mechanics!

to find the equivalent mass m' we equalise the moment of inertia about the point wer the spring is attached to the wall

\int_{0}^{l}{\frac{m}{l}dx}=m'l^2

giving m'=m/3

so...¹/₂(M+^m/₃)v²=¹/₂kx²

giving the maximum compression....[1][1][1]

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 n^th 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, n\frac{dx}{dt}=v
so, \frac{dx}{dt}=\frac{v}{n}

so the velocity of (n-1)th element is v_n-1=(n-1)\frac{dx}{dt}=\frac{n-1}{n}v

similarly, velocity of pth element, v_p=p\frac{dx}{dt}=\frac{p}{n}v

thus, v_p=\frac{v}{n}p
thus, v_p=(constant)p
thus, v_p\propto p

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=\int_{0}^{m}{\frac{1}{2}u^2dm}=\int_{0}^{L}{\frac{1}{2}(\frac{xv}{L})^{2}(\frac{mdx}{L})}=\int_{0}^{L}{\frac{1}{2}\frac{mv^2x^2dx}{L^3}}
=\frac{mv^2}{2L^3}\int_{0}^{L}{x^2dx}=\frac{mv^2}{2L^3}\frac{L^3}{3}
=\frac{1}{2}(\frac{m}{3})v^2

K.E. of attached body==\frac{1}{2}Mv^2

so total K.E.==\frac{1}{2}(M+\frac{m}{3})v^2

thus, \frac{1}{2}ka^2=\frac{1}{2}(M+\frac{m}{3})v^2(a=extension)

so, maximum extension=a=\sqrt{\frac{M+\frac{m}{3}}{k}}v

[1][1][1][1]

same answer by both processes prove genuinity of both the processes..[1][1][3][3]

akari see ur doubt has been solved...[1][1][1][1]

welcs................[3][3]