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A particle of mass m is in the ground state of a harmonic oscillator with parabolic potential energy:\(V(x) = \frac{1}{2} m\omega_{0}^{2}x^{2}\)

At time t =0, when the wave function is: finalized_jax">\(\Psi(x,0)=\psi(x)\) the "spring constant" of the oscillator is suddenly increased 5.6 times, so that the new potential energy is: \(V(x) = 2.8m\omega_{0}^{2}x^{2}\)

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