http://targetiit.com/iit-jee-forum/posts/dk-wpe-sum-12687.html
it has been discussed here.
http://targetiit.com/iit-jee-forum/posts/dk-wpe-sum-12687.html
it has been discussed here.
hey dude , eureka has picked the solution from wikipedia ,
http://en.wikipedia.org/wiki/Effective_mass_%28spring-mass_system%29 (hey eureka! , no offense indeed)
but i also have the same doubt from that solution , why are we taking u=vy/l ?
to find the equivalent mass m' we equalise the moment of inertia about the point wer the spring is attached to the wall
∫0llmdx=m′l2
giving m'=m/3
so...1/2(M+m/3)v2=1/2kx2
giving the maximum compression....[1][1][1]
wats da logic behind this satement
"
to find the equivalent mass m' we equalise the moment of inertia about the point wer the spring is attached to the wall
"
just a whim that came to mind and solved the sums in irodov jo ab tak nahi ho raha tha
anyways i thought the equivalent mass m' depends upon the distribution of mass abt the point of suspension...so bring such a mass whch gives equivalence to the distributed mass at a point...so wat better than balancing moment of inertia...[1][1][1]
another good reason is that this is tru coz it gives correct answers...[1][1][1]
got the method by which it is done in wikipedia...
akari this one is for you..
suppose there is a rubber string fixed at one end to a wall..
the other end is being pulled with constant velocity v..
this is extending all part of the rubber string equally
now suppose the rubber string is divided into n equal divisions each of length x...
so total length=nx
now suppose due to pulling each of the x length element stretches by length dx
so first element increases to x+dx
second element also increases to x+dx
... ... .....
.... .... ....
.... ...... .....
the nth element increases to x+dx
total string increases to nx+ndx
increase in length=ndx
now let the above change take place in time dt..
thus from above post displacement of the end point of string=ndx
thus, ndtdx=v
so, dtdx=nv
so the velocity of (n-1)th element is vn-1=(n−1)dtdx=nn−1v
similarly, velocity of pth element, vp=pdtdx=npv
thus, vp=nvp
thus, vp=(constant)p
thus, vp
this proves approximately that the velocity of a point on an extendable string is proportional to the distance from fixed end...[1][1]
this can be applied to springs also...
so most probably answer's akari's doubt...[1][1]
now another solution follows..[1][1]
total kinetic energy of the system initially is to be found out...
finallyy this gets converted to the spring energy at maximum extension of spring..
for simplicity's sake i hav considered the spring to be an elastic band...[3][3]
we will first find out the K.E of the spring...of mass m...then add the K.E. of the attached mass...[1][1]
mass per unit length of spring(λ)=m/L
now as it is at a dist x from the fixed end and the free end moves with velocity v, the element considered moves with velocity u=xv/L..as discussed in the posts above..[1][1][1]
mass of element = λdx=mdx/L
thus K.E. of the element=∫u2dm=∫0m21u2dm=∫0L21(Lxv)2(Lmdx)=∫0L21L3mv2x2dx
=2L3mv2∫0Lx2dx=2L3mv23L3
=21(3m)v2
K.E. of attached body==21Mv2
so total K.E.==21(M+3m)v2
thus, 21ka2=21(M+3m)v2(a=extension)
so, maximum extension=a=√kM+3mv
[1][1][1][1]
same answer by both processes prove genuinity of both the processes..[1][1][3][3]