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Anonymous
Topics Covered: a) SHM as UCM b) Same Ampltitude Same Mean Position Double SHM c) Same Mean Different Amplitude SHM (Crossing + Overtaking +Meeting <---- 3 cases)
Anonymous
Imp: Maximum distance between particles in double SHM having constant phase difference occurs when they are in a st.line parallel to diameter along which projection is being taken.
Anonymous
04:46
Anonymous
@Yashas Have a look at this question ^ I'm confused. When an identical body is attached to the bob at distance $l\alpha/2$ won't the $A$ change? If the $A$ changes then even though the $\omega$ remains same the angle between that middle position and mean won't be $\pi/4$. Shouldn't the time taken be $\frac{\pi/4}{\omega}+\frac{\sin^{-1}(\frac{l\alpha/2}{A'})}{\omega}$
Anonymous
Where $A'$ is the new amplitude
05:54
@blue From where you are writting this
Anonymous
@Koolman Etoos...
@blue does they provide study material also
Anonymous
Yes
nice
Anonymous
@Koolman You want to solve an SHM question?
Anonymous
06:05
I've got a good one
I will try it
send
Anonymous
Okay 1 min
Anonymous
Anonymous
Find angular frequency!
Anonymous
The block m is pulled down a bit and left
Anonymous
06:07
What is the angular frequency of oscillation of the system?
Anonymous
(No slipping anywhere)
ok
mass of cylinder or sphere
Anonymous
Cylinder
Anonymous
$M$
mass of block and cylinder is different
Anonymous
06:10
Mass of block is $m$
Anonymous
And cylinder is $M$
@blue Try diffrentiating: (1/2)((k(2x)^2+m(v/2)^2+Mv^2+Iw^2)
Anonymous
@TheDeadLegend I know the solution. This is a quiz. And I want the solution by the force method.
my x term is not coming
Anonymous
@Koolman There will be around 5 or 6 equations. You need to check your calculations properly
Anonymous
06:21
It must come
Anonymous
You can write your equations here
Force method. Okay. Complications.
Anonymous
Energy method doesn't provide one the insight that force method provides.
Anonymous
It is more of a shortcut
mg-T=ma
T-kx-f=Ma
$TR/2+fR-kxR=(1/2)MR^2 \alpha$
$\alpha = a/R$
@blue is these are correct
Anonymous
06:29
@TheDeadLegend Secondly that equation is wrong. If block goes down by $x$ spring doesn't extend by $2x$.
Anonymous
@Koolman Where are the eqm conditions?
Anonymous
@Koolman This is wrong
Those would be similiar like this
@blue why
Anonymous
There is an initial extension in the spring
Anonymous
You didn't take that into account
06:31
$x=x_0 + x'$
$x_0$ at equilibrium
Anonymous
Okay, then?
$f=\mu Mg$
at equilibrium we don't know about friction
Anonymous
You don't need to...just take it as $f_o$
and tension will also be changed
Anonymous
Yes, even tension will change
06:35
At equilibrium $2f' +T'=2kx_0$
torque equation
Anonymous
@Koolman How did you get the $2$ there?
multiplying the whole equation by 2 as $TR/2$
Anonymous
Oh, that's the torque equation
Anonymous
Fine
Thats it . These are the equation I know
Anonymous
06:41
Your mistake is while writing $x$. Relate the distance x by which block goes down with the distance by which spring is extended
Anonymous
You messed up by writing $x=x_o+x'$
Anonymous
Take initial extension as $h$ and extra distance by which block moves down as $x$
finally I am getting $2mg-4kx' =3(M+m)a$
Anonymous
How is $a$ coming to be proportional to $x$ ?
Anonymous
06:58
@Koolman Will I write my equations?
09:12
Hi
Yes I too feel force method gives a feel of the problem
09:26
blue: Please write your equations..
Anonymous
09:39
@samjoe @Koolman The answer is $$\sqrt{\omega=\frac{16k}{9m+6M}}$$.
Anonymous
Force Method:
Anonymous
$\frac{3}{2}mg=2kh_o$
Anonymous
$k(h_o+\frac{4}{3} x) + f - (ma+mg) = M (\frac{2}{3} a)$
Anonymous
$k(h_o+\frac{4}{3} x) R -fR - (ma+mg)\frac{R}{2} = (\frac{1}{2} MR^2)(\frac{2}{3}a)$
11:13
0
Q: Length of shortest path

user123733What is the length of shortest path that begins at the point (-1,1) , touches the x axis and then ends at point on the parabola $(x-y)^2 =2(x+y-4)$ . I put $x=x-y$ $y=x+y-4$ then the starting point would be $(-4,-2)$ . can I proceed by this methos of changing the coordinates After that I g...

