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ayc
12:00 AM
@sammygerbil its like this: put one dot on each.Now you have two dots left.you can either put the two dots as (1 on each dice) on two dices or put two dots on one dice.In the former case its the same as choosing 2 out of 4 and in the latter choosing 1 out of 4.And hence 4c2+4c1
 
@ayc ok I understand that. But I still haven't grasped why my calculation 4x4/2=8 is wrong.
 
 
1 hour later…
1:22 AM
@Abcd Q18 Two cars in wind, one blowing horn : As you suggest, work in frame of wind. Then A moves at u = 20m/s to right and B at 30 m/s to left. A is the observer, A' (the image of A reflected in car B, which acts as a mirror) is the source. A and A' both approach the mirror B at a relative speed of 50m/s, so A' moves relative to air/wind at v = 50+30 = 80m/s.
Both source and observer are moving so frequency heard by A is $$f'=\frac{c+u}{c-v}f=\frac{330+20}{330-80}\times 300Hz=\frac{35}{25}\times 300Hz$$
Wavelength of sound received by driver of car A is $$\lambda'=\frac{c}{f'}=\frac{330}{300}\times \frac{25}{35}=\frac{11}{14}m$$
Not one of the options. :(
 
1:46 AM
@sammygerbil are you available right now?
 
@Jasmine yes
 
How can i fogure out if 2 springs are in series or in parallel to take out keff as I sometimes go wrong.
 
@Jasmine If the springs have the same extension they are in parallel; if they have the same tension they are in series.
2
 
ayc
@sammygerbil What you have done is permutation not combination.I.e your answer is 4P1+4p2=16...What I have done is combination 4C1+4C2.We need combination!
 
@Jasmine Based on my explanation, what do you think about this diagram? Are these springs in series or in parallel?
 
1:53 AM
@sammygerbil series
 
@Jasmine Correct. Easy isn't it?
 
Thank you for telling me this otherwise I was using a very wrong concept
@sammygerbil so the $ K_eff $ here is 2K/3?
 
@Jasmine I think not.
 
@sammygerbil as it depends on the way the mass is attached?
 
Yes. How I solved the problem is to divide it in half, because the left and right sides are symmetrical.
The middle pulley does not rotate.
The effective spring constant is $4k/3$.
 
2:04 AM
@sammygerbil did you divide mass also?
 
However if the mass is displaced by x the effective spring extends by 2x.
@Jasmine Yes we need to divide the mass by 2 also.
 
@sammygerbil so basically two systems which are in series their effective spring constant and then parallel combination of those two?
 
However, because of the pulley the mass is twice as effective in extending the spring, so we have an effective mass of $2\times m/2=m$.
So I think the answer should be C. Is that correct?
 
I don't have the keys :(
 
@Jasmine Oh yes that sounds like a good way of explaining it. I didn't think of that. Does it give the same answer?
 
2:11 AM
I am getting confused due to mass
I got B
 
@Jasmine Yes. I get B also using your method. My method might be wrong. It is easy to get confused.
 
Effective spring constant for the half of symmetrical system is (4/3)K but it's for 'm/2' and the parallel combination is for 'm'?
 
@Jasmine However, I think your method does not take account of the fact that if the mass is displaced by x the springs extend by 2x.
 
Yes
 
So the acceleration of the mass does not match with the extension of the effective spring.
 
2:16 AM
Yes I think C should be correct I will get the keys very soon
Thank you!
 
If we wrote the equation of motion we would have $$(\frac12 m)\ddot x=-k_{eff}(2x)$$ $$\omega^2=\frac{4k_{eff}}{m}$$
 
@sammygerbil what is x and then two dots on it?
 
Sorry $$\omega^2=\frac{4k_{eff}}{m}=\frac{16k}{m}$$
Option D!
@Jasmine $\ddot x$ is acceleration. 2nd time derivative of $x$.
Now that I have written equation of motion I feel more confident about D.
Did you follow?
 
