# Tag Info

## New answers tagged physics

1

The first equation comes from Newton's second law, $F=ma$. You should write it in the form $$my''(t)=-g-by'(t)$$ where the left side is $ma$ and the right side is the force. You have implicitly defined negative to mean down (by the sign on your $g$, and thus up is positive. Well, when the ball is moving up, the air resistance pushes it down, not up. And ...

0

From $\varphi=\dot\theta,$ we have \begin{align*} \ddot{r}&=r\varphi^2\\ r\dot\varphi&=\dot{r}\varphi \\ \varphi&=Cr \quad\text{solve second equation for } \varphi \\ \ddot{r}&=C^2r^3 \\ \ddot{r} \dot{r}&=C^2r^3\dot{r} \\ \frac{\dot{r}^2}{2}&=\frac{C^2r^4}{4}+B. \end{align*} The first result follows from solving for $\dot{r},$ ...

0

You are correct that $\ddot{r} \dot{r} = r \dddot{r}/3$. But it's not quite enough to show that a solution of $\dot{r} = \sqrt{A r^4 + B}$ satisfies this equation, you'd have to show that every solution of $\ddot{r} \dot{r} = r \dddot{r}/3$ satisfies $\dot{r} = \sqrt{ A r^4 + B}$ for some constants $A$ and $B$. And if $A$ and $B$ are supposed to be real, ...

0

Requiring $h$ to be any function is not necessary. All you need is $h$ with support on arbitrary intervals, their size is irrelevant. I recommend you go over the proof of the lemma and convince yourself.

0

You have the speeds $v_1$ and $v_2$ on each of the two parts of the route, you have the total flight time $t$ for both parts of the route, you have that the distance of each part of the route was half the total distance, and you want to know the total distance $x.$ So if the times on the first and second parts were $t_1$ and $t_2$ and the distances were $... 0 We have: $$\frac d2\left(\frac 1 {v_1}+ \frac 1 {v_2}\right)=t\implies d=\frac {2v_1v_2}{v_1+v_2}t.$$ 1$t_1 = \frac{x}{2v_1}t_2 = \frac{x}{2v_2}t_1 + t_2 = \frac{17}{4}$You know$v_1$and$v_2$Find x comes to$1099.6875 \approx 1100$miles$\frac{x}{2v_1}+\frac{x}{2v_2} = \frac{17}{4} x\left(\frac{v_1+v_2}{2v_1v_2}\right) =\frac{17}{4}x = \frac{180\times460\times17\times2}{640\times4} = 1100$1 It works here because you have two variations, the angle and deflection. You can generalize the fundamental lemma like so: If two smooth functions$f$and$g$satisfy $$\int [f(x)\phi(x) + g(x)\psi(x)]~dx = 0$$ for all smooth compactly supported functions$\phi$and$\psi$, then$f = 0$and$g=0$. This follows easily from the standard version of the ... 0 Moment from the bus is$5000\times2\times g=10000g(NM)$Moment from the gold is$1000\times12\times g=12000g(NM)$Moment from the people is$70\times8\times n\times g=560ng(Nm)$Now, the total moment is$(22000-560n)g(NM)$0 You are summing up the moments about the edge of the cliff. The gold provides a clockwise torque of$(1000)(g)(12)$. The people provide a counterclockwise torque of$(70n)(g)(8)$. Then for the bus, for problems like this, we assume the mass is evenly distributed across its full length. And when that is the case, you can calculate the torque as if the mass ... 3 The angular velocity is the same at all latitudes. Then, the rotational speed is proportional to radius, meaning that$v=v_0\cos(\theta)$. 4 The time required for one full round trip is the same everywhere:$2\pi r/v$. What you need to know is the distance traveled: it is obviously$2\pi r$at the equator and$0$at the pole. At a general latitude$\phi$, trigonometry will tell you that the distance is$2\pi r \cos(\phi)$, so the speed is $$\frac{2\pi r \cos(\phi)}{2\pi r/v}=\cos(\phi) v$$ 5 At a latitude of$\alpha$, you uwill have a velocity of$v\cos\alpha$. 0 Considering $$L=\frac{1}{2}m\dot{q}^2-\frac{k}{2}(q^2+3q^2\dot{q})$$ we have$H = p\dot q - L$and here$p = \frac{\partial L}{\partial \dot q} = m\dot q-\frac 32 k q^2$so $$H(q,\dot q) = \frac 12(m \dot q^2+k q^2)$$ now eliminating$\dot qinto $$H = \frac 12(m \dot q^2+k q^2)\\ p = m\dot q-\frac 32 k q^2$$ we get $$H(p,q) = \frac{1}{8m}\... 1 The second Lagrangian is not equivalent to the given Hamiltonian. Let$$L=\frac{1}{2}m\dot{q}^2-\frac{k}{2}q^2-\frac{3}{2}kq^2\dot{q}$$Then the momentum is$$p=\frac{\partial L}{\partial \dot{q}}=m\dot{q}-\frac{3}{2}kq^2Which means that the Hamiltonian is \begin{align} H &=p \dot{q}-L\\ &=m\dot{q}^2-\frac{3}{2}kq^2\dot{q}-\frac{1}{2}m\dot{q}^2+\... 0 The Direct Method Denote the given integral as I, itrodice a=3-i and use 1/x=\int_{0}^{1} t^{x-1} dtI=\int_{0}^{1} \frac{dt}{t} \int_{0}^{\infty} e^{-ax} \sin x~ t^x ~dx=\int_{0}^{1} \frac{dt}{t} \int_{0}^{\infty} e^{-x(\ln (t^{-1}e^a))} \sin x ~dx\Rightarrow I=\int_{0}^{1} \frac{dt}{t} \int_{0}^{\infty} \exp[-x(\ln (t^{-1}e^a))]~ (2i)^{-1} [... 3I(a) = \int_0^\infty \frac {e^{-ax}\sin x}{x} \ dx$We need to find$I(3-i)$Differentiation under the integral sign.$I'(a) = \int_0^\infty -e^{-ax}\sin x \ dx\\ I'(a) = \frac {e^{-ax}(a\sin x + \cos x)}{a^2+ 1}|_0^\infty\\ I'(a) = -\frac {1}{a^2+ 1}\\ I(\infty) - I(3-i) = \int_{3-i}^\infty -\frac {1}{a^2+ 1}\ da =\arctan (3-i) - \frac{\pi}{2}\\ I(3-i) ...

