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|dw:1328424655814:dw|

The two pulleys are fixed...

The period doesn't depend on the displacement.

well if u see in the formula of time period there is not saying abu distance i guess

T=2pie under root m/k

means it is independent from distance

But there is k which is different.

i agree it will be different i m just telling in response to that distance u were saying

Heena is right. Period of oscillation is isochorous.
The spring constant doesn't change either.

heena cm there

Well if you still insist your answer is not there in the options.

See the spring constant may be same but it depends on the force The force is NOT 2kx.

Thats all is given. i think slant can be kept constant.

Is this an option?\[T = 2\pi \sqrt{2 \over 2 k \cos(\theta)}\]

Rather. \[T = 2 \pi \sqrt{m \over 2k \cos(\theta)}\]

cos theta is outside the sqrt.

I would go with that one.
Sorry for the bad answer.

Could you explain?

as spring is extended by x it got comression too x but here if u see in figure there is string too

t=F right look at this figure that i m hoping abu this ques to be
|dw:1329037050794:dw|

2Tcos thta-mg=ma
right?

now if we arrege it to form of shm equation T yeilds

how u take 2T i mean here box applies T' and pulley is apllying T

is a out come of vectors also T=T'

T in newton formula is tntion & next is period of time

don't get my goal yea?

na?

what is mean?

lolz its a word of hindi just tongue slip forget it :P

If we take the spring's extension be x, then the block moves by \[x \cos \theta\]

*bookmark. I have to go now, but if this is still unanswered when I'm back I'll help.

Ok thanks a lot...