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anonymous
 4 years ago
SHM question: Find time period:
anonymous
 4 years ago
SHM question: Find time period:

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328424655814:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The two pulleys are fixed...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The period of a simple setup is expressed as\[T = 2 \pi \sqrt{m \over k}\]Assuming the mass is constrained to vertical motion, the period is expressed as\[T = 2 \pi \sqrt{m \over 2k}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0How 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The period doesn't depend on the displacement.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well if u see in the formula of time period there is not saying abu distance i guess

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0T=2pie under root m/k

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0means it is independent from distance

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But there is k which is different.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i agree it will be different i m just telling in response to that distance u were saying

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Heena is right. Period of oscillation is isochorous. The spring constant doesn't change either.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yea and acc. to the ques i saw they didnt said K1 or K2 means it can be same also if there is no stratement

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well if you still insist your answer is not there in the options.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0See the spring constant may be same but it depends on the force The force is NOT 2kx.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Is that figure complete? We can analyze the kinematics but we'll need more information. The angle theta changes with displacement.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thats all is given. i think slant can be kept constant.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Is this an option?\[T = 2\pi \sqrt{2 \over 2 k \cos(\theta)}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Rather. \[T = 2 \pi \sqrt{m \over 2k \cos(\theta)}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0cos theta is outside the sqrt.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I would go with that one. Sorry for the bad answer.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dont 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Here 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0as spring is extended by x it got comression too x but here if u see in figure there is string too

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0nah i m sayin if u extend by x then the compression will bw of x luk in diagramdw:1329036256383:dw A=expansion bcontraction

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0True 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0t=F right look at this figure that i m hoping abu this ques to be dw:1329037050794:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.02Tcos thtamg=ma right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now if we arrege it to form of shm equation T yeilds

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how u take 2T i mean here box applies T' and pulley is apllying T

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0even 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is a out come of vectors also T=T'

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0T in newton formula is tntion & next is period of time

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0don't get my goal yea?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lolz its a word of hindi just tongue slip forget it :P

Mani_Jha
 4 years ago
Best ResponseYou've already chosen the best response.0If we take the spring's extension be x, then the block moves by \[x \cos \theta\]

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0*bookmark. I have to go now, but if this is still unanswered when I'm back I'll help.

Mani_Jha
 4 years ago
Best ResponseYou've already chosen the best response.0For 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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