The ratio of the time periods of small oscillation of the insulated spring and mass system before and after charging the masses is
(a) ≥ 1
(b) > 1
(c) ≤ 1
(d) = 1
The Attempt at a Solution
First I calculated the time period of neutral masses .
Let us consider origin to be at the left of m1 .Take the x axis to be along the length of the spring and call the left mass be m1 at coordinate x1 and right mass be m2 at coordinate x2 .
d2x/dt2 = d2(x2-x1)/dt2
EOM for mass m1 is m1d2x1/dt2 = kx
EOM for mass m2 is m2d2x2/dt2 = -kx
After manipulating the above two equations we arrive at d2/dt2 + (k/μ)x = 0 ,where μ = m1m2/(m1+m2) .
The time period is 2π√(μ/k) .
Now when the masses are charged ,
EOM for mass m1 is m1d2x1/dt2 = kx - keq2 / (x+L)2
EOM for mass m2 is m2d2x2/dt2 = -kx + keq2 / (x+L)2 .
Is this the correct way to approach the problem ? If yes ,how should I proceed .
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