A particle of mass m, and angular momentum is moving in a circular orbit of radius r_0 under the influence of an attractive force F (r) = - k r^2 r . Keeping its angular momentum unchanged, the particle is displaced radially by a small distance r r_0, due to which its radial distance varies periodically. The corresponding time period is:
A. 2 ^3 m k^2
The equation of motion for the radial distance r is given by: m d^2r dt^2 = F(r) + ^2 mr^3 = - k r^2 + ^2 mr^3 For a circular orbit of radius r_0, the radial acceleration is zero: - k r_0^2 + ^2 mr_0^3 = 0 r_0 = ^2 mk Let the particle be displaced by a small distance x such that r = r_0 + x. The restoring force is: m d^2x dt^2 = - k (r_0+x)^2 + ^2 m(r_0+x)^3 Using the binomial expansion for x r_0: m d^2x dt^2 - k r_0^2 (1 - 2x r_0 ) + ^2 mr_0^3 (1 - 3x r_0 ) Since k r_0^2 = ^2 mr_0^3 , the constant terms cancel out: m d^2x dt^2 ( 2k r_0^3 - 3 ^2 mr_0^4 ) x Substituting ^2 mr_0^4 = k r_0^3 into the equation: m d^2x dt^2 = ( 2k r_0^3 - 3k r_0^3 ) x = - k r_0^3 x This represents simple harmonic motion with angular frequency = k mr_0^3 . The time period is T = 2 = 2 mr_0^3 k . Substituting r_0 = ^2 mk : T = 2 m k ( ^2 mk )^3 = 2 m ^6 k m^3 k^3 = 2 ^3 m k^2 Answer: 2 ^3 m k^2
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