0

a) You will find it easier to work in Euclidean space so the integrand becomes $Exp(S(\phi)/\hbar + J \phi)$ (your action is negative) , which falls off rapidly with $\phi$ and also gets the proper vacuum state. b) The right side of eq. 2 is the integral of a total derivative, so it is 0. c) The n comes from $\partial^n/\partial J^n (J exp(J \phi))$ on the ...

2

Following the dimensionaly correct equation on wikipedia, you have $$\frac{d^2 u}{d\theta^2}+u=\frac{r_s c^2}{2h^2}+\frac{3r_s}2 u^2$$ with $$r_s=\frac{2G\mu}{c^2}$$ Now $u$ is inverse distance, so in order to make it dimensionless you need to multiply it with a constant with units of distance. The obvious choice would be $r_s$. Then for $x=ur_s$, you ...

3

These are phasor notations. In $A = M\angle{ (\omega t+ \theta)}$, $M$ is the amplitude and $\theta$ is the phase shift. This can be equivalently written as $Me^{j(\omega t+\theta)}$, if angular frequency $\omega$ is considered. So, let $A = 9\angle37^o = 9e^{j37^o}$ $B = 3\angle20^o = 3e^{j20^o}$ So, $$A\cdot B = 9e^{j37^o}3e^{j20^o} = 27 e^{j 57^o}... 1 Let's first solve for the launch velocity. If the landing is at the same level as the cannon, the range of a projectile (assuming no air drag etc) is$$ d = \frac{v^2}{g} \sin{2 \theta} $$where v is the launch velocity, g is the acceleration of gravity and \theta is the launch angle. Solving for v, we obtain$$ v_{\text{launch}} = \sqrt{ \frac{dg}{...

3

The easiest way is to change into the center-of-mass frame. The kinetic energy contributions split into a center-of-mass component (all mass at CoM, moving at averaged velocity) plus the kinetic energy of relative motion: $$KE_{total}=KE_{CoM}+KE_{rel}$$ The $KE_{CoM}$ is unchanged in the collision by conservation of momentum. The relative KE is ...

