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class7b

# class7b - 1 Linear Harmonic Oscillator Consider the case of...

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1 Linear Harmonic Oscillator Consider the case of small energies, when the second derivative of U ( x ) at x = 0 is a constant k . What is the nature of the motion? From the last class, the potential energy is U ( x ) = kx 2 / 2, where k = d 2 U dx 2 0 . Since the force is the negative gradient of U , F ( x ) = - kx (a linear restoring force ), and Newton’s second law becomes: m ¨ x = - kx (1) or ¨ x + ω 2 0 x = 0 (2) where we have introduced ω 0 = k/m . How do we solve this equation? The general form of this equation is a second-order ordinary di ff erential equation ( y + ay + by = f ( x )) that you will learn how to solve in a ODE class. Luckily, since y = 0 and f ( x ) = 0, we can take advantage of a special solution. First, multiply both sides by 2 ˙ x : 2 ˙ x ¨ x = - 2 ω 2 0 x ˙ x (3) and integrate ˙ x 2 = - ω 2 0 x 2 + C (4) where C = ω 2 A 2 from initial conditions. This leads to dx A 2 - x 2 = ω dt (5) which can be integrated to get arcsin x A = ω 0 t + φ (6) thus x = A sin( ω 0 t + φ ) (7) The two arbitrary constants must be determined from initial conditions. ω 0 is the angular

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class7b - 1 Linear Harmonic Oscillator Consider the case of...

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