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system shown of mass m is displaced in its plane by a small angle θ, find the period of oscillation 
I wud essentuially want to know the approach to this type of questions? ANS : 2Ï€√2(√2)l/3g
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60 Answers
62@Avni
I will take total mass as 2m for simplicity
The method for energy conservation is quite simple
Rotate it by angle theta (very small)
First find I about the point of rotaion. (it is already done by users above)
Then find Kinetic energy = 1/2 I ω2
Now gravitational potential energy = gml/2 { cos(θ+45) + cos(45-θ)}
= gml/2 { sin(θ+45) + sin(45-θ)}
= mgl cos θ/√2
Now the total energy is constant
hence
mgl cos θ/√2 + 1/2 I ω2 = c
theta is small...
now differentiate wrt θ
mglsinθ {dθ/dt}/√2 + 2/3 m l2ω dω/dt = 0
thus, (removing mωl from both sides and sin )
gθ/√2 + 2/3Ldω/dt = 0
hence you have the differential equation..
I am not sure if what priyam said that force method is easier here is actually true ;)
1but kaise? me still not gettin, sorry guyz troublin u a lot[2]
and bcos of that angular displacement ∂θ applied wud not the COM shift as well?
1OHHHHHHHH yah!!!!!!!
main kab l/√2 le rahi thi [2].........
THNX!!! n SORRY
btw COM mein θ se koi change aaya to v neglect it rite?
21Hi skym
Can u tell how/why did priyam take weight component as "m"gsin@
shudnt v consider vectr comp. or sumthing lyk dat 4 mass as well?
Or else tell me in ur questn wa tshud be the mass factor on RHS?
21I = ml^2
resulatant of lengths = √5L at an angle 60° to the length L wala component....
aage sichne do
1in RHS it is the total mass of the sytem and the distance of the com from the point of suspension .
well thank yo for pointing that out...
i missed dat out ...
so my ans is:
2pi√6L/g√41
21this time round not getting the same ans [2]
2pi √0.55L/g aa raha hai....
21PHY EXPERTS!!!!!!!1 pl. post ur ans/soltn to sky's modified question........
1u r right priyam ,messed up with a factor of 2 twice,since i did it twice i still got the answer :D.
@avni , i believe the rod is L - shaped(2 identical rods at 90 deg angle), so centre of mass will be l/2√2 below the hinged pt by simple symmetry
1Then find Kinetic energy = 1/2 I ω2
Now gravitational potential energy = gml/2 { cos(θ+45) + cos(45-θ)}
= gml/2 { sin(θ+45) + sin(45-θ)}
= mgl sin θ/√2
Now the total energy is constant
hence
mgl cos θ/√2 + 1/2 I ω2 = c
how cum this change frm sin to cos?
OTHERWISE : The method is awesome!!! [1]
this wud b an ideal way to solv SHM questn involving θ
but wat in other ques when no θ is ther how 2 use energie method? as u will not hav ω in the equation only [2]
then how will u get TIME PERIOD?
THNX A LOT FOR UR HELP [1]
62@avni.. sorry it was cos at the first place..
I am fixing that
tapan... that is not very tough either.. I think eureka had giventhis sum before..
please try it some more :)
21No Sir, I feel I hav got da questn and hence hav posted my Ans. so wanted to know if dat is crrct or not?
