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Mechanics

New kid on the Block

 Joined: 10 Feb 2007 Post: 10
10 Feb 2007 23:11:19 IST
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bipin sir still little confused !!
Engineering Entrance , Medical Entrance , NEET , JEE Main , AIIMS , JEE Main & Advanced , Physics , Mechanics

what is the potential energy of a particle performing wave motion ???
i mean shouldn't it be equal to the work done by the tension on that particle =
(Tension) (sin z ) dy , where z is the angle which tension makes with the x axis and y = A sin(wt-kx)

but the asnwer's coming as 1/2 * (tension) * (dy/dx)^2;

Blazing goIITian

Joined: 4 Jan 2007
Posts: 493
11 Feb 2007 09:59:12 IST
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First you send complete question.

Forum Expert
Joined: 23 Jan 2007
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11 Feb 2007 23:31:56 IST
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The answer which you have given has dimensions of force and not of energy.
So give the complete and correct question if you please.

Best Wishes

Blazing goIITian

Joined: 9 Feb 2007
Posts: 1347
12 Feb 2007 10:24:47 IST
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sir am somewhat confused in it.what os the expression for potential energy of a wave ?

Forum Expert
Joined: 23 Jan 2007
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12 Feb 2007 17:22:44 IST
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Take any portion of the string dx
Let it be displaced upwards by dy due to tension
Then change in length dl = (dx2+dy2)  -  dx
= dx[1+(dy/dx)2]1/2 - dx
Expanding binomially and neglecting higher order terms we get
dl = (1/2)(dy/dx)2.dx

Hence the work done due to tension = T.dl
= (T/2)(y/x)2.dx
since y/x = -Akcos(wt-kx)            = (T/2)(Ak)2cos2(wt-kx).dx

Integrating it over time period.

Best Wishes

New kid on the Block

Joined: 10 Feb 2007
Posts: 10
12 Feb 2007 23:41:26 IST
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but sir, can't we say that the potential energy of that particle will be equal to the work done by tension on it , if i do by this i am obtaining 2 times the expression u have obtained above , please help !

Forum Expert
Joined: 23 Jan 2007
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13 Feb 2007 10:32:10 IST
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First of all your answer does not have the dimensionns of energy.
So tell me the steps how u came to this answer then only I can help u.

Best Wishes

Blazing goIITian

Joined: 9 Feb 2007
Posts: 1347
13 Feb 2007 10:50:28 IST
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sir, i did it like this ::=

considering a wave propogating on the x axis ::
first of all , i chose any element on  the wave. so, the potential energy of this particle will be equal to the work done by tension force(T) on this element
dw = T Sina  * dY
where a is the angle which the tangential tension force makes with the x axis.
and Y = A sin(wt - kx) , differentiating this we get dY =  - A k cos(wt - kx ) dx

hence, the work done = potential energy of the element =

du =  -T * sina * A k cos(wt-kx)dx

or du/dx = - T* sina * A k cos (wt - kx)-------------------------- (1)

taking 'a'  to be very small , sina  tan a = dy/dx

and dy/dx = -Akcos(wt - kx) --------------------------- (2)

subst. (2) in (1) ;

we get : du/dx = T * A2 k2 cos2(wt - kx) which is differing by the expression u derived by a factor of 1/2 .
i am not able to get the reason of my error. please advise and help

I HAD one more quest. also :- when we choose an axis of rotation, does that mean that the observer is now observing the body from the frame of axis of rotation and thus, necessary pseudo forces also must be acompanied ??

thank u !!

Forum Expert
Joined: 23 Jan 2007
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13 Feb 2007 11:12:37 IST
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You have taken sina = tan a = dy/dx
This happens only when a is small.
here dy/dx = -Akcos(wt-kx) is not small everwhere.
Hope u got it.

Best Wishes

Blazing goIITian

Joined: 9 Feb 2007
Posts: 1347
13 Feb 2007 11:18:22 IST
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sir, in the derivation also which u have done above, the higher binomail terms have been neglected . this means dy/dx is small or a is small !
I HAD one more quest. also :- when we choose an axis of rotation, does that mean that the observer is now observing the body from the frame of axis of rotation and thus, necessary pseudo forces also must be acompanied ??

Forum Expert
Joined: 23 Jan 2007
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13 Feb 2007 11:32:38 IST
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Sorry yaar i went wrong.
Actually u have taken work to be T Sina * dY
but tension is lifting the centre of mass by dy/2.

Your thinking is good and i praise it.

Best Wishes

Blazing goIITian

Joined: 9 Feb 2007
Posts: 1347
13 Feb 2007 11:36:51 IST
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fantastic sir !!!!!

ok, now i understand , this is the same cse as that of constrained motion , isn't it ? that's why ----- when the element is lifted by a distance dy then strings from both side will come into the picture and hence the acceleration of center of mass will be halved ????
am i thinking correct ??

Blazing goIITian

Joined: 9 Feb 2007
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13 Feb 2007 11:40:36 IST
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i mean i am a little confused regarding this dy/2, can u explain in little detail please ??

Forum Expert
Joined: 23 Jan 2007
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13 Feb 2007 11:51:26 IST
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Just draw the figure and u will find that centre of that elemnet would be dy/2 upper from the normal position.

Blazing goIITian

Joined: 9 Feb 2007
Posts: 1347
13 Feb 2007 11:55:05 IST
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well, another confusion as well !!!

when the element is moving upwards by dy, then the center of mass should also move with the element and hence , it shuold also move dy !!

Forum Expert
Joined: 23 Jan 2007
Posts: 7958
13 Feb 2007 12:33:44 IST
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Consider a rod kept on the ground.

when u straighten it up vertically then its lower remain as it is but the upper part is displaced by the lenght of the rod in the vertical direction.
And the centre of mass is at height of length/2.

Blazing goIITian

Joined: 9 Feb 2007
Posts: 1347
13 Feb 2007 16:00:23 IST
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thank you very much sir, i have got the concept !

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