91影视

According to Newton鈥檚 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鈥檚 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 \( \pm \frac{1}{2}{x_0}\).