Question

A block suspended from a spring is oscillating vertically with a frequency of 4 Hz and an amplitude of 7 cm. A very small rock is placed on top of the oscillating block just as it reaches its lowest point. Assume that the rock has no effect on the oscillation. At what distance above the block’s equilibrium position does the rock lose contact with the block? (hint this occurs when the rock’s acceleration equals the value of gravity) What is the speed of the rock when it leaves the block? What is the greatest distance above the block’s equilibrium position reached by the rock? (Let t = 0 be when the rock is placed on the block)

Answer 1

Answer:

v = - 1,715 m / s ,   x = 0.0156 m

Explanation:

This is an oscillatory movement exercise, which is described by the expression

          x = A cos (wt + Ф)

we can assume that the block is released from its maximum elongation, so the phase constant (Ф) is zero

As we are told that the stone does not affect the movement of the spring mass system, the amplitude and angular velocity do not change, in the upward movement the stone is attached to the mass, but in the downward movement the mass has an acceleration greater than g leave the stone behind, let's look for time, for this we use the definition of speed and acceleration

         v = dx / dt

         v = - A w sin wt

          a = - Aw² cos wt

         a = -g

         -g = - Aw² cos wt

          wt = cos⁻¹ (g / Aw²)

          t = 1 / w cos⁻¹ (g / Aw²)

       

angular velocity and frequency are related

        w = 2π f

        w = 2π 4

         w = 8π   rad / s

remember that the angles are in radians

          t = 1 / 8π cos⁻¹ (9.8 / (0.07 64π²))

          t = 0.039789 1.3473

          t = 0.0536 s

let's find the speed for this time

          v = - A w sin wt

          v = - 0.07 8π sin (8π 0.0536)

          v = - 1,715 m / s

the distance is

           x = A cos wt

           x = 0.07 cos (8π 0.0536)

           x = 0.0156 m