Add examples for de moivre and euler

mechanical/mmme1026_maths_for_engineering.md

@@ -42,14 +42,14 @@ $$\bar{z} = z -iy$$
 
 #### Example
 
-\begin{align*}
+> \begin{align*}
   z_1 &= 5 + i \\
   z_2 &= 1 -i \\
   \\
   \frac{z_1}{z_2} &= \frac{5+i}{1-i} \cdot \frac{\bar{z_2}}{\bar{z_2}} = \frac{(5+i)(1+i)}{(1-i)(1+i)} \\
     &= \frac{5 + i + 5i -1}{1 + 1} \\
     &= \frac{4 + 6i}{2} = 2 + 3i
-\end{align*}
+> \end{align*}
 
 ### Algebra and Conjugation
 
@@ -111,7 +111,7 @@ always hold true as there are many solutions.
 2. Find a value of $\theta$ such that $\tan \theta = \frac y x$ and check that it is consistent.
    If it puts you in the wrong quadrant, add or subtract $\pi$.
 
-# Lecture 2 // Complex Numbers (2021-10-11)
+# Lecture 2 // Complex Numbers (2021-10-12)
 
 ## Exponential Functions
 
@@ -146,15 +146,60 @@ $$e^{i\theta} = \cos\theta + i\sin\theta$$
 
   **Note**: $$\bar z = r\cos\theta - ir\sin\theta = re^{-i\theta}$$
 
-  Example:
+### Example 1
+
+Write $z = -1 + i$ in exponential form
 
-  > Write $z = -1 + i$ in exponential form:
-  >
 > $\arg z = \frac {3\pi} 4$
 > $|z| = \sqrt 2$
 > 
 > So $z = \sqrt2e^{i\frac{3\pi} 4}$
 
+### Example 2
+
+The equations for a mechanical vibration problem are found to have the following mathematical
+solution:
+
+$$z(t) = \frac{e^{i\omega t}}{\omega_0^2-\omega^2 + i\gamma}$$
+
+where $t$ represents time and $\omega$, $\omega_0$ and $\gamma$ are all positive real physical
+constants.
+Although $z(t)$
+is complex and cannot directly represent a physical solution, it turns out that the real and
+imaginary parts $x(t)$ and $y(t)$ in $z(t) = x(t) + iy(t)$ can. Polar notation can be used to extract
+this physical information efficiently as follows:
+
+a. Put the denominator in the form
+
+   $$ae^{i\delta}$$
+
+   where you should give explicit expressions for $a$ and $\delta$ in terms of $\gamma$, $\gamma_0$,
+   and $\gamma$.
+
+
+   > \begin{align*}
+     a &= \sqrt{\gamma^2 + (\omega_0^2 - \omega^2)^2} \\
+     \delta &= \tan^{-1}\frac \delta {\omega_0^2 - \omega^2}
+   > \end{align*}
+
+b. Hence find the constants $b$ and $\varphi$ such that
+
+   $$x(t) = b\cos(\omega t + \varphi)$$
+
+   and write a similar expression for $y(t)$.
+
+   > \begin{align*}
+     z &= \frac{e^i\omega t}{ae^{i\delta}} = \frac 1 a e^{i (\omega t - \delta)} \\
+     x + iy &= \frac 1 a \cos(\omega t - \delta) + \frac 1 a \sin(\omega t - \delta) \\
+     \therefore \Re z &= x = \frac 1 a \cos(\omega t - \delta), \\
+     \Im z &= y = \frac 1 a \sin(\omega t - \delta) \\
+     \\
+     b &= \frac 1 a = \frac 1 {\sqrt{\gamma^2 + (\omega_0^2 - \omega^2)^2}} \\
+     \varphi &= -\delta = \tan^{-1}\frac \delta {\omega_0^2 - \omega^2}\\
+     \\
+     y(t) &= \frac 1 a \sin(\omega t - \delta) \\
+   > \end{align*}
+
 ## Products of Complex Numbers
 
 Suppose we have 2 complex numbers:
@@ -193,7 +238,7 @@ z^n &= (re^{i\theta})^n \\
 
 Write $1+i$ in polar form and use de Moivre's theorem to calculate $(1+i)^{15}$.
 