1

Because $x$ is not measured from $A$ or $B$, but from $C$. The midpoint of $AB$ is at $0.6$ meters from $A$ (or $B$). The position of the end of the unextended spring is $0.4$ meters from $A$, which means that is $0.6-0.4=0.2$ meters from $C$.

0

Note that the kinetic energie $E_{kin}$ gained by the acceleration is $E_{kin} = \frac{m}{2}v^2$ For this energy it holds $E_{kin} = Pt$ Note, that $1 \mbox{horse} = 735.49875 \frac{kg\cdot m^2}{s^3}$ Putting all this together you get $$E = Pt\Rightarrow t = \frac{E}{P}= \frac{mv^2}{2P} = \frac{1800\cdot 17.5^2}{2\cdot 75\cdot 735.49875}s \approx 5 s$$...

0

I can highly recommend section 5-2 of Eisberg and Resnick's Quantum Physics of Atoms, Molecules, Solids, Nuclei, and Particles. I'll give a brief overview of the argument. I missed the part about middleschoolers in the OP. But this is a discussion that any high school student should be able to follow, and it shows why complex numbers are not just ...

2

Your mistake is you assumed the acceleration is constant. The acceleration is proportional to the force, which is not constant. Since the power $P$ is constant and since (as you know) $F = \frac PV,$ as $V$ increases (from $0$ to $17.5$) the force must decrease. You are better off to ignore the force entirely. Instead, you can use other facts about power. ...

0

A complex number represents a rotation and a scaling and translation of another complex number. That is about as physical as it gets in mathematics. Given two complex numbers, a+b is translating a by b. And a*b is rotating a by the angle of b, then scaling the result by the magnitude of b. A strange thing happens because the same value b represents both ...

-1

I don't know how seriously to take the suggestion, but the novel Smilla's Sense of Snow by Peter Høeg suggests that the crystal formation of ice is one such example. In one passage, the protagonist embarks on a long monologue about numbers: Do you know what the foundation of mathematics is? The foundation of mathematics is numbers. If anyone asks me ...

0

Here is a method working from a definition you may be using, or a feature of the centre of mass, that it acts as the balance point for a collection of masses. So you are working on a line and you have masses $m_1$ and $m_2$ placed at points $x_1$ and $x_2$. Suppose $x_1\lt x_2$, then we are looking for the centre of mass $c$, a point between $x_1$ and $x_2$ ...

1

The center of mass for an object, or group of objects, is the point about which the entire mass of the system is equally distributed. The formula given here is for the center of mass $$\text{center of mass}= \frac{\text{sum of all (position \times mass)}}{\text{sum of all masses}}$$ The concept of the center of mass is that of an average of the masses ...

3

${m_1 \over m_1+m_2}$ is the fraction of weight at $x_1$ and ${m_2 \over m_1+m_2}$ is the fraction of weight at $x_2$, so the balance point will be at $x^*={m_1 \over m_1+m_2} x_1 + {m_2 \over m_1+m_2} x_2$.

-1

Why not use space dimensions and cartesian coordinate frames to clarfy complex numbers? that is the most intuitive way to understand them that I came across when I was at middle school. Imagine an ant in a 1-D world. The ant can travel only backwards or forwards and its position can be described by one real number x. In a parallel universe, we have ants ...

-1

A physical example that should be comprehensible to bright 14 year olds. Fourier series and transforms can be used to describe functions and naturally produce complex numbers. I think complex numbers are easiest introduced by defining the solutions to $x^2+1=0$, analogously to negative numbers defining the solution to $x+1=0$. But it's good to have a ...

-2

Not 100% sure if it is a good example, but afaik the time-dependent Schrödinger equation contains i (https://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation#Time-dependent_equation).

0

I am not sure whether this counts but it may be useful. There is a simple, well-known formula for solving quadratic equations. This is frequently useful. There is a less well-known and more complex formula for cubic equations. One quirk of this is that you may need to take square roots of negative numbers even if they final answer is real. Today, ...

-1

I was amazed when I realised for the first time that osclilating events, like sound waves, are best described and calculated by complex numbers. That was almost twenty years after I learned about them at school. That was also the first real-world application of complex numbers that I knew. My teachers failed to give us such an application. So at school, I ...

4

I recommend the geometric algebra approach to complex numbers. We start with vectors in 2D. Vector addition and scalar multiplication are obvious. Vector multiplication is where it gets interesting. We assume that the product of vectors is associative, $\vec a(\vec b\vec c)=(\vec a\vec b)\vec c$, and that it distributes over addition. We also assume that a ...