I can't find wrong in logic of that symmetrical system
The way you previously did
We didn't use that 1/2m is getting displaced
And also the fact that displacement is twice
Ok so that's why we didn't get 4keff/m
@sammygerbil I am confused with all this
 
@Jasmine Is there anything you would like me to explain?
 
2:31 AM
If there were even number of springs then what would we do?
One extra spring with no connection with other springs
But connected with the mass
 
@Jasmine If all the springs were attached to the rope which winds round the pulleys then they would still be in series. We would calculate effective spring constant in same way.
 
@sammygerbil ok
@sammygerbil so the acceleration of (1/2m) will be equal to $w^2$ ?
 
They could be in any position - eg all in the left hand rope (no. 1), or all in rope no. 2, or any combination.
Their positions would make no difference.
However, we could no longer say that middle pulley does not rotate. We would have to deal with whole system.
We could not split it in two.
 
2 mins ago, by Jasmine
@sammygerbil so the acceleration of (1/2m) will be equal to $w^2$ ?
 
@Jasmine No. Why do you think this?
 
2:39 AM
21 mins ago, by sammy gerbil
If we wrote the equation of motion we would have $$(\frac12 m)\ddot x=-k_{eff}(2x)$$ $$\omega^2=\frac{4k_{eff}}{m}$$
 
Perhaps you are trying to say the acceleration would be $-\omega^2 x$?
 
I understand the 1st part of your statement
@sammygerbil yes
But I think it should be $ 2a=-w^{2}x$
 
@Jasmine Rearrange the equation of motion into the form for SHM : $$\ddot x+\omega^2 x=0$$
@Jasmine No.
 
What I mean is the acceleration we calculated is for m/2 but we have m so it's acceleration can be double of what we got for m/2
Am I wrong?
 
@Jasmine No. If we divide the system in two then the left side does exactly the same as the right side. So each half of the mass moves with the same acceleration as the whole mass.
If it would help I could explain without dividing the system in two.
 
2:47 AM
@sammygerbil ok ok ok it was a silly doubt
There are 2 points of connection
Had there been only one point of Connection and then we replaced m by m/2 then 'a' would have doubled
 
Going back to what I said about the springs, they can be in any order and position. Regardless of how they are arranged, the mass will stay level.
So first we find the effective spring constant.
We could also have more pulleys.
So the second step is to find out the relation between displacement of mass and extension of effective spring.
In the system we have, when mass is displaced by x the springs extend by 2x.
Here effective spring constant is given by $1/K=1/4k+1/2k+1/2k+1/4k=6/4k=3/2k$
$K=2k/3$
 
But here we got option D by calculating Keff for 1/2 part of the system right? And then we used the fact that the acceleration for 1/2m ,m would have been same and so as $w$
 
Whoops! I made a mistake. In whole system if mass is displaced by x the total extension of springs is 4x, because each side extends by 2x.
 
57 secs ago, by Jasmine
But here we got option D by calculating Keff for 1/2 part of the system right? And then we used the fact that the acceleration for 1/2m ,m would have been same and so as $w$
 
Third step is to write equation of motion : $m\ddot x=-K(4x)=-\frac{8k}{3}x$$
Now that gives answer B again!
 
3:00 AM
@sammygerbil so I was correct initially?
 
@Jasmine Sorry I am getting confused myself. I am getting tired so I had better sleep for a while and sort it out when I wake. Goodnight.
 
@sammygerbil fine! Good night
 
 
2 hours later…
4:56 AM
@sammygerbil @Jasmine I have the key. Answer is A
@sammygerbil its my doubt too. My method was:
Let spring be stretched by x.
Now write energy equation:
$U = \dfrac 12 (2k)x^2 + \dfrac 12 (2k)x^2 + \dfrac 12 (4k)x^2 + \dfrac 12 (4k)x^2 + \dfrac 12 mv^2$
$U = 2k x^2 + 4k x^2 + \dfrac 12 mv^2 $
$U = 6kx^2 + \dfrac 12 mv^2$
$\dfrac{dU}{dt} = 12kxv+ mva$
$\implies 0 = 12kx + ma$
$\implies a= \dfrac{-12k}{m}$
$\implies f = \dfrac{1}{2\pi} \sqrt {\dfrac {12k}{m}}$
@sammygerbil What's my mistake?
@sammygerbil wow btw, you killed it with this method.
 
