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26 Dec 2008 08:41:37 IST

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What exactly is energy of oscillation for an object in SHM?

None

For a system of objects shouldn't the total energy before the starting of oscillation be same as the energy of oscillation?

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sajal kumar

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26 Dec 2008 11:38:03 IST

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of course not.....the total energy before starting is zero. when we start shm we provide it certain energy which is stored as potential energy of the spring....this energy oscillates....got it??????

Abhijith

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26 Dec 2008 13:00:08 IST

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Yep got it. But thats not wat I xactly meant. Suppose the system initially has got potential energy (gravitational or elastic) and then it starts SHM. Then is initial P.E=Energy of oscillation?

tka

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26 Dec 2008 15:13:03 IST

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your term 'energy of oscillation' is unclear. in oscillations, the total energy (PE + KE) is USUALLY conserved because the restoring force is usually a conservative force.

Abhijith

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26 Dec 2008 16:37:25 IST

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Well I'm talking about a situation like u drop a block of mass m on to an unstretched spring of spring const k lying on a horizontal table (Assume the block sticks to the spring and there is an SHM). what will be the total energy ( which I presumed to be energy of oscillation)?

tka

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26 Dec 2008 17:20:34 IST

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here, 'energy of oscillation' is the energy of the spring + block system after the block has stuck, as i assume that there is loss of energy in the sticking process

Abhijith

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26 Dec 2008 17:45:13 IST

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Well take all these stuff as ideal (no energy loss in sticking no air drag no contact friction)

Will energy of oscillation (PE+KE of oscillating system block & spring)=PE of block before dropping it

tka

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26 Dec 2008 17:48:22 IST

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yes

Abhijith

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26 Dec 2008 18:01:50 IST

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Here when drop the block on the spring(unstretched) the energy of oscillation= mg(h+mg/2k)

(According to a book)

How? How did mg/2k become the remaining length??

tka

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26 Dec 2008 18:08:02 IST

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dude, notice that mg/k ( i don't think it's mg/2k) is the equilibrium position of the shm. they have assumed this to be their reference so that it simplifies the problem later on

Abhijith

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26 Dec 2008 18:15:48 IST

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no paI .now I got the way!

It is mg/2k -The maximum compression that can be produced when the block is just placed on the spring (not dropped). They took mg/2k as reference (cuz mg/2k is the amplitude and at amplitude energy of oscillation is completely elastic PE.

tka

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26 Dec 2008 18:18:02 IST

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no, that gives you 2mg/k and not mg/2k

Abhijith

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26 Dec 2008 18:22:42 IST

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oops sorry ur rite

I double checked it

they have given mg/2k+h

tka

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26 Dec 2008 18:31:44 IST

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i'm quite sure that amplitude must be calculated from the position of equilibrium. which is mg/k and thus it makes sense to take this point as reference

Abhijith

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26 Dec 2008 22:01:11 IST

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Oh.

So do you mean that a block dropped (from height h) on a spring and block of same mass placed on the spring (same) will give SHM of same amplitude??

tka

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27 Dec 2008 09:24:40 IST

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oh i said amplitude? pardon me, it's equilibrium position. although, time period is the same for both cases. anyways, let's write down equations:

mg (h+x) = kx²/2

gives me the two extreme positions of shm

kx² -2mgx - 2mgh = 0

the roots of which are

----- ( 1 )

Which clearly means that the quantity representing amplitude is:

----------------- ( 2 )

this is the required amplitude, which is not mg/2k + h. note that equation (1) also tells us that the equilibrium position is mg/k.

Navi arora

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27 Dec 2008 18:38:23 IST

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Simple Harmonic Motion ....

to n fro motoion

http://www.icbse.com

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