October 2022 – Physicspupil

October 28, 2022November 2, 2022

202210280936 Problem 1.24

Expand $f(t)=\sin^2t\cos^3t$ in Fourier series.

_{Extracted from Hwei Piao Hsu. (1984). HBJ College Outline of Applied Fourier Analysis.}

Roughwork.

One can make use of the identities below:

$\begin{aligned} e^{\pm\mathrm{i}n\theta} & = \cos n\theta\pm\mathrm{i}\sin n\theta \\ \cos n\theta & = \frac{e^{\mathrm{i}n\theta}+e^{-\mathrm{i}n\theta}}{2} \\ \sin n\theta & = \frac{e^{\mathrm{i}n\theta}-e^{-\mathrm{i}n\theta}}{2\mathrm{i}} \\ \end{aligned}$

and thus

$\begin{aligned} & \begin{cases} \sin^2t = \displaystyle{\bigg(\frac{e^{\mathrm{i}t}-e^{-\mathrm{i}t}}{2\mathrm{i}}\bigg)^2 }\\ \cos^3t = \displaystyle{ \bigg(\frac{e^{\mathrm{i}t}+e^{-\mathrm{i}t}}{2}\bigg)^3 }\\ \end{cases} \\ \Longrightarrow & \begin{cases} \sin^2t = \displaystyle{\frac{e^{2\mathrm{i}t}-2e^{\mathrm{i}t}e^{-\mathrm{i}t}+e^{-2\mathrm{i}t}}{-4}} \\ \cos^3t = \displaystyle{\frac{e^{3\mathrm{i}t}+3e^{\mathrm{i}t}+3e^{-\mathrm{i}t}+e^{-3\mathrm{i}t}}{8}} \\ \end{cases} \\ \end{aligned}$

Or one can make use of the identity

$\sin^2\theta + \cos^2\theta = 1$ .

and hence

$\begin{aligned} f(t) & = (1-\cos^2t)\cos^3t \\ & = \cos^3t-\cos^5t \\ \end{aligned}$

where $\cos^5t$ is as expanded

$\begin{aligned} \cos^5t & = \bigg(\frac{e^{\mathrm{i}t}+e^{-\mathrm{i}t}}{2}\bigg)^5 \\ & = \frac{e^{5\mathrm{i}t}+5e^{3\mathrm{i}t} +10e^{\mathrm{i}t} +10e^{-\mathrm{i}t}+5e^{-3\mathrm{i}t}+ e^{-5\mathrm{i}t}}{32}\\ \end{aligned}$

so that

$\begin{aligned} f(t) & = -\frac{e^{5\mathrm{i}t}}{32}+\bigg(\frac{1}{8}-\frac{5}{32}\bigg)e^{3\mathrm{i}t} + \bigg(\frac{3}{8}-\frac{10}{32}\bigg) e^{\mathrm{i}t} \\ & \qquad + \bigg(\frac{3}{8}-\frac{10}{32}\bigg) e^{-\mathrm{i}t} + \bigg(\frac{1}{8}-\frac{5}{32}\bigg) e^{-3\mathrm{i}t} - \frac{e^{-5\mathrm{i}t}}{32}\\ & = -\frac{e^{5\mathrm{i}t}}{32}-\frac{e^{3\mathrm{i}t}}{32} + \frac{e^{\mathrm{i}t}}{16} + \frac{e^{-\mathrm{i}t}}{16} - \frac{e^{-3\mathrm{i}t}}{32} -\frac{e^{-5\mathrm{i}t}}{32} \\ & = \frac{1}{16}\Bigg( 2\bigg(\frac{e^{\mathrm{i}t}+e^{-\mathrm{i}t}}{2}\bigg) - \frac{e^{3\mathrm{i}t}+e^{-3\mathrm{i}t}}{2} - \frac{e^{5\mathrm{i}t}+e^{-5\mathrm{i}t}}{2} \Bigg)\\ & = \cdots \\ \end{aligned}$

Answer. $\displaystyle{\frac{1}{16}(2\cos t-\cos 3t-\cos 5t)}$ .

No slightest disrespect to de Moivre’s and Euler’s formulae but in due respect of Fourier’s analysis, this problem had need be treated again.