5:22 AM
@JohnRennie good morning
Are you free
 
@harambe morning, yes, I'm around for several hours this morning
 
Q17 doubt
I found the value of rest mass of Ra* from the bottom reaction
 
OK ... ?
 
I added the rest mass of Ra with energy of beta to calculate it
 
Yes, that's the correct approach
 
5:26 AM
Then I calculated the Q value of the above reaction
But the Q value is not matchhing with the answer
Is this method correct
 
What is the Q value?
Energy released in the reaction?
 
Yes
 
Did you get too high a KE for the alpha particle?
 
The energy available as kinetic energy as my book states in alpha decay
Yes I got more than than kinetic energy. Guess it's calculation mistake
 
Remember that the Ra* nucleus carries away some of the energy because momentum has to be conserved.
 
5:31 AM
Okay
 
If we use M and V for the mass and velocity of the Ra* nucleus and m and v for the alpha particle then we get:
$$ MV = mv $$
and:
$$ \tfrac{1}{2}MV^2 + \tfrac{1}{2}mv^2 = E $$
 
@JohnRennie Hi, morning. Please ping when done with Harambe. Thanks.
 
where E is the energy released in the reaction.
@Abcd will do
Solve these two equations to get the KE of the alpha particle.
 
Okay
 
ayc
@JohnRennie Morning!
 
5:41 AM
@harambe so ... does that give the correct answer?
@ayc morning :-)
 
@JohnRennie not getting answer
 
@Abcd we got everything but not A :p
 
Give me a moment to make a coffee and I'll have a go
 
@Jasmine they are designed in such a way that the options have all the possible answers obtained through different wrong methods
 
@harambe I get:
$$ \tfrac{1}{2} m v^2 = \frac{E}{1 + m/M} $$
@harambe hello? You still there?
 
5:57 AM
@JohnRennie I got the a answer now
It was a calculation mistake
 
Cool :-)
@Abcd you there?
 
@JohnRennie hi
4 hours ago, by Jasmine
user image
@JohnRennie please tell me if you want a clearer pic. I have it.
1 hour ago, by Abcd
@sammygerbil its my doubt too. My method was:
1 hour ago, by Abcd
Let spring be stretched by x.
59 mins ago, by Abcd
$U = \dfrac 12 (2k)x^2 + \dfrac 12 (2k)x^2 + \dfrac 12 (4k)x^2 + \dfrac 12 (4k)x^2 + \dfrac 12 mv^2$
@JohnRennie Please tell me if my energy equation is correct. That's my doubt.
 
You've assumed that all the springs will be stretched by the same amount, and I suspect that is not the case.
 
@JohnRennie Oh,even I was doubtful about that.
So by how much will each spring be stretched?
 
The tension must be the same everywhere in the string/springs, and that means the extension of the 4k springs will be half that of the 2k springs.
 
6:04 AM
@JohnRennie didnt get
 
Let me draw a quick diagram ...
@Abcd Suppose you have four springs connected by string as shown above and you pull on the end to stretch the whole system.
 
@JohnRennie but its attached to wall at other end too
 
The 2k springs will stretch more than the 4k springs because they aren't as stiff.
 
@JohnRennie ya
$(4k)x_1 = 2k(x_2)$
$x_2 = 2x_1$
 
Cool :-)
This is what happens in your question
 
6:12 AM
@JohnRennie wait, let me write energy equation now and check the answer.
Wait.
@JohnRennie Now x in terms of x_1 and x_2 will be? I think this is the important part of the problem.
Where x is displacement of plank.
 
If you pull the plank down a distance $x$ the string has to increase in length by $4x$
 
ya
 
So that would be $x = 2x_1 + 2x_2$
 
so ultimately
$x = 6x_1$
Now:
 
No, wait, I got the initial equation wrong!
 
