IIT JEE , AIEEE Forum - Physics , Mechanics

bhuwanaroracorroded

m1 m2 connected by light spring(k)

placed on frictionless table

initially spring stretched through x

after releasing from rest

find d distance moved by d 2 masses b4 they againcom 2 rest?

crd_123

I think the masses will never come to rest...

anant_c

check using the work energy theorem

bhuwanaroracorroded

tried every thing u give d sol.

crd_123

pl tell me how to draw diagrams in the editor.....i'll explain then....

anant_c

yes it is POSSIBLE for them to come to rest....this can be seen by the principle of conservation of momentum...At any instant the momentum of the system is zero..

So it can be that both objects come to rest

bhuwanaroracorroded

@ anant

give d distnce moved

anant_c

if x1 and x2 are the coordinates of the masses wrt.. a coordinate system...then x =x2-x1....ill try to tell u the distance 2morow...meanwhile try with this

bhuwanaroracorroded

@ crd

i too dont know how to draw all these

u just write d equations i will unerstand myself

svj29

as during oscillations of the 2 block system, dere is no xtrnal frce prsnt. thus we can state that the cente of mass of the 2 blck systm will remain at rest durin the oscillation. make d diagram and consider C as the COM of the blocks . l is the natural length of the system, then we can say that in this situation point C will remain at rest with respect to this point m1 and m2 will oscillate independently. u can also split the spring in 2 parts of lenghts l1 and l2 which r given by

l1=m2l/m1+m2

l2=m1l/m1+m2

now if these 2 springs are assumd to be fixed at point C separately, the case still reamins same as in absence of external forces COM of system C remains at rest. the respective forces are:

k1=kl/l1=k(m1+m2)

similarly for k2

now the angular frequency and time period of the oscillation can be drctly given frm that of a spring blck system with one end of the spring fixed as

W(omega)=underroot k/m

for mass m1

W=underroot k1/m1=underroot (m1+m2)k/m1m2

similarly u can write for m2

both are same thus both the blks oscillates with same angular frequency given in the above eqns

therefore time period can be given as 2pi underroot m1m2/(m1+m2)k

hence u get the time period from which u can calculate the dist. travelled....

plz do rate is satisfied...

koustuv

as no external force acts on the system of two masses the acc of system wrt grnd frame=0,, hence cm at rest

so let m1 and m2 traverses a distances x1 and x2 at any instant. and instantaneously velocities v1 and v2 respectively . now conserving the energies (x' is instaneous compression/elongation of sp)
1/2 kx^2=1/2m1(v1)^2+1/2m2(v2)^2+1/2k(x')^2

v1 and v2 are dependent parameters as m1v1=m2v2;
so,at the instant when each of them become equal to 0,
x'=x . nearest posn when spring compressed by x

so we may write

x1+x2-x=x
=>x1+x2=2x

again m1x1=m2x2;
x2=(m1/m2)x1;
x1+m1/m2x1=2x
(m1+m2)x1=2x(m2)
=>x1=2x(m2)/(m1+m2)

likewise determine x2

I HOPE I AM CORRECT. PLEASE CONFIRM

bhuwanaroracorroded

@ kostuv

how u have written d eqn x1+x2-x=x

koustuv

IF i am not mistaken i think svj29 has actually gone into the simple harmonic aspect of the problem
but actually its there also the bodies will attain some instantaneous rest

and then it would continue its motion becuse of inertia

koustuv

see the spring is initially elongated by x,
now when left at rest the blocks first traverse through the natural length of spring. (at this instant suppose x1'+x2'=x)
the blocks now compress the spring more by x
so(x1''+x2''=x)

together taking these 2 int considerartion

(x1'+x1")+(x2'+x2'')=x
=>x1+x2=x;

koustuv

thank u bhuwanarora for ur rates. these let me get off the mark

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