]> Exercise 4

# Definitions and Basics, Exercise 4

## Question

According to Newton's mechanics for one dimension along the x axis, a point with mass m has an acceleration a: $\begin{array}{l}a=\frac{F}{m}\text{ }\text{where}\text{ }F=\text{force}\\ a=\frac{\text{d}v}{\text{d}t}\text{ }\text{where}\left\{\begin{array}{l}v=\frac{\text{d}x}{\text{d}t}=\text{velocity}\\ t=\text{time}\end{array}\end{array}$

In this particular case the force is: $F=C\mathrm{sin}\left(\omega t\right)$ where C and ω are known constants.

1. The following rule is necessary for the solution of the exercise: $\begin{array}{l}\text{If}\text{ }\int f\left(x\right)dx=I\left(x\right)\\ \text{then}\text{ }\int f\left(bx\right)dx=\frac{I\left(bx\right)}{b}\end{array}$ where b is a constant. Prove it!
2. Obtain the velocity as a function of the time!
3. Obtain the position x as a function of the time!
4. In the expression of x as a function of time that you found, prove that all the terms have the correct physical dimensions!

## Reminder

 $\left\{\begin{array}{l}f\left(x\right)=\frac{\text{d}I}{\text{d}x}\\ I\left(x\right)=\int f\text{d}x\end{array}\right\}$ (2.1.1.6)

The function I(x) from (2.1.1.6) is called the indefinite integral of fdx , and its meaning is that its derivative is the function f(x).

 $\int {x}^{p}\text{d}x=\frac{{x}^{p+1}}{p+1}+C$ (2.1.1.10)
 $\left\{\begin{array}{l}\int \mathrm{cos}x\text{ }\text{d}x=\mathrm{sin}x+C\\ \int \mathrm{sin}x\text{ }\text{d}x=-\mathrm{cos}x+C\\ \int \left(1+{\mathrm{tan}}^{2}x\right)\text{d}x=\int \frac{\text{d}x}{{\mathrm{cos}}^{2}x}=\mathrm{tan}x+C\\ \int \left(1+{\mathrm{cot}}^{2}x\right)\text{d}x=\int \frac{\text{d}x}{{\mathrm{sin}}^{2}x}=-\mathrm{cot}x+C\end{array}\right\}$ (2.1.1.14)

## Parts 1-2

Solution of question 1

1. According to (2.1.1.6) one has to prove that $\begin{array}{l}\text{If}\text{ }\frac{\text{d}I\left(x\right)}{\text{d}x}=f\left(x\right)\\ \text{then}\text{ }\frac{\text{d}I\left(bx\right)}{b\text{d}x}=f\left(bx\right)\end{array}$
2. From the first expression of part 1 it follows that

$\frac{\text{d}I\left(u\right)}{\text{d}u}=f\left(u\right)$ where u is any variable.
But by choosing u=bx and taking in account that du=bdx we obtain $\frac{\text{d}I\left(bx\right)}{b\text{d}x}=f\left(bx\right)$ and the statement that $\begin{array}{l}\text{If}\text{ }\int f\left(x\right)dx=I\left(x\right)\\ \text{then}\text{ }\int f\left(bx\right)dx=\frac{I\left(bx\right)}{b}\end{array}$ is proven.

## Parts 3-5

Solution of question 2

1. from the data presented in this exercise we have:

$\frac{\text{d}v}{\text{d}t}=a=\frac{F}{m}=\frac{C\mathrm{sin}\left(\omega t\right)}{m}$ and the velocity v is $v=\frac{C}{m}\int \mathrm{sin}\left(\omega t\right)\text{d}t$
2. From (2.1.1.14) formally: $\int \mathrm{sin}\left(t\right)\text{d}t=-\mathrm{cos}t+C$ and therefore from part 2 we obtain: $\int \mathrm{sin}\left(\omega t\right)\text{d}t=-\frac{\mathrm{cos}\omega t}{\omega }+C$
3. Finally we obtain from parts 3 and 4 that the velocity is:

$v=-\frac{C\mathrm{cos}\omega t}{m\omega }+{C}_{1}$ where C1 is a constant of the integration.

## Parts 6-7

Solution of question 3

1. Since $\frac{\text{d}x}{\text{d}t}=v$ the position x should be obtained from $x=\int v\text{d}t=-\frac{C}{m\omega }\int \mathrm{cos}\left(\omega t\right)\text{d}t+{C}_{1}\int \text{d}t$

2. By following the footsteps of parts 4 and 5 we obtain finally

$x=-\frac{C}{m{\omega }^{2}}\mathrm{sin}\left(\omega t\right)+{C}_{1}t+{C}_{2}$

where C2 is a constant of integration.

## Parts 8-10

Solution of question 4

1. For the solution of question 4 we'll use the notation of

$\left[z\right]=\text{physical dimensions of}\text{\hspace{0.28em}}\text{}z\text{}$ In this case $\begin{array}{l}\left[x\right]=\Lambda \text{ }\left(\text{length}\right)\\ \left[t\right]=\Theta \text{ }\left(\text{time}\right)\end{array}$ From the data supplied to this exercise: $\left[a\right]=\left[\frac{F}{m}\right]=\left[\frac{C}{m}\right]=\Lambda {\Theta }^{-2}$
2. From the argument of sine and cosine: $\left[\omega \right]={\Theta }^{-1}$ From this together with part 8 one obtains: $\left[x\right]=\left[\frac{C}{m{\omega }^{2}}\mathrm{sin}\left(\omega t\right)\right]=\Lambda$which takes care of the correct dimensions of the first right term in part 7.
3. From part 5 $\left[{C}_{1}\right]=\Lambda {\Theta }^{-1}$ and from part 7 $\left[{C}_{2}\right]=\Lambda$ which take care of the remaining terms.

## Score

By questions:

Questions 1,3 give credit of 2 points each.
Questions 2,4 give credit of 3 points each.

By parts:

All parts are worth 1 point each.