6:16 AM
Oh
@JohnRennie which one
 
2 mins ago, by John Rennie
So that would be $x = 2x_1 + 2x_2$
It should be:
$4x = 2x_1 + 2x_2$
 
Oh so now ultimately
$4x = 6x_1$
 
Yes
 
$U = \dfrac 12 (4k){x_1}^2 + \dfrac 12 (2k){x_2}^2+ \dfrac 12 (2k){x_2}^2 + \dfrac 12 (4k)x_1^2$
$U = 4kx_1^2 + 2kx_2 ^2$
$U = 4kx_1 ^2 + 8kx_1 ^2$
$U = 12kx_1^2$
$x_1= 4/6 x$
$x_1 = \dfrac 23 x $
$U = 12 k \times \dfrac 4 9 x$
$U = 16 k \dfrac {x^2} 3 $
$\dfrac{dU}{dt}= 0 $
I have missed kinetic energy in energy equation:
$U =\dfrac {16}{3}k x^2 +\dfrac 12 mv^2 $
 
@JohnRennie can this question be solved by effective spring method?
 
6:22 AM
@Jasmine Please give me 2 minutes , then you can ask
 
Sorry to interrupt go on :)
 
$\dfrac{dU}{dt} = \dfrac{32kx v}{3}+ mv a$
$\implies \dfrac{32 kx }{3}= -ma$
$\implies \omega^2 = \dfrac{32k }{3m}$
Wow right answer !!!
Done!!
 
BOOM!
 
Thanks @JohnRennie !!!
 
@Jasmine yes
 
6:24 AM
@JohnRennie how?
 
That's your effective spring
Work out the force constant for the effective spring.
But you need to remember that when you displace the plank a distance $x$ the extension of the effective spring is $4x$.
 
@JohnRennie yes it's 4k/6
So by equating forces $ma=Keff(4x)$ @JohnRennie ?
 
Hmm, that's going to end up with 8/3 k and give an answer a factor of four too small ...
 
i think this force constant method isnt taking intricate details into account.
 
@JohnRennie that's what I am wondering
 
6:33 AM
@Abcd if you look at your result from above:
$$ U = 16 k \dfrac {x^2} 3 $$
(ignoring KE)
 
OK?
 
and we know $U = \tfrac{1}{2}K_{eff}x^2$
 
No
$U = \dfrac {1}{2}m \omega^2 A^2$ for SHM, where A is amplitude
 
No?
 
@JohnRennie I mean am talking about total energy
 
6:34 AM
Ah, you're using $U$ for total energy?
 
Yeah.
@JohnRennie So you mean you are getting the wrong effective spring constant?
9 mins ago, by Jasmine
@JohnRennie yes it's 4k/6
 
But my argument still holds because your total energy is $U = 16 k \dfrac {x^2} 3 + KE$ so the PE is $16 k \dfrac {x^2} 3$
 
yes, so Jasmines's effective spring constant is wrong ...
 
Yes, while the $K_{eff}$ you got is the correct value of 32/3 k
 
@JohnRennie I understand that my answer is wrong but I don't know how will I solve it where to make changes and get right?
 
6:38 AM
I must admit that I don't immediately see where the missing factor of four is coming from ...
 
Ok
 
@Jasmine my immediate instinct is to pull the plank down a distance $x$ and calculate the resulting tension in the string.
AHA!!
@Jasmine If the tension in the string is $T$ the force on the plank is $4T$.
The effective spring constant is 2/3k, but when we displace the plank a distance $x$ we stretch the spring by $4x$, so we have to multiply this force constant by four. So the tension in the string is:
$$ T = \tfrac{8}{3} x $$
@Jasmine OK so far?
We lost Jasmine ...
Anyhow, the force on the plank is $4T$ so we end up with:
$$ F = \tfrac{32}{3} k x $$
And that tallies with @Abcd's answer.
 
@JohnRennie oh
I got Thank you!!
 
