Question 1.3: Using the law of conservation of energy, establish the equat...
Using the law of conservation of energy, establish the equation of motion of a body of mass m attracted to the origin O with a force F = -Kx. What is the frequency of oscillations if m = 100 g and K = 4.0 \times 10³ N/m ? What is the maximum displacement if the mass is brought to the position x = 5 cm and thrown with a velocity 10 m/s in the positive x direction?
The total energy of this oscillator is U_{(T)} = ½m \dot{x}^2 + ½Kx^2 . As U_{(T)} is constant, the time-derivative of the equation U_{(T)} = constant gives
½ m(2 \dot{x} \ \ddot{x}) + ½ K(2x \ \dot{x} ) = 0 \ \ \ \ \Rightarrow \ \ \ \ \dot{x} (\ddot{x} + ω^2x) = 0,\ \ \ \text{ where} \ \ ω^2 = K/m.
This equation is identically satisfied if \dot{x} = 0 (which corresponds to the body at rest) or if \ddot{x} + ω^2x = 0 (which is the equation of simple harmonic oscillations). Thus, the oscillation frequency is \widetilde{v} = \sqrt{K/m}/2π = 32 Hz .
If the body is thrown from point x(0) = 5 cm with a velocity of \dot{x}(0) = 10 m/s, its total energy is U_{(T)} = ½ m \ \dot{x}^2 + ½ Kx^2 = 10 J. As it reaches its maximum displacement x_m , its velocity vanishes and its energy is totally in the potential form. Thus, at this position, ½ Kx_m^2 = U_{(T)} \text{ and } x_m = \sqrt{2U_{(T)}/K} = 7.1 cm. This result may also be obtained by writing the explicit solution, which corresponds to the given initial conditions (see section 1.2). Equation [1.8] gives the amplitude of the
oscillations A = \sqrt{x(0)^2 + \dot{x}(0)^2 /ω^2} = 7.1 cm which is equal to x_m.
A = \sqrt{u(0)^2 + \dot{u}(0)^2 /ω^2} [1.8]