In physical chemistry, the van der Waals forces (or van der Waals' interaction), named after Dutch scientist Johannes Diderik van der Waals, are the residual attractive or repulsive forces between molecules or atomic groups that do not arise from a covalent bond, or electrostatic interaction of ions or of ionic groups with one another or with neutral molecules. The resulting van der Waals forces can be attractive or repulsive.
The term includes:
force between permanent dipoles (Keesom force)
force between a permanent dipole and a corresponding induced dipole (Debye force)
force between inst...
@DeNiSkA Aha, so we're looking at a soliton propagating along the string. Yes?
@DeNiSkA That was of my attempt at giving you hint, but I'm not sure it's really useful. Ultimately, consider what happens to a given point on a string as the wave passes.
@JohnRennie : Permission to speak? Arnold gave some good references, but when you actually read them it's all fairy tales and mathematical handwaving that totally ignores what a gravitational field actually is. And then his "dry facts" add up to electrons and positrons pop into existence spontaneously like worms from mud. It's just not enough.
@DeNiSkA Yes, that's why I said it :P OK, focus on a point on the string that's 5 inches from the left end of the string. Now, as the wave hits it, what is its y position (this actually isn't easy, but since I'm not doing the math, it's fun)
@DeNiSkA Perhaps more specifically, what's the velocity of that specific point on the string.
@DeNiSkA I might mean acceleration (force = tension)
@JohnRennie Come back, I have so much love to give.
@DeNiSkA You know two things about a given point on the string: 1) it goes from y=0 to y=1 and back to y=0, 2) unless the string is stretchy, a given point remains a constant distance from one end of the string.
I love giving hints that imply I know the answer, when I don't.
This question came in my exam.
My Attempt: I thought that the tension will not vary in the string because since we know that velocity of wave in a string is given by
$$v=(T/m)^{1/2} $$
where $v$ is the velocity of the progressive wave in string and $T$ is tension in the string and $m$ is ma...
This may or may not be helpful.
The diagram above shows the following in the planet frame:
The solid blue line is Earth, which remains 8 light years distant from
the planet at all times. I've drawn dots for t=-8 to t=10.
The solid green line is the planet, which remains at x=0 at all
t...
Since David Z wants us to write a trivial comment's worth of a remark as an answer:
1) Plenty of such functions exist. The radial distance of a planet as a function of angle is one. In general all periodic functions are non-injective.
2) No, since it's not true. I am not even sure how one woul...
Easier example of non-injective function in physics: The motion of a vibrating string. Since it (the position of one point on the string as a function of time) is periodic in time, it cannot be injective.
@barrycarter $$\ddot y= -\sqrt{8} \pi \cos(\pi x+ \pi t)$$ and this equation tells me at yellow point tension is maximum and at orange it is zero! which is totally opposite to the correct answer (*i think i am going wrong somewhere*) picture............
@ACuriousMind May I ask another question? May you explain this sentence for me: "Since David Z wants us to write a trivial comment's worth of a remark as an answer"
@lucas I'm not sure what you're saying. Whenever I have an expression a = b*c/d, a is a function of b,c and d. So when I write x = sin(wt), I have written the position as a function of time (and angular frequency), and as a function of time, it is not injective.
@DeNiSkA Not exactly, though. OK, if you have a piece of rope/string handy, mark one spot on the rope and then pass a wave through it. Notice the spot goes closer to the origin briefly.
@DeNiSkA I'm talking about a single fixed point on the string, which is a little different from the string as a whole.
@lucas I would guess that you are fairly young and just learning physics, and having difficulty communicating because your English skills are (admittedly) not that great.
@lucas I actually didn't see your question and don't have an opinion about whether it's stupid. In general, I used your comments to determine your age, not vice versa.
@lucas However, do keep in mind that homework questions are frowned upon here, so you might try another place to ask.
@lucas Hmmm. See, the issue here is that you speak a different language, and don't appear to know much about physics, so the questions you ask could be both difficult to read and not make much sense. That's probably why we won't be able to help you much.
@DeNiSkA Umm, no. Because x[t]^2 + y[t]^2 is fixed, remember?
@DeNiSkA Any point on the string is constrained to a fixed distance from the origin.
@barrycarter I don't need my questions answers. I left the science since 7 years ago. I want to help the others. My questions are useful for thinking people, but they bother people only have studied and don't like thinking.
hullo, $F=ma$ this equation doesn't give any physical meaning similarly $OPL=\mu x$ (where $\mu$ is refractive index and $x$ is path length) doesn't give any physical meaning or sense! i want to know what does optical path length acutally mean?
ok sir, tell me how to do this : in young's double slit experiment if a thin sheet of thickness t and RI $\mu$ is introduced find change in optical path
@lucas "I want to help the others" - It's been my observation that those that genuinely help others simply do it and that the phrase "I want to help the others" is almost always used by those with something else in mind.