TITLE: Determining length of tube from acoustic reflection
QUESTION [0 upvotes]: I am interested in determining the length of a closed ended tube. I would like to do this using an acoustic signal, like a sine wave. Depending on the length of the tube, I believe the reflected signal will arrive back at the source with a different phase. The actual length of the tube is on the order of 2-3cm. I am interested in sending an acoustic wave that has a frequency of 18-20kHz.
However, I am not sure how to relate the phase to the length of the tube. Is there an intuition or equation I could use to help calculate this?

REPLY [0 votes]: Assuming the tube behaves ideally, the simplest possible answer is to use $v=f\lambda$: If the tube has some length $L$, it will take $2L/v$ seconds for the sound to return in which $2Lf/v = 2L/\lambda$ cycles will have occurred.
In principle, supposing you have a source of sound with pressure wave $P(t) = P_0 \sin(\omega t)$ the measured signal at that precise point in space will be $$P_0\sin(\omega t) + \alpha P_0 \sin(\omega t + \phi)$$ where $\phi = 2\pi \frac{2L}{\lambda}$ is the relative phase offset and $\alpha < 1$ is some attenuation from loss of sound.
By using some tedious trigonometry, it can be shown that the net measured acoustic intensity will become
$$ P' \sin(\omega t + \delta)  $$
where $P'=P_0\sqrt{1 + \alpha^2 + 2\alpha \cos(\phi)}$ and $\delta = \text{atan2}(\alpha \sin(\phi), 1 + \alpha \cos(\phi))$. (Note that $\phi$ depends on $\omega$ as well, since $\phi = 2Lv/\omega$)
Physically, this is sensible: the detected intensity is higher than the emitted intensity, and in the limit $\alpha\to0$, $P' \to P_0, \delta \to 0$.
In principle, we're done - By (electronically) measuring the phase offset between the applied signal and measured signal, it ought to be possible to recover $\delta$ and hence $L$. However, the electronics required for such a process are not trivial to design.
In terms of engineering such a system practically, distance measurements like this are usually achieved by sending a very short pulse of sound and measuring the time until it echoes. Many commercial modules are available that achieve this.


TITLE: Potential difference between point on surface and point on axis of uniformly charged cylinder
QUESTION [3 upvotes]: Question:
Charge is uniformly distributed with charge density $ρ$ inside a very long cylinder of radius $R$.
Find the potential difference between the surface and the axis of the cylinder.
Express your answer in terms of the variables $ρ$, $R$, and appropriate constants.
$Attempt:$
I am struggling with determining which Gaussian surface to use. If I use a cylinder, then the cylinder would have an infinite area, right? How can I deal with that? If I use a sphere (since I am trying to find the potential difference between only two points, one on the surface and one on the axis), what will be the charge inside the sphere?
If I use a sphere as my Gaussian surface, I get:
$$\int \overrightarrow{E}.d\overrightarrow{A}=\frac{Q }{\epsilon _{0}}$$
$$\Delta V = -\int_{i}^{f}\overrightarrow{E}.d\overrightarrow{s}$$
$$E = \frac{\rho  }{4\pi R^{2}\epsilon _{0}}$$
$$\Delta V = \frac{\rho  }{4\pi R^{2}\epsilon _{0}} \int_{0}^{R}dR=\frac{\rho  }{4\pi R\epsilon _{0}}$$
But this is wrong.

REPLY [2 votes]: All the above answers are correct, although none gives you an answer as to WHY you should NOT use a sphere and none addresses your "infinite gaussian surface" problem and you seem to be a bit confused on how things actually work (you have to understand how things work before you delve into the mathematics part).
So, if you use a sphere, then the integral in Gauss' law will not be easy because most of the electric field vectors will not be perpendicular (and thus you can get $E$ out of the integral). You are simply not using the symmetries of the system.
You must use your intuition to guess where are the electric field vectors are pointing (here outwards cylindrically because you have an infinite cylinder), so you must use a cylinder. Think about the symmetries of the sources to figure out the symmetries of the fields.
Now, about your infinite gaussian surface problem: You only have to create a gaussian surface of finite size, because Gauss' law gives you the NET (total) electric field -created from all its surroundings- on one side of the equation (in the integral) but you can find it by only the enclosed charge (other side of the equation). So, you must not have an infinite gaussian surface.


