DQW

hey

 

i am not sure whether i am getting this problem correct...

 

i reckon the answer would be:

 

f(n) = nv/{lamda1} ; where n = 1' date=' 3, 5, 7, . . .

 

{lamda1} being the fundamental harmonic's wavelength,

and

v being the speed of sound in air

 

well anyway, thats what i think the answer is but yeah i am new to thise stuff, so just thought i'd check

 

-Sarah[/quote']

Your answer is perfectly correct.

 

I hope you understand exactly why this is true - that's the important part.

 

All you have to do is draw a couple of pictures, and the rest follows automatically.

No immediate thoughts on how that came about. But as I said, I know very little about strobes, so perhaps I should read up something about how they work. I'll do that, if I find some time tomorrow. It's late now.

 

Here's my thinking. The LCM of 1/6 and 1/8 is 1/2. This means that every 1/2 second, (or at every third frame) the phase of the wave will be the same. So, I guessed that the apparent period is 1/2 second or the apparent frequency is 2 Hz. This gave me a wavelength of 6cm.

I get an answer of 6 cm but I'm not certain about it yet. What is the answer in the book?

If a wave with wavelength 1.5 cm and frequency 8Hz' date=' is viewed through a stroboscope with 6Hz, what's the change of its wavelength?

My answer is not identical to what in my text book.

My thought is :there is no change of the wavelength, as the time given is enough for the wave to move over one wavelength.[/quote']I'm not terribly familiar with the working of stroboscopes, but I'm pretty sure that you do not have enough information ot say that there is sufficient time to view an entire wavelength. The question says that the strobe frequency is 6 Hz. This means that there are 6 viewign windows each second. The length of each viewing window could be any number from say, a nanosecond to just under one-sixth of a second. My guess is that you must assume the viewing windows to be infinitesimally small.

 

PS : What is the answer by the textbook ?

 

Edit : Do not answer that yet...let me make a suggestion first.

deleted...no patience now for long post

Any thoughts or comments?

Yes, it's a load of hooey !

I know I'm screwing up somewhere, but Q2 (day 1) looks trivial to me. Anyone else looked at this ?

would you be able to point me to somewhere where i can find this inequality written out properly (i.e. a maths webpage or something?)

I doubt you'd find it written out anywhere. All it takes to prove it, is writing out explicitly, each term of the expression on either side of the inequality and multiplying out the terms on the RHS.

[math]eg:~~\sum_n |a_n| = (|a_1|~+~|a_2|~+~|a_3|~+~\dots ) [/math]

The inequality holds; though it should be "strictly less than", unless all the terms are 0. But you can not just "make it up" and hope it's true, so understand why it is true.

The binomial expansion is the Taylor expansion about x=0

I think Dave meant to say that he got 3/2 by working out :

 

[math] \lim_{k\to\infty} \left| \frac{a_k}{a_{k+1}} \right| < 1[/math] ?

i drew a triangle on the complex plane to find the tring functions for an angle. i started with 45 degrees. the hypotenuse was zero.

I can't make any sense of this. Can you attach a picture ?

Assuming you are a 48 lb. arthritic senior citizen!

Hey !!! I was throwing in some room for the benefit of doubt.

Dear reyveta,

 

WAKE UP !

Yes, that's correct. In fact, between "substituting" and "eventually" is hardly one step.

Ice, why do you want to make zinc sulfide ?

Look up Faraday's Laws of Electrolysis.

1. Typical bar magnets have a field strengths of 0.05 to 0.2T. NdFeB and SmCo5 type supermagnets are the only magnets you will find that have field strengths of order 1T (only at the magnet poles - the field decays outside the magnet, and the field along the sides of the magnet is much smaller than at the poles).

 

2. Mumetal saturates at about 0.8T. There are other alloys that saturate much higher, which have permeabilities in the many thousands.

 

3. Unless you have a VERY powerful magnet (> 3 tesla pole strength!!!), mumetal will achieve better than 90% (and in most cases, I guess better than 99%) shielding for the geometry described in the OP - at a spot that's half a magnet length away, along the equatorial direction. This is my opinion.

 

4. We have come a long way, from saying that DC magnetic shielding is a myth to making it look like it was meant all along, that you were talking about DC fields much in excess of 1T.

 

5. This statement : "In these cases, supposing the placement of a strong permanent magnet the size of a golf-ball (no lightweight item), you are dealing with shielding field-strengths [imath]\frac{1}{10^{10}}[/imath] of the strength at the source" shows that you do not understand the equation that you began an earlier post with. In that equation, Bo is not the magnetic field at the source. It is the field at the surface of the shield. The result quite clearly shows that the shielding efficiency is independent of the distance or size of the magnet (as long as Bo < B(sat)).

But how did you calculate it ? That's the important part.

Solving jointly all of the equations with regard to the integration constants, we finally obtain:

What equations ? What integration constants (having solved enough and more Laplacians, I can guess) ??? Holy cow ! If you are going to throw equations at us to "prove" your "point" you ought to do better than using fancy words which mean absolutely NOTHING. I hope you realize this reeks of scientific disingenuity.

