IIT JEE , AIEEE Forum - Physics , Mechanics - CHALLENGE-3

frenied

ahh...so u break the spring anyway u want and that doesnt change its spring constt??as far as i knw the spring constt depends on length...and if it has been broken then from where has it been broken???

karthik2007

See, don't think that we are actually breaking the spring into two! That's not the case. He just considered them as 2 cases. Its quite simple to visualize. He's considered the action of the two forces individually.

frenied

but how can u do it for the simplification sake??i mean there must be some logic behind doing it....and wat 2 forces??there's just one spring and not 2 springs...

karthik2007

See man!! When you look at the right part of the spring, let its elongation be x1, BUT IT IS ACTED UPON BY THE WHOLE SPRING, so spring constant remains the same!!! If you read the problem correctly, there are 2 forces acting on the spring???

frenied

besides there's rooney's confusion...why do we not consider the elongation of x2 ?? basically guys ill let out this fact myself...this question is not going to be solved by the conventional methods of force and energy conservation.....i also had to ask my teacher for one or 2 of the most interesting and unworldly steps....

karthik2007

well if you know the right solution, why dont you post the unwordly solution then?

frenied

ill post the solution if u want......i dunno how u can digest the fact that spring constt doesnt change...but i m simply 2 uncomfortable with that solution....there's no logic in "looking" at the spring in a certain manner..u can't "look" at thngs

frenied

oohhh....u are getting angry...see i have 2 methods....one is the unworldly one which was told by my sir...and one is the one that i use to solve this problem....but im not really confident of my method

karthik2007

No one is getting angry. Post your method.

frenied

Let us consider that F2>F1.now let m1 be displaced to the ryt by x1 and m2 to the ryt by x2.therefore net displacement of m1 wrt m2=x1-x2.this is also the elongation of the spring.so we get the concept that the relative disp. of the 2 blocks is the elongation of the spring.let us denote it by Srel.then dSrel/dt=Vrel of the two blocks.in position of max.elongation the Vrel=0.now consider the fbds of the 2 blocks.
T-F1=m1a1.
F2-T=m2a2.
solving a1 and a2 we get a1=T-F1/m1
a2=F2-T/m2
therefore rel accn.=a2-a1.

frenied

now write above accn.=VdVrel/dSrel and put T=kSrel....integrate with limtis Srel=0, Vrel=0 and Srel=x,Vrel=0 where x is max elongation.

rhd92781

hmmm.....
dere's no flaw in ur method...think its correct
regarding my soln...ya dere r some
actually not taking it frm the shm point of view was my biggest mistake
but i wish to see the unworldly soln too ...if u can post it plz

karthik2007

Nice one friened.

rooney

good one

priyesh

gooooooood solution frienied

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