Background.

A function $f(t)$ satisfying $f(t+T)=f(t)$ is called a periodic function, and the smallest $T$ its period. A periodic function $f(t)$ can be represented by the trigonometric Fourier series

$\displaystyle{f(t)=\frac{1}{2}a_0+\sum_{n=1}^{\infty}(a_n\cos n\omega_0t+b_n\sin n\omega_0t)}$

$\displaystyle{f(t)=C_0+\sum_{n=1}^{\infty}C_n\cos (n\omega_0t-\theta_n)}$

where $\omega_0=2\pi /T$ . The coefficients of the Fourier series are found by using the orthogonality properties of sine and cosine functions over a period:

$\begin{aligned} \begin{Bmatrix} a_n \\ b_n \end{Bmatrix} & = \frac{2}{T}\int_{-T/2}^{T/2}f(t)\begin{Bmatrix}\cos n\omega_0t \\ \sin n\omega_0t \end{Bmatrix}\,\mathrm{d}t \\ C_0 & = \frac{1}{2}a_0 \\ C_n & = \sqrt{a_n^2+b_n^2} \\ \theta_n & = \tan^{-1}\bigg(\frac{b_n}{a_n}\bigg) \\ \end{aligned}$

_{Text on pg. 11}

For $f(t)=\sin^2t\cos^3t$ ,

$\begin{aligned} T & = 2\pi \\ \omega_0 & = 1 \\ \begin{Bmatrix} a_n \\ b_n \end{Bmatrix} & = \frac{1}{\pi}\int_{-\pi}^{\pi}\sin^2t\cos^3t\begin{Bmatrix}\cos nt \\ \sin nt\end{Bmatrix}\,\mathrm{d}t\\ \end{aligned}$

When $n=0$ :

$\begin{aligned} a_0 & = \frac{1}{\pi}\int_{-\pi}^{\pi}\sin^2t\cos^3t\,\mathrm{d}t \\ \cdots\enspace\textrm{Let }& u=\sin t\textrm{ s.t. }\mathrm{d}t=\frac{\mathrm{d}u}{\cos t}\enspace\cdots \\ & = \frac{1}{\pi}\int_{t=-\pi}^{t=\pi}u^2\cos^2t\,\mathrm{d}u \\ & = \frac{1}{\pi}\int_{t=-\pi}^{t=\pi}u^2(1-u^2)\,\mathrm{d}u \\ & = \frac{1}{\pi}\int_{t=-\pi}^{t=\pi}(u^2-u^4)\,\mathrm{d}u \\ & = \frac{1}{\pi}\bigg[ \frac{u^3}{3}-\frac{u^5}{5} \bigg]\bigg|_{t=-\pi}^{t=\pi} \\ & = \frac{1}{\pi}\bigg[ \frac{\sin^3t}{3}-\frac{\sin^5t}{5} \bigg]\bigg|_{t=-\pi}^{t=\pi} \\ & = 0 \\ \end{aligned}$

When $n=1$ :

$\begin{aligned} a_1 & = \frac{1}{\pi}\int_{-\pi}^{\pi}\sin^2t\cos^4t\,\mathrm{d}t \\ \dots\enspace &\textrm{By WolframAlpha}\enspace\dots \\ & = \frac{1}{\pi}\bigg[\frac{1}{192}(12t+3\sin (2t)-3\sin (4t)-\sin (6t))\bigg]\bigg|_{-\pi}^{\pi} \\ & = \frac{1}{\pi}\bigg(\frac{24\pi}{192}\bigg) \\ & = \frac{1}{8} \\ \end{aligned}$

a bit tiring an exercise.

(to be continued)

October 26, 2022January 19, 2023

202210261208 Dynamics Figures (Elementary) Q8

215. A cylindrical tube with a radius $r$ is connected by means of spokes to two hoops with a radius $R$ . The mass of both the hoops is $M$ . The mass of the tube and the spokes in comparison with the mass $M$ can be neglected. A string passed over a weightless pulley is wound around the tube. A weight with a mass $m$ is attached to the end of the string.

