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4

$A$ is constant in the approximate solution at large $\xi$. So Griffiths makes the ansatz that for general $\xi$, the solution is of the form for some function $A(\xi)$ that becomes "constant" compared to the exponentation at large $\xi$. It's an ansatz, it is not derived. Look at the condition on $K$ that terminating the sequence starting from a non-zero ...


3

It's not a coincidence! You can see the reason even in classical mechanics: if you take a charge and shake it sinusoidally at frequency $\omega_q$, it makes light with equal frequency $\omega_{\gamma} = \omega_q$. If you quantize light wave emission into individual photons, so that $E = \hbar \omega_{\gamma}$, the spacing between harmonic oscillator energy ...


3

The equations of motion for the position determine the accelerations: they are second-order differential equations in time: $$\vec F = m\vec a = m\ddot{\vec x }$$ So the acceleration, the second derivative of the location in time, has to be determined from the state of the physical system in some way. Typically, it is determined using the $F=ma$ formula ...


2

Assuming an incompressible liquid, Bernoulli for instationary flow (neglecting friction) is $$ \int_1^2 \frac{\partial c}{\partial t} \, \mathrm{d}s + \tfrac12 (c_2^2-c_1^2) + g(z_2-z_1) + \frac1{\rho}(p_2-p_1)=0 $$ with velocity $c$, gravitation accceleration $g$, height $z$, density $\rho$ and pressure $p$ and $1$ and $2$ denoting the two positions ...


2

Suppose you do a force balance on the portion of the fluid situated between elevations z and $z +\Delta z$ in the left column. You get: $$p(z+\Delta z)S-p(z)S+\rho g S\Delta z=\rho S\Delta z \frac{dv}{dt}\tag{1} $$where $v$ is the downward velocity in the left column:$$v=-\frac{dx}{dt}\tag{2}$$ The latter equation is correct because the fluid is ...


2

To expand on Andrei's answer a bit, start with your equation: $$ \ddot x + \omega^2 x = C $$ and rewrite it as: $$ \ddot x + \omega^2 \left(x - \frac{C}{\omega^2}\right) = 0 $$ Then define a new variable $y$ by: $$ y = x - \frac{C}{\omega^2} $$ and differentiate twice to get: $$ \ddot{y} = \ddot{x} $$ Finally substitute into your original equation to ...


2

Just add a constant say $C$ to your solution. $x=A\cos(\omega t)+B\sin(\omega t)+C$ Taking the second derivative, the term with $C$ will be 0. But you still have $\omega^2C=-V\omega/c$ so $C=-V/(\omega c)$. Therefore $x=A\cos(\omega t)+B\sin(\omega t)-V/(\omega c)$


2

A photon is produced by a transition between two levels and by definition of "photon" its energy is $h\nu$, where $\nu$ is the frequency of the classical electromagnetic wave that will emerge from a great number of the same energy photons. So it is a matter of coincidence only because Maxwell's equations have sinusoidal solutions for the electromagnetic ...


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I think you made an mistake in the algebra. Let's focus on the PDE $$p \frac{\partial \rho }{\partial q} - q \frac{\partial \rho }{\partial p} = 0.$$ You claim that $\rho(p,q) = \exp^{\, f(p+q)}$ is a solution to this equation for any function $f$, but that's simply not true. If you plug that Ansatz in the PDE, you get $$ (p-q)\; f'(p+q) = 0 $$ which is ...


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In fact you can think of freely propagating electromagnetic field as an infinite set of harmonic oscillators. To see this, note that (with appropariate choice of unit system) energy density of electromagnetic field is proportional to $\vec E ^2 + \vec B ^2$. This expression is sum of two quadratic terms, which closely resembles Hamiltonian of harmonic ...


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You don't actually have to use harmonic motion,you can solve this question by using equations of motion. Given that the collision are elastic (coefficient of restitution is 1),i.e,when the ball collides with the building its velocity is reversed. The point to note here is only the horizontal component is affected by the collision. Along the ...



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