Can anybody explain me how to with my method
Anonymous
11:32
@user123733 Take reflection of (-1,1) on x axis. Find shortest distance of that point from parabola! Done!
Anonymous
@user123733 Oh sorry, I didn't notice you already got an answer.
12:26
@blue can we not do it by changing the coordinate system
@blue And in that how we get point (2,2)
Anonymous
@user123733 Why do you want to change the coordinate system? That would unnecessarily complicate it.
Anonymous
@user123733 You can get (2,2) by observation
Anonymous
Or else, if you really want you can put x-y=Y and x+y-4=X and shift the origin. Find shortest distance using calculus.
12:43
@blue I think would become very complicated
@blue by observation ?
Anonymous
@user123733 You can find the axis of the parabola. Can't you?
Anonymous
You will see that (2,2) and (-1,-1) both lie on the axis
@blue Physics galaxy is very helpful
Anonymous
@Programmer Indeed
12:58
I am atleast able to hear some terms
Anonymous
Did you learn any chapter ?
Anonymous
From there?
Yes. Mechanics first part
@blue okay thanks
 
1 hour later…
14:28
Hey guys thanks for the inorganic tricks! What tricks do you use to learn names of ORES and MINERALS? E.g ZnS : Zinc Blende, FeCO3 : Siderite ?
I usually keep looking at the names once in a while, but still forget some of them..
Hey one more question, in JEE 2015, one question came in the paper (organic):
14:42
I made the answer as B
But answer is D
They shifted the double bond to more substituted dbl bond, but how can we decide that this will occur or the option B will?
Sid
Sid
15:05
This is a good question. The answer is "D" because it is the more stable product.(B is kinetically-controlled product and D is thermodynamically controlled product) In addition of HBr in room temperature(Since no temperature is mentioned, we assume room temperature), thermodynamically-controlled product is obtained. This is a very typical information.
However, if there was Br2 addition, then, B would be the major product because in that case, Kinetically-controlled product is obtained.
Thanks a lot sid!
In which chapter do we find this information?
Sid
Sid
You are welcome! In no chapter, I saw this on a youtube video.
I don't quite remember where...
Probably the answer solutions to 2015 JEE advanced..
Anonymous
@samjoe This is basically because (D) is more stable due to maximum hyperconjugation. 6 alpha hydrogens are created due to the double bond shift.
Anonymous
You'll can easily predict this if you have learnt GOC well
Anonymous
(Remember that you have only 1 eq of HBr)
15:12
I get confused? What to see Hyper conjugation or the stability of carbocation (that too is stablised by HC i guess?)
Sid
Sid
@blue I don't quite agree because in case of Br2, the opposite would happen. B would be the option. Organic seems easy once you know the answer, yet predicting what is the correct answer is rather tough.
Anonymous
@Sid No. In case of Br2 the mechanism is quite different. That takes place through a non classical carbocation mechanism.
Where do you people learn all this advanced things?
Sid and blue
Anonymous
Anonymous
This is the mechanism that occurs for Br2 addition. ^
Sid
Sid
15:15
Advanced is the real deal. If you want to study, you will enjoy. All other exams are boring. For example, JEE main was so boring. Quite a few ambiguous questions and it is just a test of who can do things fast. Advanced is far more conceptual.
Anonymous
@samjoe You need to follow proper study materials/books.
Anonymous
School books don't teach them in detail.
Sid
Sid
@blue do you have one for HBr addition? I know that works on the basis of carbocations too. I am talking about conjugate addition.
Anonymous
@Sid "I know that works on the basis of carbocations too." It doesn't. Source?
Anonymous
Anonymous
15:19
This is the basic mechanism for HBr.
Sid
Sid
What? Isn't that carbocation formation?
I was talking about that
Anonymous
@Sid That is for HBr.
Anonymous
This is called classical carbocation mechanism.
Anonymous
For Br2 non classical mechanism is followed.
Anonymous
Check Morrison Boyd or LG Wade for details.
Sid
Sid
15:21
Of course, it is for HBr.
@blue how was JEE main? Did you give the offline one or the online one?
Anonymous
@Sid Offline. It was okay. Yours?
Sid
Sid
Bad. Could have done a lot better. But then, everyone feels like that..
I secured 180.
Anonymous
Ah, well. 180 isn't bad! :)
Anonymous
You rank should be around 12k I expect?
Sid
Sid
Yeah. Around that. But, I made some stupid mistakes. Like, forgetting to take the transpose while finding the adjoint of matrix and then that strongest-reducing agent question. I was feeling rather smug about solving that for some reason and in my excitement, I ticked Cr3+ instead of Cr.
Anonymous
15:30
Well, even I did that mistake. Ticked Cr(3+) in excitement :P Ended up getting 225. I was expecting 250+. But meh :/. In maths I got only about 60.
Sid
Sid
Yeah. Same here, Maths-57. I was unhappy with my low Chemistry score. I actually left a lot of questions in Chemistry which could have been solved...
Anonymous
Oh, chemistry was quite easy. It was from NCERT mostly.
Anonymous
Maths went bad for me.
Anonymous
Physics was okay. Got 85.
Sid
Sid
Cool. 85 is pretty good. How long ago was this room created?
Anonymous
15:36
About 2 months I suppose.
Sid
Sid
That long? Weird. I don't remember seeing this room the last time I came on Physics.
Anonymous
Check the info on that link ^
Anonymous
It states the date of first message.
Sid
Sid
@blue Are you appearing for BITSAT as well?
Anonymous
15:49
Yes. But not preparing...
Anonymous
as such
Sid
Sid
On?
Anonymous
After JEE
Anonymous
2 days later
Sid
Sid
23rd?
Anonymous
15:50
yea
Sid
Sid
Anyway, I don't really think preparation for BITSAT is necessary if someone is preparing for Advanced.
Anonymous
English...
Sid
Sid
I have seen previous year questions, they are boring. Just a speed test. If you know a bit of English and some usual Logical Reasoning, you would do fine.
 
2 hours later…
 
1 hour later…
19:08
@user123733 I think the question means that the distance is increasing at L m/s. This is just the radial component of the runner's velocity, with Father's position as the origin. Draw a diagram.

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