How Is Work Calculated for Masses on a Stretched Spring?

[vissh]

Hi :D
1) Two equal masses (let a and b) are attached to the ends of a spring of spring constant k. The masses are pulled out symmetrically to stretch the spring by a length x over its natural length. The work done by the spring force on each mass is :
(a)1/2kx² (b)-1/2kx² (C) 1/4kx² (d) -1/4kx²


I tried to solve in following way [may be wrong :)]
=> the force acting on 'a' and 'b' will be -kx . Dividing the displacement in 2 parts and thus, the displacement of 'a' or 'b' is x/2 from their initial positions. Considering one of the masses - let take mass 'a' :-
dW = -kx.dx [For any small displacement dx of 'a'] . So to get the total work done by spring , integrate dW from 0 to W and the right side from 0 to x/2 . I got W = -1/8kx² .
But the solution in book says -1/4kx² .
I think i am wrong somewhere can you please point it out for me .

Thanks in advance (^.^)

 

[tiny-tim]

welcome to pf!

hi vissh! welcome to pf!

… Considering one of the masses :-
dW = -kx.dx . So to get the total work done by spring integrate dW from 0 to x and the right side from 0 to x . I got W = -1/8kx² .

ah, if you're integrating from 0 to x, you must write ∫ kx.(dx/2)

(alternatively, if you let r be the right-hand displacement, you can write ∫ 2kr.dr, and integrate from r = 0 to x/2)

but even easier is to use the work-energy theorem, and just take half the energy!

 

[Doc Al]

I tried to solve in following way [may be wrong :)]
=> the force acting on the each mass will be -kx. Dividing the displacement in 2 parts and thus, the displacement of each mass is x/2 from initial. Considering one of the masses :-
dW = -kx.dx .

x is the total amount of stretch in the spring, not the displacement of each mass. So: dW = -kx.dx is not correct. How would you write the displacement of each mass in terms of x?

Edit: Tiny-tim beat me to it!

 

[vissh]

Thanks for the welcome :D
First, I did some mistakes [in writing above],so can u see it again :D
Second,About my taking displacement of 'a' or 'b' to be 'x/2' :-
As i think , Let the 'centre' of spring [pt at equal dist from both sides] be the origin and 'a' is on left and 'b' on right. Let we pull a and b .When the spring is stretched symmetrically by x , both 'a' and 'b' moves by same distance on their respective sides.At the end ,'a' has moved by distance 'x/2' on left and 'b' by 'x/2' on right from their respective intial positions . Thus,an expansion of 'x' units in spring.
So,I think 'x/2' is the displacement of the 'a' and 'b'.
[if i did wrong , pls explain.I will love to hear it :D]

To tiny-tim :-
Hm The answer came right from the 2 ways you told .But how you wrote that ??[Can you pls explain :) I will love to see the explanation (^.^)]
And how can Work-energy theorem be applied on it ! There are two forces acting on 'a' or 'b' - one spring force and the other 'external force 'F' pulling these masses' .As 'F' is unknown How can Work done by spring could be found out

 

[Doc Al]

Thanks for the welcome :D
First, I did some mistakes [in writing above],so can u see it again :D
Second,About my taking displacement of 'a' or 'b' to be 'x/2' :-
As i think , Let the 'centre' of spring [pt at equal dist from both sides] be the origin and 'a' is on left and 'b' on right. Let we pull a and b .When the spring is stretched symmetrically by x , both 'a' and 'b' moves by same distance on their respective sides.At the end ,'a' has moved by distance 'x/2' on left and 'b' by 'x/2' on right from their respective intial positions . Thus,an expansion of 'x' units in spring.
So,I think 'x/2' is the displacement of the 'a' and 'b'.
[if i did wrong , pls explain.I will love to hear it :D]

Yes, x/2 is the displacement of each mass. But you didn't make use of that in setting up your expression for work!

You wrote: dW = -kxdx, which implies that the displacement is dx. As tiny-tim explained, the displacement is really dx/2.

And how can Work-energy theorem be applied on it ! There are two forces acting on 'a' or 'b' - one spring force and the other 'external force 'F' pulling these masses' .As 'F' is unknown How can Work done by spring could be found out

You know (or should know) the energy stored in a stretched spring. That must equal the total work done by the spring. So the work done on each mass must be half that.

 

[vissh]

(^_^) Doc Al , Thanks a lot :D Got it now.
When there is further dx displacement in spring, displacement in 'a' or 'b' is dx/2 .

The elastic potential energy which get stored on 'x' comp or extension is 1/2kx².
And this is also the magnitude of work done by the spring on the agencies at end points.
Thus,work done on 'a' or 'b' is -1/4kx² as the work done got divided in two parts.

Did I wrote right this time ! :)

 

[Doc Al]

Now you've got it!