The ratio of the time periods of small oscillations of the insulated spring and mass system before and after charging the masses is:

Question

Arun · Answer

At equilibrium keq2/(x2e-x1e)2 = k(x2e-x1e-L) ,where x2eand x1e are respective equilibrium positions . Or, keq2/xe2 = k(xe-L) ,where xe is the separation between the masses at equilibrium. Is it correct ? EOM for mass m1 is m1d2x1/dt2 = kx - keq2 / (x+L)2 EOM for mass m2 is m2d2x2/dt2 = -kx + keq2 / (x+L)2 .

Vikas TU · Answer

Dear student

Charge of small bodies in eqm = (1/4*pie*epselon) q^2/(lo+x0)^2 = kx0
F = -[k(x+x0) - (1/4*pie*epselon) q^2/(lo+x0)^2 ] = -kx , since x
a = -2k/m * x
F = mew *a
mew =m/2
w = sqrt(2k/m)
T = 2*pie sqrt(m/2k)
T0 = 2*pie sqrt(m/2k)
T/T0 = 1

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The ratio of the time periods of small oscillations of the insulated spring and mass system before and after charging the masses is:

The ratio of the time periods of small oscillations of the insulated spring and mass system before and after charging the masses is:

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