1
akari
·2010-01-31 04:01:20

by reduced mass system formula
wew have
ω=√3k2m
xmax can be found using two equations
momentum conservation
4v=3v'
v'=4v/3
and
energy conservation
kxmax2=2mv'2
now
x=xmaxsinωt
49
Subhomoy Bakshi
·2010-01-31 04:26:30
hey AKARI...y 4v=3v'???
shudn't it be mv=2mv', so v=2v'??
no no...instead of asking this i should have asked what are v and v'??
106
Asish Mahapatra
·2010-01-31 22:17:34
but akari..
why in the equation u get xmax why not xmin ??
As I see it using that eqn #2 .. will (should, may) give xmin
23
qwerty
·2010-02-01 04:45:51
asish ,
let m move to the right by x1, and 2m by x2 to the right
ten distance between m and 2m is( L+ x2 - x1)
using momentum conservation and conservation of KE , as akari has done , we can find
x1 - x2 = ± (2mv2/3k)1/2
x1 - x2 will be negative when x1 is to the left , i.e negative of wat we assumed earlier ,
and hence it will account for the condition of max distance
dus distance max = L + (2mv2/3k)1/2
distancemin = L - (2mv2/3k)1/2
and w ecan find eqn by the method wich akari gave
49
Subhomoy Bakshi
·2010-02-01 07:47:27
was my q so faltu?? wat r v n v'???
23
qwerty
·2010-02-01 19:46:46
v is initial velocity, v' is velocity at the reqd condition
106
Asish Mahapatra
·2010-02-01 20:10:02
wat abt range of speeds then..
agreed abt the previous questions...
1
ABHI
·2010-02-01 20:34:17
akari, can u plz explain the reduced mass concept?