-\begin{align*}
+> \begin{align*}
   r &= |1+i| = \sqrt2 \\
   \theta &= \arg{1+i} = \frac \pi 4 \\
   \\
@@ -203,7 +248,7 @@ r &= |1+i| = \sqrt2 \\
              &= 2^{\frac 15 2} \left(\frac 1 {\sqrt2} - \frac i {\sqrt2}\right) \\
              &= 2^7 (1 - i) \\
              &= 128 - 128i
-\end{align*}
+> \end{align*}
 
 ### Example 2
 
@@ -215,13 +260,28 @@ Use de Moivre's theorem to show that
 \sin{2\theta} &= 2\sin\theta\cos\theta
 \end{align*}
 
-Let $n=2$:
+> Let $n=2$:
 
-\begin{align*}
+> \begin{align*}
   (\cos\theta+i\sin\theta)^2 &= \cos^2\theta + 2i\sin\theta\cos\theta - \sin^2\theta \\
   \text{Real part: } \cos^2\theta - \sin^2\theta &= \cos{2\theta}\\
   \text{Imaginary part: } 2\sin\theta\cos\theta &= \sin{2\theta}
-\end{align*}
+> \end{align*}
+
+### Example 3
+
+Given that $n \in \mathbb{N}$ and $\omega = -1 + i$, show that
+$w^n + \bar{w}^n = 2^{\frac n 2 + 1}\cos{\frac{3n\pi} 4}$ with Euler's formula.
+
+> \begin{align*}
+  r &= \sqrt{2} \\
+  \arg \omega = \theta &= \frac 3 4 \pi \\
+  \\
+  \omega^n &= r^n(cos{n\theta} + i\sin{n\theta}) \\
+  \bar\omega^n &= r^n(cos{n\theta} - i\sin{n\theta}) \\
+  \omega^n + \bar\omega^n &= r^n(2\cos{n\theta}) \\
+                          &= 2^{\frac n 2 + 1}\cos{\frac {3n\pi} 4}
+> \end{align*}
 
 ## Complex Roots of Polynomials
 
@@ -231,45 +291,45 @@ Which complex numbers $z$ satisfy
 
 $$z^3 = 8i$$
 
-1. Write $8i$ in exponential form,
+> 1. Write $8i$ in exponential form,
-
+> 
-   $|8i| = 8$ and $\arg{8i} = \frac \pi 2$
+>    $|8i| = 8$ and $\arg{8i} = \frac \pi 2$
-
+> 
-   $\therefore 8i = 8e^{i\frac \pi 2}$
+>    $\therefore 8i = 8e^{i\frac \pi 2}$
-
+> 
-
+> 
-2. Let the solution be $r = re^{i\theta}$.
+> 2. Let the solution be $r = re^{i\theta}$.
-
+> 
-   Then $z^3 = r^3e^{3i\theta}$.
+>    Then $z^3 = r^3e^{3i\theta}$.
-
+> 
-3. $z^3 = r^3e^{3i\theta} = 8e^{i\frac \pi 2}$
+> 3. $z^3 = r^3e^{3i\theta} = 8e^{i\frac \pi 2}$
-
+> 
-   i. Compare modulus:
+>    i. Compare modulus:
-
+> 
-      $r^3 = 8 \rightarrow r = 2$
+>       $r^3 = 8 \rightarrow r = 2$
-
+> 
-   ii. Compare argument:
+>    ii. Compare argument:
-
+> 
-       $$3\theta = \frac \pi 2$$
+>        $$3\theta = \frac \pi 2$$
-
+> 
-       is a solution but there are others since 
+>        is a solution but there are others since 
-
+> 
-       $$e^{i\frac \pi 2} = e^{i \frac \pi 2 + 2n\pi}$$
+>        $$e^{i\frac \pi 2} = e^{i \frac \pi 2 + 2n\pi}$$
-
+> 
-       so we get a solution whenever
+>        so we get a solution whenever
-
+> 
-       $$3\theta = \frac \pi 2 + 2n\pi$$
+>        $$3\theta = \frac \pi 2 + 2n\pi$$
-
+> 
-       for any integer `n`
+>        for any integer `n`
-
+> 
-       - $n = 0 \rightarrow z = \sqrt3 + i$
+>        - $n = 0 \rightarrow z = \sqrt3 + i$
-       - $n = 1 \rightarrow z = -\sqrt3 + i$
+>        - $n = 1 \rightarrow z = -\sqrt3 + i$
-       - $n = 2 \rightarrow z = -2i$
+>        - $n = 2 \rightarrow z = -2i$
-       - $n = 3 \rightarrow z = \sqrt3 + i$
+>        - $n = 3 \rightarrow z = \sqrt3 + i$
-       - $n = 4 \rightarrow z = -\sqrt3 + i$
+>        - $n = 4 \rightarrow z = -\sqrt3 + i$
-       - The solutions start repeating as you can see
+>        - The solutions start repeating as you can see
-
+> 
-       In general, an $n$-th order polynomial has exactly $n$ complex roots.
+>        In general, an $n$-th order polynomial has exactly $n$ complex roots.
-       Some of these complex roots may be real numbers.
+>        Some of these complex roots may be real numbers.
-
+> 
-4. There are three solution
+> 4. There are three solutions

mechanical/mmme1048_fluid_mechanics.md

@@ -151,3 +151,95 @@ The -ve sign indicates that as $z$, height, increases, $p$, pressure, decreases.
 
