F ma du dv f m dx dt d2x m kx dt d2x m 2 am 2 sin

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Unformatted text preview: mass ; ω: frequency Energy x(t ) = A sin(ω t + φ ) 1 1 E = mv 2 + kx 2 2 2 Newton Eq. F = ma dU dv F=− =m dX dt d2x m = − kx dt d2x m 2 = − Amω 2 sin (ω t + φ ) = F = − kx = − kA sin (ω t + φ ) dt → ω2 = k m 2 1 ⎛ dx ⎞ 12 E = m ⎜ ⎟ + kx 2 ⎝ dt ⎠ 2 1 1 2 2 E = m (ω A cos (ω t + φ )) + k ( A sin(ω t + φ )) 2 2 12 1 ⎡sin 2 (ω t + φ ) + cos 2 (ω t + φ ) ⎤ = kA 2 E = kA ⎣ ⎦2 2 The larger is the amplitude A, the larger is the classical energy. If the amplitude is zero the oscillator energy is zero (an OK solu)on) Quantum Oscillator Hamiltonian: d k2 H =− +x 2 2 m dx 2 2 2 Boundary condition: ψ ( x) → 0 x → ±∞ Quantum harmonic oscillator 2 d 2ψ 1...
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