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6

In this equation $\omega$ does not refer to the speed of angular motion, but the frequency of oscillation when measured in angular terms (usually radians/sec, but it can be degrees/sec). Frequency is usually measured in cycles per second (Hertz), but it is sometimes more conveniently measured in angular terms, when it is called angular frequency. The angle ...


5

Imagine a point $P$ moving on a circle of radius $R$ with angular velocity $\omega$. The projection of $P$ onto the $y$-axis is: $$y=R\sin \theta=R\sin \omega t$$ The point $P'$ is in simple harmonic oscillation (SHO). For the spring mass system it just so happens that: $$x=A\sin \omega t$$ where: $$\omega=\sqrt{\frac{k}{m}}$$ So although there is no ...


3

The equation for the period $T$ comes by using Newton's second law $F=ma$ to obtain the equation of motion of the spring-mass system $$-kx =ma \Rightarrow a=-\frac k m$$ where $x$ is the displacement from a fixed point and $a$ is the acceleration. This equation is of the form $a=-\omega^2 x$ where $\omega$ is a constant of the simple harmonic motion. It ...


3

The annihilation operator is a linear operator. A linear operator can be applied to ANY state. And yes, it returns zero when applied to the ground state. You can really take this as a definition. The definition of the momentum operator $\hat{p}$ is the operator such that $\langle x|\hat{p}|\psi\rangle=-i\hbar \psi'(x)$. One could write this as $\langle ...


3

Similarly to AccidentalFourierTransform I am not sure to understand well your issue. However there is a crucial missed point in your argument, usually absent in many textbooks on these topics. It is true that decomposing $H$ as $H= \hbar\omega( a^\dagger a + \frac{1}{2})$ and taking the relations (following from CCR) $[a, a^\dagger] =I$ into account one ...


3

Well, I'm not sure I understood your question so I'm going to write what I think and let's see if it's useful :-) The algebra $[a,a^\dagger]=1$ is all you need to diagonalise $H$, but this is because what $H$ looks like: $$ H=\omega a^\dagger a $$ The important observables, namely $H,P,X$, can be written as polynomials in $a,a^\dagger$: \begin{aligned} ...


1

Let's look at what happens right when the falling jumper passes that equilibrium point (EP). At that point, as you correctly pointed out, the force on them (up, from the bungee) is equal to the force down (gravity). So, there is a total net force of zero. However, they already have momentum from falling. Newton's first law tells us that if there's no net ...


1

Now the real problem I'm having is trying to decide what the forces are acting on the system in order to come up with my differential equation? As there are no external forces the linear momentum of the system will be constant of motion and the constant can be taken as zero as well. If m1 and m2 are the masses at any time described by position ...


1

To do the equations of motion you need positions from an inertial reference frame. Say a wall far away. Call positions of the two masses $x_1$ and $x_2$. The spring force (tension is positive) is $$ F = k (x_2-x_1 ) $$ and the two equations of motion $$\begin{align} F & = m_1 \ddot{x}_1 \\ -F & = m_2 \ddot{x}_2 \end{align} $$ All this is trivial. ...


1

Assuming the spring is "ideal" (massless) you actually have 2 masses. You can describe your problem as the motion of the center of mass, and either of the masses. And if no external force is exerted on your system, you are only left with the motion of 1 mass relative to the center of the mass of the system. Let's say your masses are m1, m2, and the spring ...


1

Standing wave or Stationary wave is a result of two waves (incident wave and reflected wave) propagating in opposite direction with same amplitude and same frequency. This phenomena is a result of interference of wave where energies of two waves add or cancel out depending on the phase. (energies add up in case of complete in-phase and cancel out in case of ...


1

The differential equation you quote is fairly standard in university physics/engineering course but definitely requires some calculus to solve. As a first step, if you know how to differentiate products and chains, you can substitute the given solution into the differential equation and verify that it is indeed a solution. It contains two arbitrary constants ...



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