 - It is usually better to use SI units
 - If in doubt, DA can be useful to check that your answer makes sense
+
+# Lecture 2 // Manometers (2021-10-13)
+
+![](./images/vimscrot-2021-10-13T09:09:32,037006075+01:00.png)
+
+$$p_{1,gauge} = \rho g(z_2-z_1)$$
+
+- Manometers work on the principle that pressure along any horizontal plane through a continuous
+  fluid is constant
+- Manometers can be used to measure the pressure of a gas, vapour, or liquid
+- Manometers can measure higher pressures than a piezometer
+- Manometer fluid and working should be immiscible (don't mix)
+
+![](./images/vimscrot-2021-10-13T09:14:59,628661490+01:00.png)
+
+\begin{align*}
+p_A &= p_{A'} \\
+p_{bottom} &= p_{top} + \rho gh \\
+\rho_1 &= density\,of\,fluid\,1 \\
+\rho_2 &= density\,of\,fluid\,2
+\end{align*}
+
+Left hand side:
+
+$$p_A =  p_1 + \rho_1g\Delta z_1$$
+
+Right hand side:
+
+$$p_{A'} = p_{at} + \rho_2g\Delta z_2$$
+
+Equate and rearrange:
+
+\begin{align*}
+p_1 + \rho_1g\Delta z_1 &= p_{at} + \rho_2g\Delta z_2 \\
+p_1-p_{at} &= g(\rho_2\Delta z_2 - \rho_1\Delta z_1) \\
+p_{1,gauge} &= g(\rho_2\Delta z_2 - \rho_1\Delta z_1)
+\end{align*}
+
+If $\rho_a << \rho_2$:
+
+$$\rho_{1,gauge} \approx \rho_2g\Delta z_2$$
+
+## Differential U-Tube Manometer
+
+![](./images/vimscrot-2021-10-13T09:37:02,070474894+01:00.png)
+
+- Used to find the difference between two unknown pressures
+- Can be used for any fluid that doesn't react with manometer fluid
+- Same principle used in analysis
+
+\begin{align*}
+p_A &= p_{A'} \\
+p_{bottom} &= p_{top} + \rho gh \\
+\rho_1 &= density\,of\,fluid\,1 \\
+\rho_2 &= density\,of\,fluid\,2
+\end{align*}
+
+Left hand side:
+
+$$p_A = p_1 + \rho_wg(z_C-z_A)$$
+
+Right hand side:
+
+$$p_B = p_2 + \rho_wg(z_C-z_B)$$
+
+Right hand manometer fluid:
+
+$$p_{A'} = p_B + \rho_mg(z_B - z_a)$$
+
+\begin{align*}
+p_{A'} &= p_2 + \rho_mg(z_C - z_B) + \rho_mg(z_B - zA)\\
+       &= p_2 + \rho_mg(z_C - z_B) + \rho_mg\Delta z \\
+\\
+p_A &= p_{A'} \\
+p_1 + \rho_wg(z_C-z_A) &= p_2 + \rho_mg(z_C - z_B) + \rho_mg\Delta z \\
+p_1 - p_2 &= \rho_wg(z_C-z_B-z_C+z_A) + \rho_mg\Delta z \\
+&= \rho_wg(z_A-z_B) + \rho_mg\Delta z \\
+&= -\rho_wg\Delta z + \rho_mg\Delta z 
+\end{align*}
+
+## Angled Differential Manometer
+
+![](./images/vimscrot-2021-10-13T09:56:15,656796805+01:00.png)
+
+- If the pipe is sloped then
+
+  $$p_1-p_2 = (\rho_m-\rho_w)g\Delta z + \rho_wg(z_{C2} - z_{C1})$$
+
+- $p_1 > p_2$ as $p_1$ is lower
+- If there is no flow along the tube, then
+
+  $$p_1 = p_2 + \rho_wg(z_{C2} - z_{C1})$$