Mechanics

New kid on the Block

Joined: 10 Feb 2007
Post: 10
10 Feb 2007 23:11:19 IST
0 People liked this
16
1646 View Post
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;



Comments (16)


Blazing goIITian

Joined: 4 Jan 2007
Posts: 493
11 Feb 2007 09:59:12 IST
0 people liked this

First you send complete question.
Bipin Dubey's Avatar

Forum Expert
Joined: 23 Jan 2007
Posts: 7958
11 Feb 2007 23:31:56 IST
1 people liked this

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
vishak p's Avatar

Blazing goIITian

Joined: 9 Feb 2007
Posts: 1347
12 Feb 2007 10:24:47 IST
0 people liked this

sir am somewhat confused in it.what os the expression for potential energy of a wave ?
Bipin Dubey's Avatar

Forum Expert
Joined: 23 Jan 2007
Posts: 7958
12 Feb 2007 17:22:44 IST
4 people liked this

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
2 people liked this

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 !
Bipin Dubey's Avatar

Forum Expert
Joined: 23 Jan 2007
Posts: 7958
13 Feb 2007 10:32:10 IST
2 people liked this

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
vishak p's Avatar

Blazing goIITian

Joined: 9 Feb 2007
Posts: 1347
13 Feb 2007 10:50:28 IST
0 people liked this

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 !!
Bipin Dubey's Avatar

Forum Expert
Joined: 23 Jan 2007
Posts: 7958
13 Feb 2007 11:12:37 IST
1 people liked this

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
vishak p's Avatar

Blazing goIITian

Joined: 9 Feb 2007
Posts: 1347
13 Feb 2007 11:18:22 IST
0 people liked this

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 ??
 
Bipin Dubey's Avatar

Forum Expert
Joined: 23 Jan 2007
Posts: 7958
13 Feb 2007 11:32:38 IST
2 people liked this

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
vishak p's Avatar

Blazing goIITian

Joined: 9 Feb 2007
Posts: 1347
13 Feb 2007 11:36:51 IST
0 people liked this

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 ??
vishak p's Avatar

Blazing goIITian

Joined: 9 Feb 2007
Posts: 1347
13 Feb 2007 11:40:36 IST
0 people liked this

i mean i am a little confused regarding this dy/2, can u explain in little detail please ??
Bipin Dubey's Avatar

Forum Expert
Joined: 23 Jan 2007
Posts: 7958
13 Feb 2007 11:51:26 IST
2 people liked this

Just draw the figure and u will find that centre of that elemnet would be dy/2 upper from the normal position.
vishak p's Avatar

Blazing goIITian

Joined: 9 Feb 2007
Posts: 1347
13 Feb 2007 11:55:05 IST
0 people liked this

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 !!
 
please answer !
Bipin Dubey's Avatar

Forum Expert
Joined: 23 Jan 2007
Posts: 7958
13 Feb 2007 12:33:44 IST
1 people liked this

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.
vishak p's Avatar

Blazing goIITian

Joined: 9 Feb 2007
Posts: 1347
13 Feb 2007 16:00:23 IST
0 people liked this

thank you very much sir, i have got the concept !



Quick Reply


Reply

Some HTML allowed.
Keep your comments above the belt or risk having them deleted.
Signup for a avatar to have your pictures show up by your comment
If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team
Free Sign Up!
Sponsored Ads

Preparing for JEE?

Kickstart your preparation with new improved study material - Books & Online Test Series for JEE 2014/ 2015


@ INR 5,443/-

For Quick Info

Name

Mobile

E-mail

City

Class

Vertical Limit

Top Contributors
All Time This Month Last Week
1. Bipin Dubey
Altitude - 16545 m
Post - 7958
2. Himanshu
Altitude - 10925 m
Post - 3836
3. Hari Shankar
Altitude - 10085 m
Post - 2217
4. edison
Altitude - 10825 m
Post - 7804
5. Sagar Saxena
Altitude - 8635 m
Post - 8064
6. Yagyadutt Mishr..
Altitude - 6330 m
Post - 1958

Find Posts by Topics

Physics

Topics

Mathematics

Chemistry

Biology

Parents Corner

Board

Fun Zone