Answer
See answers.
Work Step by Step
a. Equation 11-5b relates the velocity and the position of a simple harmonic oscillator.
Set v equal to half the maximum speed, and solve for the displacement.
$$v=v_{max}\sqrt{1-x^2/x_0^2}=\frac{1}{2}v_{max}$$
$$ \sqrt{1-x^2/x_0^2}=\frac{1}{2} $$
$$ 1-x^2/x_0^2=\frac{1}{4} $$
$$ x^2/x_0^2=\frac{3}{4} $$
$$x=\frac{\sqrt{3}}{2}x_0$$
b. For a mass on a spring, the magnitude of the force is proportional to the distance from equilibrium, F = -kx, so the acceleration is also proportional to the displacement. This may be expressed as $a=-kx/m$. Therefore, the acceleration reaches half its maximum value when the displacement also reaches half its maximum value.
The distance is $x=\frac{1}{2}x_0$.