SHM question: Find time period:

- anonymous

SHM question: Find time period:

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- schrodinger

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- anonymous

|dw:1328424655814:dw|

- anonymous

The two pulleys are fixed...

- anonymous

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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## More answers

- anonymous

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.

- anonymous

The period doesn't depend on the displacement.

- anonymous

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.

- anonymous

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

- anonymous

T=2pie under root m/k

- anonymous

means it is independent from distance

- anonymous

But there is k which is different.

- anonymous

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

- anonymous

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

- anonymous

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

- AravindG

heena cm there

- anonymous

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

- anonymous

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

- anonymous

Is that figure complete? We can analyze the kinematics but we'll need more information. The angle theta changes with displacement.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

cos theta is outside the sqrt.

- anonymous

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

- anonymous

Could you explain?

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

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?

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

2Tcos thta-mg=ma
right?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

is a out come of vectors also T=T'

- anonymous

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

- anonymous

don't get my goal yea?

- anonymous

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

- anonymous

na?

- anonymous

what is mean?

- anonymous

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

- Mani_Jha

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

- JamesJ

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

- anonymous

Ok thanks a lot...

- Mani_Jha

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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