Find the acceleration $a=|\mathbf{a}|$ of the weight, the tension $T=|\mathbf{T}|$ of the string and the force of friction $f=|\mathbf{f}|$ acting between the hoops and the surface. (Assume that the hoops do not slip.) Also determine $k$ the coefficient of friction at which the hoops will slip.

_{Extracted from B. Bukhovtsev et al. (1978). Problems In Elementary Physics.}

Setup.

The kinetic energy $T$ of the dumbbell $M$ is in two parts, translational (/linear) and rotational (/angular), i.e.,

$\begin{aligned} \textrm{KE}_{\textrm{translational}} & = \frac{1}{2}Mv_M^2 \\ \textrm{KE}_{\textrm{rotational}} & = \frac{1}{2}I\omega^2 \\ \end{aligned}$

where the velocity of its centre of gravity (CG) is $v_M$ , and its moment of inertia for discs $I=MR^2$ . It has no potential energy $V$ of its position $h=0$ .

The kinetic energy $T$ of the weight $m$ is given by $\textrm{KE}=\frac{1}{2}mv_m^2$ , and its potential energy $V$ by $\textrm{PE}=mgh$ .

The Lagrangian $\mathcal{L}$ of a system is obtained from

$\mathcal{L}=T-V$ ,

whereby the Euler–Lagrange equation is given as

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{\partial\mathcal{L}}{\partial \dot{x}}\bigg)-\frac{\partial\mathcal{L}}{\partial x}=0}$

Note the relationship between $r$ , $R$ , $v_M$ , $v_m$ , and $\omega$ , i.e.,

$\begin{aligned} v_m & = r\omega \\ v_M & = R\omega \\ \end{aligned}$

Also beware that forces $F$ of any kinds, e.g., weight $\mathbf{W}=m\mathbf{g}$ , tension $\mathbf{T}$ , normal reaction $\mathbf{N}$ , and friction $\mathbf{f}$ , have not been taken into consideration.

(to be continued)

October 25, 2022January 19, 2023

202210251204 Dynamics Figures (Elementary) Q7

126. Two carts are pushed apart by an explosion of a powder charge $Q$ placed between them. The cart weighing $100\,\mathrm{kg}$ travels a distance of $18$ metres and stops. What distance will be covered by the other cart weighing $300\,\mathrm{kg}$ ? The coefficients of friction $k$ between the ground and the carts are the same.

_{Modified from B. Bukhovtsev et al. (1978). Problems In Elementary Physics.}

First, have a picture in mind:

Applying the law of conservation of linear momentum at the instant of impact,

$\begin{aligned} m_A\mathbf{u}_A+m_B\mathbf{u}_B & = (m_A+m_B)\mathbf{0} \\ -100u_A+300u_B & = 0 \\ u_A & = 3u_B \\ \end{aligned}$

and the law of conservation of mechanical energy after then,

$\begin{aligned} \textrm{KE}_A & = W_{f_A} \\ \frac{1}{2}m_Au_A^2 & = km_Ags_A \\ u_A & = 6\sqrt{kg} \\ u_B & = 2\sqrt{kg} \\ \cdots\cdots\cdots\cdots\enspace & \enspace\cdots\cdots\cdots\cdots \\ \textrm{KE}_B & = W_{f_B} \\ \frac{1}{2}m_Bu_B^2 & = km_Bgs_B \\ \frac{1}{2}(300)(2\sqrt{kg})^2 & = k(300)gs_B \\ s_B & = 2\\ \end{aligned}$

$\therefore$ The other cart will travel a distance of $2\,\mathrm{m}$ .

October 20, 2022October 24, 2022

202210201704 Solution to 1971-CE-AMATH-II-XX

A particle is projected vertically from the ground so that its velocity (in $\mathrm{ft/sec}$ ) in the first $4$ seconds is given by

$V_1=128-32t\qquad (0\leqslant t\leqslant 4)$

and after the fourth second by

$V_2=3t^2-32t+80\qquad (4\leqslant t)$

where $t$ is the time (in seconds) after projection.

i. Show that $V_1=V_2$ when $t=4$ .
ii. Calculate the height of the particle when $t=4$ .
iii. Calculate the height of the particle when $t=5$ .
iv. Find the value of $t$ when the acceleration of the particle is $40\,\mathrm{ft/sec^2}$ .

