3since part i is solved lemme try 2 ...........
x2 - x1 max = 2(m1 - m2)/m2 A .......................
1/2 of it shud be the natural length of spring as x2-x1 min =0
m1-m2/m2A = Lo
Two point masses m1 and m2 are coupled by a spring of spring constant k and uncompressed length L0. The spring is fully compressed and a thread ties the masses together with negligible speration between them. the tied assembly is moving in the +x direction with uniform speed v0.
At a time say t=0, it is passing the origin and at that instant the thread breaks. The masses, attached to the spring start oscillating.
The displacement of mass m1 is given by x1(t)=v0t-A(1-cosωt) where A is a constant
Fidn
1) displacement x2(t) of mass m2
ii) the relationship between A and L0
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3 Answers
1v1(t)=v0-Aωsin(ωt)
m2v2(t)+m1v1(t)=(m1+m2)v0
m2v2=m1v0+m2v0-m1v0+m1Aωsin(ωt)
v2=v0+m1/m2Aωsin(ωt)
dx/dt=v0+m1/m2Aωsin(ωt)
0∫xdx=0∫tv0+m1/m2Aωsin(ωt)dt
x2(t)=v0t-m1/m2Acos(ωt)+m1A/m2
x2(t)=v0t+m1/m2A(1-cos(ωt))
3oh!!!!! thanx for postin ques. bhiyya didnt see it bfr and i went offline ..............
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