SHM question: Find time period:

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SHM question: Find time period:

Physics
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|dw:1328424655814:dw|
The two pulleys are fixed...
The period of a simple set-up is expressed as\[T = 2 \pi \sqrt{m \over k}\]Assuming the mass is constrained to vertical motion, the period is expressed as\[T = 2 \pi \sqrt{m \over 2k}\]

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How do you say that? If you move the block down by x. The springs dont extend by the same x. So F is not equal to 2kx.
The period doesn't depend on the displacement.
if extension in springs is not x then force is not 2kx. which clearly implies the constant isn't 2k. If the constant isn't 2k how do you say that period remains same.
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.
yea and acc. to the ques i saw they didnt said K1 or K2 means it can be same also if there is no stratement
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.
Is that figure complete? We can analyze the kinematics but we'll need more information. The angle theta changes with displacement.
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?
dont u think shankvee its a mixing question of NLM and SHM i mean here spring is going down means pulling is rolling down so tension is also there as COME i think we should compare them
Here tension=force of spring. There is no mixing of NLM and all... I just don;t get if you pull the mass by x how much the springs get extended. Eashmore: Is the extension in the springs x cos theta or is it x/ cos theta , Could you explain this, i never get this.
as spring is extended by x it got comression too x but here if u see in figure there is string too
No it is not x. If you pull in one direction by x then the extension in another direction can't be x how is it possible?
nah i m sayin if u extend by x then the compression will bw of x luk in diagram|dw:1329036256383:dw| A=expansion b-contraction
True but the expansion isn't what i'm saying is.|dw:1328429223949:dw| If x is amount pulled downward x is not equal to y.
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
even i m not sure how to solve i only know how to make diagrams and wat can be done here but how to calculate for that case i say help :P
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?
ok thenn so by this we can say na that angle between them is120 degree ?? if manitude same and resulatant is also same then the angle must be 120degree
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...
For the block of mass m, \[2T \cos \theta=-ma\] T=kx So, \[2k \cos thetax+ma=0\] This is a differential equation in x, a=d2x/dt2 So, its solution will be \[T=2\pi \sqrt{m/2k \cos \theta}\] You have to just take the coefficients of a and x. That should be the answer as far as i think

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