1let the mass be m1 and m2
k(x)=(m(1) +m(2))g
x=(m(1)+m(2))g/k
by enrgy conservation
1/2 (k)*(x2-y2) =m(1)gy
we know x and hence y can be evaluated
y=\frac{g}{k} (\sqrt{m_{1}^{2}+ (m_{1}+m_{2})^{2}}-m_{1})
IF TWO BLOCKS GLUED TOGETHER ARE ATTACHED TO A SPRING & ONE OF THEM FALLS INSTANTANEOUSLY ,,,,,,,,,,,THEN FIND AMPLITUDE OF OSCILLATION OF FIRST
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4 Answers
111See the force acting on it in equilibrium when the blocks is intact is
F = (m1 + m2)g
kx = F
x = (m1+m2)g/k
Now when the block leaves the force downward will be = m1g
Therefore the new equilibrium position will be
x = m1g/k
Therefore change in equilibrium length is m2g/k
Therefore the amplitude will be m2g/k
11I think xyz is right.
virang, F=kx is not correct for amplitude cases, bcoz this is true for all x, not only the amplitude.
