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10 Feb 2007 23:11:19 IST
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bipin sir still little confused !!
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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;
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yahiyafirdous
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Joined: 4 Jan 2007
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11 Feb 2007 09:59:12 IST
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First you send complete question.
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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
sincey/x = -Akcos(wt-kx) = 
(T/2)(Ak)2cos2(wt-kx).dx
Integrating it over time period.
Best Wishes
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).dxIntegrating it over time period.
Best Wishes
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
tan a = dy/dxand 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 !!
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 !
please clarify !!
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 ??
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 ??
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 ??
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