A particle of mass m is moving in a potential well, for which the potential energy is  given by U(x)=U01−sinbx where U0 and a are constants. Then for small  oscillations

U(x)=U0(1−sin⁡bx)F=−dUdx=−aU0sin⁡bx For small oscillations, sin⁡bx≈bx⇒ma'=−a2U0x⇒a'=−a2U0mx . Since, a'α−x , oscillations are simple harmonic.Hence, T=2πmb2U0 .Speed is maximum when a’ = 0. It happens at x = 0.dvdt=−a2U0xdvdxdxdt=−a2U0xvdv=−a2U0xdx∫max0 vdv=−a2U0∫0A xdxSince, vmax is unknown, A cannot be found.