Roughwork.

i.

When $t=4$ :

$\begin{aligned} V_1 & = 128-32(4) \\ & = 0 \\ V_2 & = 3(4)^2-32(4)+80 \\ & = 0 \\ V_1 & = V_2 \\ \end{aligned}$

ii.

$\begin{aligned} h(t\in [0,4]) & = \int_{0}^{4}V_1\,\mathrm{d}t \\ & = \int_{0}^{4}(128-32t)\,\mathrm{d}t \\ & = \bigg[ 128\bigg(\frac{t^{(0)+1}}{(0)+1}\bigg) - 32\bigg(\frac{t^{(1)+1}}{(1)+1}\bigg)\bigg]\bigg|_{0}^{4} \\ & = [128(4)-16(4)^2]-[128(0)-16(0)^2] \\ & = 256\,\mathrm{ft} \\ \end{aligned}$

iii.

$\begin{aligned} h(t\in [0,5]) & = h(t\in[0,4]) + h(t\in[4,5]) \\ & = 256\,\mathrm{ft} + \int_{4}^{5}V_2\,\mathrm{d}t \\ & = 256\,\mathrm{ft} + \int_{4}^{5}(3t^2-32t+80)\,\mathrm{d}t \\ & = 256\,\mathrm{ft} + \big[ t^3-16t^2+80t\big]\big|_{4}^{5} \\ & = 256\,\mathrm{ft} -3\,\mathrm{ft} \\ & = 253\,\mathrm{ft} \\ \end{aligned}$

iv.

$\begin{aligned} a(t\in [0,4]) & = \frac{\mathrm{d}}{\mathrm{d}t}V_1=-32 \\ a(t\in [4,T]) & = \frac{\mathrm{d}}{\mathrm{d}t}V_2=6t-32 \\ \end{aligned}$

$\begin{aligned} 6t-32 & = 40 \\ t & = 12\,\mathrm{s}\\ \end{aligned}$

October 20, 2022October 28, 2022

202210201648 Solution to 1972-CE-AMATH-II-XX

It is given that $\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}=4x-\frac{3}{2}x^2}$ and that $y=10$ when $x=2$ . Find $y$ in terms of $x$ .

Roughwork.

$\begin{aligned} y & = \int\frac{\mathrm{d}y}{\mathrm{d}x}\,\mathrm{d}x \\ & = \int\bigg( 4x-\frac{3}{2}x^2\bigg)\,\mathrm{d}x \\ & = \frac{4x^{(1)+1}}{(1)+1} -\frac{3}{2}\bigg(\frac{x^{(2)+1}}{(2)+1}\bigg) + C\textrm{ for some constant} \\ & = 2x^2+\frac{1}{2}x^3 + C \\ \end{aligned}$

$\begin{aligned} 10 & = 2(2)^2+\frac{1}{2}(2)^3+C \\ C & = -2 \\ \end{aligned}$

In conclusion,

$y(x)=2x^2+\frac{1}{2}x^3-2$ .

October 20, 2022October 20, 2022

202210201553 Solution to 1977-CE-AMATH-I-XX

In the figure below, the complex numbers $z_0$ , $z_1$ , $z_2$ , $z_3$ , and $z_4$ are represented in the Argand diagram by the vertices of a regular pentagon with centre at the origin $O$ . If $z_0=2$ , write $z_2$ in polar form and calculate the value of $(z_2)^5$ .

Recall.

A complex number $z$ in Cartesian form $z=a+\mathrm{i}b$ can also be expressed in the polar form

$z=re^{\mathrm{i}\varphi}=r(\cos\varphi +\mathrm{i}\sin\varphi )$

where $r=\sqrt{a^2+b^2}$ is called the modulus $|z|$ , and $\varphi$ the argument $\textrm{arg}(z)$ , of $z$ .

Roughwork.

Observe that

$|z_0|=|z_1|=|z_2|=|z_3|=|z_4|=2$ .

and

$\displaystyle{\textrm{arg}(z_2)=2\pi\cdot\frac{2}{5}=\frac{4\pi}{5}}$ .

