Question

particle on a spring moves in simple harmonic motion along the x axis between turning points at x1 = 85 cm and x2 = 160 cm. (i) At which of the following positions does the particle have maximum speed? 85 cm 103.75 cm 122.5 cm at none of those positions (ii) At which position does it have maximum acceleration? 85 cm 103.75 cm 122.5 cm at none of those positions (iii) At which position is the greatest net force exerted on the particle? 85 cm 103.75 cm 122.5 cm at none of those positions

Answer 1

(i) 122.5cm

(ii) 85cm

(iii) 85cm

Step-by-step explanation:

A particle undergoing simple harmonic motion around an equilibrium point has its maximum velocity at its equilibrium point. This happens because the particle is accelerated, in this case by a spring, from a turning point towards the other one. When it passes the equilibrium point, the acceleration becomes negative since the spring exerts its force in the opposite direction, so the speed starts decreasing, having its maximum value at the equilibrium point.

The equilibrium point is located exactly halfway between the turning points, so, we first calculate the distance between them (160cm-85cm=75cm), then divide it by two (75cm/2=37.5cm) and sum it to the distance to the nearest turning point (37.5cm+85cm=122.5cm). This is the point in which the particle has maximum speed, 122.5cm (i).

Now, since the net force is proportional to the acceleration, they have their maximum values in the same places. The spring force is proportional to the distance to the equilibrium point (|F|=kx), so the maximum value for the force and the acceleration is reached where the distance to the equilibrium is higher. This occurs in the turning points (85cm and 160cm). As the 160cm is not between the possible answers, the correct answer for (ii) and (iii) is 85cm.