 

[math]S = \frac{H^I}{H_O} = \frac{4\mu_s r^2_e}{r^2_e(1 + \mu_s)^2 - r^2_i (1 - \mu_s)^2} = \frac{4}{\mu_s} \frac{1}{1-\frac{r^2_i}{r^2_e}} [/math]

I'll accept this equation without proof, for now. But next time you come up with something like this, as "proof" I'm going to insist that you show the derivation or else provide a link to the place you cut-and-pasted from. How are all those symbols supposed to make sense to someone who has no idea what they mean ?

 

Nevertheless, let's start from here :

 

 

[math]S = \frac{4\mu_s r^2_e}{r^2_e(1 + \mu_s)^2 - r^2_i (1 - \mu_s)^2} = \frac{4}{\mu_s} \frac{1}{1-\frac{r^2_i}{r^2_e}} [/math]

 

I make the substitutions, [imath]r_e \equiv r~;~~r_i = r-t~,~~t<<r [/imath] where r is the outer radius and t is the wall thickness. Then,

 

[math]{1-\frac{r^2_i}{r^2_e} = {1-\frac{(r-t)^2}{r^2} = \frac {r^2 - (r-t)^2}{r^2} = \frac {2rt - t^2}{r^2} \approx \frac{2t}{r} [/math]

 

Mumetal® can have a relative permeability of 230,000 or higher. For now, I'll use a conservative number [imath]\mu _s = 2 \cdot 10^5 [/imath]. Now if the radius of your spherical shell (or whatever the heck it is you wrote that equation for) is 10cm (about the size of a bowling ball) and has a wall thickness of 10mm, this gives you S = 1/10000 !!! Even if you had giant sphere of radius 2m (a 13 ft dia enclosure, bigger than my kitchen) with a wall thickness of only 4mm (less than a quarter inch, hardly enough for just structural strength) you have an impressive shielding ratio of S = 1/200 !

 

In other words, when you say...

Even if the shield's magnetic permeability [math]\mu_s > 1[/math] and the shield thickness is small, [math] ( r_e = r_i ) [/math], the shield effectiveness is not very good. Only very thick shields with large permeability provide protection from static and low-frequency magnetic fields.

...you are TALKING THROUGH YOUR HAT.

 

And what on earth does "Even if the shield's magnetic permeability [math]\mu_s > 1[/math]" mean ? I'll tell you what : it is misleading the casual reader into believing that you are making a rare exception by permitting the shield material to have a relative permeability greater than 1. For every metal that you can name with [imath]\mu _s < 1 [/imath], you know I can name tens of metals with [imath]\mu _s >> 1 [/imath]. You must know that diamagnetic susceptibilities are several orders of magnitude smaller than ferromagnetic susceptibilities ! Yet you make a statement like this ? You know that most shielding materials have [imath]\mu _s = 100,000 ~to~1,000,000 [/imath] or more, and you have the gall to make it look like having [imath]\mu _s > 1 [/imath] is a rarity among shielding metals ?

 

And then you go on to obfuscate the matter even more (Hoping what ? That nobody will check the details ?), rather than accept that you were wrong.

 

Mumetal is a very good DC magnetic shield, and there's NO DENYING THAT.

 

http://www.mushield.com/

http://www.goodfellow.com/csp/active/STATIC/E/Mumetal.HTML

http://farside.ph.utexas.edu/teaching/jk1/lectures/node52.html

 

"Magnetic Shielding for a Spaceborne Adiabatic Demagnetization Refrigerator (ADR)", Brent A. Warner, Peter J. Shirron, Stephen H. Castles, Aristides T. Serlemitsos, Adv. Cryo. Engg., 37, 907 (1992)

 

" An analysis of magnetic shielding against DC power lines based on homogenization", Waki, H; Igarashi, H; Honma, T, Int J for Comp. Math. in Electrical and Electronic Eng., 24, 2, 566-580 (2005)

More generally, notice that :

 

[math]1-r^n = (1-r)(1+r+r^2+ \dots +r^{n-1})[/math]

[Math]rS-S=S(r-1)=-1/ (r-1)[/Math]

That is incorrect...or so it looks to me. I think you meant :

[Math]rS-S=S(r-1)=-1[/Math]

[math]\implies S = 1/(1-r) [/math]

At time t=0

(i) the ray of light just enters the top of the scope, at L, and

(ii) the bottom of the scope is at P and moving to the right at speed v

 

At time t=T

(i) the ray of light reaches the bottom of the scope at Q, and hence

(ii) the bottom of the scope must be at Q

 

So, in the time it takes for the ray to travel down the scope, the scope bottom has moved from P to Q.

On something like a 1/4"-20 screw, you can generate of order an inch-pound of torque (converts to about 0.12 Nm, I think). Pascal is the wrong unit for expressing a torque.

It is flux-pinning that greatly stabilizes the SC in a typical levitation demo.

 

http://superconductors.org/terms.htm#ferr

 

http://www.spacedaily.com/news/future-99f.html

 

http://www.iop.org/EJ/abstract/0953-2048/12/9/306