TITLE: Why is $\alpha(t)=(t^3,t^2)$ not an immersion?
QUESTION [3 upvotes]: $$\alpha: \mathbb{R}\to\mathbb{R}^2\quad \alpha(t)=(t^3,t^2)$$
The condition for an immersion is that $\alpha'(t)=(3t^2,2t)$ is injective for all $t\in\mathbb{R}$.
The book seems to imply that there is a problem at $t=0$ but I don't understand why. At $t=0$ the tangent is undefined because if you take the limit from both sides you get opposite results. However that does not make $\alpha'$ non-injective.
What am I missing?

REPLY [1 votes]: Keep in mind that a map between manifolds $f:M\to N$ is called an immersion at the point $p\in M$ when the differential $f_{*p}:T_pM\to T_{f(p)}N$ is $1-1$, that is when $\ker(f_{*p})=\{0\}$. $f$ is called an immersion when it is an immersion on every point of the domain manifold $M$.
Let's see if we can apply this to your map $a:\mathbb{R}\to\mathbb{R^2}$ with $a(t)=(t^3,t^2)$. For a point $p$ of $\mathbb{R}$, it is $a_{*p}:T_p\mathbb{R}\to T_{(p^3,p^2)}\mathbb{R}^2$ with $\displaystyle{a_{*p}\big{(}\frac{d}{dt}\vert_p\big{)}=\frac{dt^3}{dt}\vert_p\frac{\partial}{\partial x}|_{(p^3,p^2)}+\frac{dt^2}{dt}\vert_{p}\frac{\partial}{\partial y}\vert_{(p^3,p^2)}=3p^2\frac{\partial}{\partial x}\vert_{(p^3,p^2)}+2p\frac{\partial}{\partial y}\vert_{(p^3,p^2)}}$,
so for any vector $\lambda\frac{d}{dt}\vert_p\in T_p\mathbb{R}$:
$\displaystyle{a_{*p}\big{(}\lambda\frac{d}{dt}\vert_p\big{)}=3\lambda p^2\frac{\partial}{\partial x}\vert_{(p^3,p^2)}+2\lambda p\frac{\partial}{\partial y}\vert_{(p^3,p^2)}}$
Therefore, at the point $p=0$ you have $a_{*0}(v)=0$ for any $v\in T_0\mathbb{R}$. Hence $\ker(a_{*0})=T_0\mathbb{R}$, and $a$ is not an immersion at point $0$. Now if you exclude 0 and consider the domain of $a$ to be the submanifold $(0,+\infty)$, $a$ is indeed an immersion.


TITLE: Borel sets in $\Bbb R (0,1]$ element of $B(\Bbb R)$
QUESTION [0 upvotes]: Determine if interval $(0,1]$ is an element of $B(\Bbb R)$ Borel real numbers.
Can anyone help me with this? I know what sigma algebras are and Borel sets.
Also if $(0,1]$ is an element of $B(0,1]$ does that mean it is also element of $B(\Bbb R)$?

REPLY [0 votes]: Note $(0,1]^c = (-\infty,0] \cup (1,\infty)$
$ (-\infty,0]$ is closed in $\mathbb R$ and $(1,\infty)$ is open in $\mathbb R$. Hence belongs borel sigma algebra.
Now by definition of sigma algebra, complement of the union of above sets belongs to sigma algebra. Hence interval $(0,1]$ is an element of borel sigma algebra on $\mathbb R$.