3

I know you marked the question as answered but Impedance is a perfect physical phenomenon to show how complex numbers arise naturally. Impedance is kind of like regular resistance, but for AC circuits. Impedance behaves like a complex number, thus making it difficult to calculate the resistance (impedance) of AC circuits without using complex numbers. Here ...

0

If you want to find an equation that describes the current position $\vec{x}(t)$, you set up an equation of motion $$m\cdot\frac{\mathrm{d}^2 \vec{x}(t)}{\mathrm{d}t^2} = \vec{F}$$ where $\vec{F}$ is a force that you have to specify (can be gravitational force, Lorentz force, etc.). I assume, for your scenario it is likely the gravitational force applied ...

14

If you want a physical phenomenon for which complex numbers greatly simplify analysis, may I draw your attention to alternating electrical current? You can either use calculus to analyze how a AC signal responds to a given circuit of resistors, capacitors, and inductors, or you can use complex numbers that turn all of this calculus into algebra.

-1

Here is an "un"-natural but possibly fun thought-experiment for middle schoolers, provided they know the circle area formula $A = \pi r^2$. Start with a x-y coordinate plane, draw a circle at the center with radius 5. Show them the radius is 5 by drawing a line from (0,0) to (5,0). What's the area? $25\pi$. Now make the radius 4: draw your line from (0,0) ...

6

I would suggest the Riemann sphere. It maps latitude and longitude onto a single number in a simple and delightful way. After you have done that, you can rotate the Earth about the North Pole by multiplying by $e^{i\theta}$, rotate it $90°$ about the equatorial points $90°W$ and $90°E$ by transforming $z$ to $\frac{1+z}{1-z}$, and even reflect it in the ...

61

I don't know a simple, physical situation where complex numbers emerge naturally but I can suggest a way to help you teach middle schoolers about the emergence of complex numbers and I want to motivate this organically. I did this once as a guest lecturer in a middle school classroom by developing a geometric interpretation of arithmetic on the ...

5

While it's not "physical", I'd suggest perhaps the one "grade school" level introduction that is not artificial would, perhaps, ironically, be the one that is also that which motivated their creation in the first place and thus fits your bill of "something people were trying to solve that required their creation": the solution of a cubic polynomial. I ...

22

The historical origin of the complex numbers is, I think, the finest approach. Consider the problem of solving cubic equations of the type $x^3+px+q=0$. For this, you have Cardano's formula:$$x=\sqrt[3]{-\frac q2+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+\sqrt[3]{-\frac q2-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}.$$But what do you do if it turns out that $\frac{q^2}{... 7 Well..it may not be much use for middle-school students, but http://www.feynmanlectures.caltech.edu/I_22.html is pretty compelling. One natural answer for middle-school students is that sine-waves look an awful lot like cosine-waves, and we have a ton of formulas tying them together in various ways, but if we introduce phase, then things get pretty. So ... 0 Instead of trying to compute the collision point directly, compute the collision time (as Rahul commented). Using the distance formula that you used, solve $$(x_0+v_{x_0}t)^2+(y_0+v_{y_0}t)^2 = c^2t^2$$ for$t$, take the least nonnegative solution, and substitute that into the equations of motion of particle 2. 0 Take a loot ak the following picture: Particle 1 starts in$A=(0.0)$, particle 2 starts in$B=(x_0,y_0)$. Particale 2 travels with velocity$\lvert v \rvert$in the direction given by$v$. Both particles meet at point$C=(x,y)$, which is unknown. Note that the the value of$\beta=\angle ABC$is fixed and can be calculated, it is the angle between the lines ... 0 Since particle 1 is traveling in an unknown direction, your approach is correct. Otherwise it would just be a linear equation. I would eliminate the square root of your final equation, as it might confuse Wolfram. It might assume complex variables and then simple rules known from real square root cannot be applied anymore. You get$\left(\frac{x - x_0}{...

2

As you said, the time required for the cars to meet is $0.8$ hours. Where are the cars $0.8$ hours later? The first car will have travelled $14.4$ kilometers, and the second $21.6$. Thus both cars are in the same spot (duh) and since the bird stays between them, the bird is there also; hence the displacement is $14.4$ kilometers towards the second car.

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