Hence

$z_2=2e^{\mathrm{i}\frac{4\pi}{5}}$ .

The remaining are left the reader as an exercise.

October 20, 2022October 20, 2022

202210201121 Solution to 1977-CE-AMATH-I-XX

Prove, by mathematical induction, that $23^n-1$ is divisible by $11$ .

Roughwork.

Let proposition $P(n)$ be that

$P(n):\enspace 11|23^n-1$ .

First, note

$\begin{aligned} & \quad\enspace P(1):\enspace 11|23^{1}-1 \\ & \Leftrightarrow P(1):\enspace 11|23-1 \\ & \Leftrightarrow P(1):\enspace 11|22 \\ & \Leftrightarrow P(1)\textrm{ is true} \\ \end{aligned}$

that $P(n)$ is true for $n=1$ .

Next, assumed be to $P(n)$ true.

$\begin{aligned} &\quad\enspace P(n+1):\enspace 11|23^{(n+1)}-1 \\ & \Leftrightarrow P(n+1):\enspace 11|23^{n}\cdot 23-1 \\ & \Leftrightarrow P(n+1):\enspace 11|23^n-1+23^n\cdot 22\\ & \dots\textrm{ by assumption }11|23^n-1\textrm{ and for }11|22 \\ & \Leftrightarrow P(n+1)\textrm{ is true} \\ \end{aligned}$

Thus, $P(n)$ is true implies $P(n+1)$ true.

From $P(1)\textrm{ true}\,\vee\, P(n)\Longrightarrow P(n+1)\textrm{ true}$ follows $P(n)$ the proposition is true for all positive integers $n\geqslant 1$ . By the principle of mathematical induction proven is that $11$ divides $23^n-1$ .

October 19, 2022October 19, 2022

202210191606 Solution to 1969-CE-AMATH-II-XX

Differentiate $2(3-4x)^{\frac{5}{6}}$ with respect to $x$ .

Roughwork.

$\begin{aligned} & \quad\, \frac{\mathrm{d}}{\mathrm{d}x}\Big( 2(3-4x)^{\frac{5}{6}}\Big) \\ & = 2\bigg( \frac{5}{6}\bigg) (3-4x)^{\frac{5}{6}-1}(-4) \\ & = -\frac{20}{3}(3-4x)^{-\frac{1}{6}} \\ & = -\frac{20}{3\sqrt[6]{3-4x}} \\ \end{aligned}$

October 19, 2022October 19, 2022

202210191554 Solution to 1969-CE-AMATH-II-XX

Find the $x$ -coordinate of the point on the curve $y=ax^2+bx+c$ where the tangent is horizontal.

Roughwork.

The gradient $y'(x)$ of the curve $y(x)$ is

$y'(x)=2ax+b$ .

When the tangent is horizontal,

$\begin{aligned} y'(x) & =0 \\ 2ax+b & =0 \\ x & = -\frac{b}{2a} \\ y & = a\bigg( \frac{-b}{2a}\bigg)^2+b\bigg( \frac{-b}{2a}\bigg)+c \\ & = \dots \\ \end{aligned}$

October 19, 2022October 19, 2022

202210191506 Solution to 1970-CE-AMATH-II-XX

Calculate the gradient of the curve $x^2+xy+y^2=1$ at the point $(1,-1)$ .

Roughwork.

Let the function $f$ of variables $x$ and $y$ be

$f(x,y):=x^2+xy+y^2-1=0$

Then,

$\begin{aligned} \Delta f=2x\Delta x+x\Delta y+y\Delta x+2y\Delta y & =0 \\ 2x + x\frac{\Delta y}{\Delta x}+y+2y\frac{\Delta y}{\Delta x} & = 0 \\ (x+2y)\frac{\Delta y}{\Delta x} & = -2x-y \\ \frac{\Delta y}{\Delta x} & = - \frac{2x+y}{x+2y} \\ \end{aligned}$

The gradient $\nabla$ at point $(x=1,y=-1)$ is

$\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}\bigg|_{(1,-1)}=-\frac{2(1)+(-1)}{(1)+2(-1)}=1}$ .