A mass attached to the end of a spring is stretched a distance \({x_0}\) from equilibrium and released. At what distance from equilibrium will it have (a) velocity equal to half its maximum velocity, and (b) acceleration equal to half its maximum acceleration?

According to Newton’s second law of motion, force is equal to the product of mass and its acceleration.

\(F = ma\)

Here,\(a\)is the acceleration and\(m\)is the mass.

The mass of the car is\(m\).

The extension of the spring is\({x_0}\).

Part (a)

Using the relation of velocity and the position for SHM is:

\(v = \pm {v_{{\rm{max}}}}\sqrt {1 - \frac{{{x^2}}}{{{A^2}}}} \)

This relation is equal to half of its velocity.

\(\begin{aligned}{c}v = \frac{1}{2}{v_{{\rm{max}}}}\\ \pm {v_{{\rm{max}}}}\sqrt {1 - \frac{{{x^2}}}{{{A^2}}}} = \frac{1}{2}{v_{{\rm{max}}}}\\1 - \frac{{{x^2}}}{{{A^2}}} = \frac{1}{4}\\\frac{{{x^2}}}{{{A^2}}} = \frac{3}{4}\end{aligned}\)

Solving further to find the position:

\(\begin{aligned}{c}x = \pm \frac{{\sqrt 3 A}}{2}\\ = \pm 0.866A\end{aligned}\)

Hence,at\( \pm 0.866A\), velocity will be equal to half of its maximum velocity.

Part (b)

Using Newton’s second law of motion, force is related to acceleration and displacement.

\(\begin{aligned}{c}F = ma\\ma = - kx\\a = - \frac{{kx}}{m}\\a \propto x\end{aligned}\)

Hence, as displacement has half of its maximum value, so does the acceleration will have half of its maximum value, at \( \pm \frac{1}{2}{x_0}\).