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Amplitude and maximum speed for a spring and mass given initial conditions x(0) and v(0).
We start with known values of the initial position and velocity of the simple harmonic oscillator mass, and we are asked to calculate the amplitude. To compute the amplitude of the SHO, we first find the total energy of the oscillator by adding the initial kinetic energy and initial spring potential energy. Next, we discuss the turning point of the harmonic oscillator: at the maximum displacement for the mass on a spring, the displacement is equal to the amplitude, the mass is turning around and it is momentarily stationary. This means the kinetic energy vanishes and all the energy is spring potential when x=A. We set the total energy equal to the spring potential and solve for A, and we've got the amplitude.
In the second part of the question, we're asked to compute the maximum velocity of the mass in the SHO. The maximum velocity occurs at the equilibrium position because the spring potential energy is equal to zero there. Thus the kinetic energy is maximized for the oscillator and we get the maximum speed. This time, we set the total energy of the oscillator equal to the kinetic energy, and we solve for the maximum speed of the oscillator.