 
9 hours later…
3:24 PM
@sammygerbil are you there ?
 
4:14 PM
@EshaManideep hello
 
@sammygerbil Hi ! Till when are you gonna stay ? 3 more hours ??
I am giving a test right now, so..
 
@EshaManideep yes I shall be available in 3 hours.
 
@sammygerbil Cool ! I will be back then :)
 
@sammygerbil Hi
 
@Abcd hi
 
4:18 PM
@sammygerbil mistake in my intensity with polariser question method?
 
@Abcd ??
 
11 hours ago, by Abcd
@sammygerbil What's my mistake?
@sammygerbil ^^
 
@Abcd What is your method?
 
@sammygerbil i told you yesterday
@sammygerbil $I_o/2$, does that remind you something?
 
@Abcd You have not taken into account the vector nature of the electric field. See my solution.
17 hours ago, by sammy gerbil
@Abcd Q27 YDSE with Polarizer : Divide the electric field of light falling onto the slits into amplitudes parallel (p) and perpendicular (s) to the polarizer. Only p gets through the polariser and emerges from slit 2 whereas p+s emerges from slit 1. Max amplitude in the interference pattern is p+s+p (slits 1 & 2 in phasee) while min amplitude is p+s-p = s.
17 hours ago, by sammy gerbil
Now p and s are orthogonal, so for the max amplitude we have a p-component of 2 units and an s-component of 1 unit, which add to as vectors to give a magnitude of $\sqrt5$.
 
4:24 PM
@sammygerbil didnt get this
 
@Abcd Why not?
 
@sammygerbil ??
 
Please be more specific.
 
@sammygerbil Is $|p|= |s|?$
 
@Abcd Yes.
 
4:25 PM
@sammygerbil y
 
@Abcd The light from the source is randomly polarised. So it can be resolved into equal amplitude components at right angles.
It contains equal amounts of horizontal and vertical polarized waves.
 
 
2 hours later…
6:37 PM
@sammygerbil hi !
 
@EshaManideep Hello
 
My teacher says that the lens is deep inside the camera because that if it was inside it would collect light form all the angles by diffraction at the lens end. Is it true ?
 
@EshaManideep Do you think your teacher is wrong?
 
@sammygerbil I get a little uncomfortable feeling
 
@EshaManideep Because?
 
6:43 PM
@sammygerbil let's consider the lens circumference. Diffraction is due to huygens principle. Now, if the circumference acts as a secondary source, and we have no way of obtaining the rays directly from the objects, we shouldn't be able to distinguish between the objects
I know its sort of vague, but I tried my best to express my 'feeling'
 
@EshaManideep It is rather vague, I don't understand what you are trying to say. This is really a concept question and I don't have any experience of camera design. I would recommend searching online for information about cameras. As far as I know the lens is usually close to the front of the camera, not "deep inside".
 
@sammygerbil Alright, thanks
Another one
 
Possibly your teacher is talking about a particular kind of camera. Best to ask him/her for more details if you find nothing online.
 
If, in a diffraction experiment, I put two to three glowing objects behind the slit, what kind of pattern is obtained ?
 
@EshaManideep What are you expecting to get and why?
 
6:50 PM
@sammygerbil Three different images for the three objects ?
 
@EshaManideep If the slit is relatively wide compared with the wavelength you would get 3 different images, as with a pinhole camera.
 
@sammygerbil For a small slit ? (narrow)
 
If the slit was about the same size as the wavelength the centre of each image would be in the same position as for a wide slit, but the rays from each object would be diffracted so much that each image is smeared out and overlaps with the others.
Assuming the 3 sources are not coherent you would not see any interference.
 
@sammygerbil That's all right ?
Then fine. Thanks !
 
@EshaManideep Are you expecting something else?
 
6:55 PM
@sammygerbil Nope, that was what I was expecting.
 
@EshaManideep Any more questions? I need to go out for about 30 minutes.
Will ping you on my return.
 
7:33 PM
@EshaManideep Back.
 

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