TITLE: Boolean algebra, unsure if I am doing it correct.
QUESTION [1 upvotes]: So I am in my second year of university, and am being shown examples of logic gates. I know that I have the output correct, but am trying to get it simplified to decrease the amount of gates needed.
Here is the expression
$$\overline{\overline{A.\overline{AB}}.\overline{B.\overline{AB}}}$$
$$\big((A\land (A\land B)^\prime)^\prime \land (B\land (A\land B)^\prime)^\prime\big)^\prime$$
I have a solution for it but am sure it's not correct; it's a lot of flipping on the operators. If anyone can post a solution with workings that would be great.
(probably easier if you do it on pen and paper then post a picture)

REPLY [0 votes]: $X=\lnot(\lnot(A\land\lnot(A\land B))\land\lnot(B\land\lnot(A\land B)))$
$C=A\land B$
$X=\lnot(\lnot(A\land\lnot C)\land\lnot(B\land\lnot C))$
$D=\lnot C$
$X=\lnot(\lnot(A\land D)\land\lnot(B\land D))$
$E=A\land D$
$X=\lnot(\lnot E\land\lnot(B\land D))$
$F=\lnot E$
$X=\lnot(F\land\lnot(B\land D))$
$G=B\land D$
$X=\lnot(F\land\lnot G)$
$H=\lnot G$
$X=\lnot(F\land H)$
$I=F\land H$
$X=\lnot I$
$\mathtt{AB⋅CDEFGHI⋅X}$
$\mathtt{00⋅0101011⋅0}$
$\mathtt{01⋅0101100⋅1}$
$\mathtt{10⋅0110010⋅1}$
$\mathtt{11⋅1001011⋅0}$
$X=((A\land\lnot B)\lor(\lnot A\land B))$
$\text{or }X=\lnot(A\equiv B)$


TITLE: linear dependence proof using subsets
QUESTION [0 upvotes]: how do you prove if a subset of a set of vectors is linearly dependent, then the entire set is linearly dependent.
i know that linear dependence means that it can be written as a linear combination.
so i can take 1 vectors from the set and write it as a linear combination of some of the vectors find it to be linearly dependent and do so for the whole and find it to be linearly dependent.
i do not know how i can prove  without actually resorting to an example and you cant prove by example.

REPLY [0 votes]: Quite formally and suitable for arbitrary diensions:
Let $\{v_i\}_{i\in I}$ be a family of vectors. Assume the subfamily $\{v_i\}_{i\in J}$ with $J\subset I$ is linearly dependant. That means by definition that there is a family $\{a_i\}_{i\in J}$ of scalars such that almost all $a_i$ are zero and at least one $a_i$ is nonzero and $\sum_{i\in J}a_iv_i=0$. Let $\{b_i\}_{i\in I}$ be the family of scalars given by
$$b_i=\begin{cases}a_i&\text{if }i\in J\\0&\text{if }i\notin J\end{cases} $$
Then one immediatyly verifies that almost all $b_i$ are zero, at least one $b_i$ is nonzero and $\sum_{i\in I}b_iv_i=0$, thus showing that  $\{v_i\}_{i\in I}$  is linearly dependant.


TITLE: Jordan measurable for closure and interior
QUESTION [2 upvotes]: Give an example of a bounded set $S\subseteq\mathbb{R}^n$ such that $\overline{S}$ and $\operatorname{Int}S$ are Jordan-measurable, but $S$ is not.
Jordan-measurability of a set $A$ is equivalent to the condition that $\operatorname{Bd}A$ has measure zero, where $\operatorname{Bd}A$ is a set of points that belong to neither the interior nor the exterior of $A$.
We have that $\operatorname{Int}S\subseteq S\subseteq\overline{S}$, so the desired condition is tricky to satisfy. I don't know what set $S$ to pick.

REPLY [3 votes]: Let's take $n = 2$, and identify $\mathbb{R}^2$ with $\mathbb{C}$ for simplicity.
Then let
$$S = \{ z : \lvert z\rvert < 1\} \cup \bigcup_{\varphi \in \mathbb{Q}} \left\{t\cdot e^{2\pi i\varphi} : t \in [1,2]\right\}.$$
Then $S$ is obviously bounded, $\operatorname{int}(S) = \{ z : \lvert z\rvert < 1\}$ and $\overline{S} = \{ z : \lvert z\rvert \leqslant 2\}$ are Jordan measurable. But $\operatorname{Bd}(S) = \{z : 1\leqslant \lvert z\rvert \leqslant 2\}$ is not a null set, so $S$ itself is not Jordan measurable.