=2\pi \text{or}t=\frac{2\pi }{\omega }$$


8. therefore the mean kinetic energy with respect to time becomes
   
   $$\begin{array}{l}{\langle K\rangle }_t=\frac{k{x}_0^2}{2}\frac{{\displaystyle \underset{0}{\overset{\frac{2\pi }{\omega }}{\int }}{\sin }^2\left(\omega t\right)}\text{d}t}{{\displaystyle \underset{0}{\overset{\frac{2\pi }{\omega }}{\int }}\text{d}}t}=\frac{k{x}_0^2}{2}\frac{{\displaystyle \underset{0}{\overset{2\pi }{\int }}{\sin }^{2}u\text{d}u}}{2\pi }=\\ =\frac{k{x}_0^2}{2}\frac{\pi }{2\pi }=\frac{1}{2}{K}_{\max }\end{array}$$

Score

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