what is the answer bhaiya,is my answer incorrect????if yes,please post the solution.You posted this question on APRIL 20, 2010.I mean now break the suspense bhaiya,please.....
The distance between the two blocks is decreased by a small distance ‘x’
And the let the increase in the spring length be ‘y’.
By pythogoras theorem
L2=[lcosA-x/2]2+[lsinA-y/2]2
Since x is small we can neglect x2 and y2
=>y=cotA*x
Fspring=ky=kcotA*x
2TsinA=Fspring
T=kcotA.cosecA.x/2
We can replace the whole setup by two masses attached to a single spring with spring const ks= (kcotA.cosecA)/2 so that ob decreasing the dist. Between the two masses by x we get a force equal to the original setup.
Time period f such a setup Is given by T=2pi(mred/ks)1/2 Mred=m1.m2/m1+m2
This comment has been removed by the author.
ReplyDeleteThis comment has been removed by the author.
ReplyDelete2 pi underroot (8k cos^2(A)/m)
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteis my answer correct?????????????????
ReplyDelete2pi[mtanA/k]
ReplyDelete[]=sqrt
yaar isme oscillations ho kaise rahi hain ???
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteis my answer correct??????
ReplyDeleteif my answer is not correct,please tell me the right answer please
ReplyDeletethis was a problem i solved during my JEE days, and i remember solving it with some difficulty.
ReplyDeletebut now, two approaches were giving me different answers:
1)The energy conservation (which seems more justified and is giving the Cos^2 type answers)
2)Forces, which are giving Cot^2A types...
any suggestions?
i have completely no idea..
ReplyDeletemaybe pi sqrt(m cotA/k)
ReplyDeletewhat is the answer bhaiya,is my answer incorrect????if yes,please post the solution.You posted this question on APRIL 20, 2010.I mean now break the suspense bhaiya,please.....
ReplyDeleteplz. read my last post...it says that i have not been able to solve this question myself...
ReplyDeleteplz. post your solution, in very brief. then i'll see wether its correct ar not.
Distance between the two blocks=2l.cosA
ReplyDeleteLength of the spring=2l.sinA
The distance between the two blocks is decreased by a small distance ‘x’
And the let the increase in the spring length be ‘y’.
By pythogoras theorem
L2=[lcosA-x/2]2+[lsinA-y/2]2
Since x is small we can neglect x2 and y2
=>y=cotA*x
Fspring=ky=kcotA*x
2TsinA=Fspring
T=kcotA.cosecA.x/2
We can replace the whole setup by two masses attached to a single spring with spring const ks= (kcotA.cosecA)/2 so that ob decreasing the dist. Between the two masses by x we get a force equal to the original setup.
Time period f such a setup Is given by T=2pi(mred/ks)1/2
Mred=m1.m2/m1+m2
are you now online bhaiya????if yes,please check my solution and please reply
ReplyDelete