Question

A 0.108-kg block is suspended from a spring. When a small pebble of mass 31 g is placed on the block, the spring stretches an additional 4.9 cm. With the pebble on the block, the block oscillates with an amplitude of 10 cm. Find the maximum amplitude of oscillation at which the pebble will remain in contact with the block.

Answer 1

Answer:

A = 0.22 m

Explanation:

The spring with the block and the pebble forms an oscillatory system, this system is described by the expression

         x = A cos (wt + φ)

         w = √ (k / m).

The data they give us is the amplitude of the movement (A = 10 cm), the oscillation mass is equal to the block mass plus the mass of the pebble

         m = m + M

         m = 0.031 + 0.108

         m = 0. 139 kg

To find the spring constant let's use Hooke's law

         F = k X

The force is the weight of the pebble and the additional elongation is x = 4.9 cm, let's calculate

         k = F / x

         k = mg / x

         k = 0.031   9.8 / 0.049

         k = 6.2 N / m

Let's look for angular velocity

        w = √ (6.2 / 0.139)

        w = 6,670 rad / s

Let's write the oscillation equation with this data

         x = 0.10 cos (6,670 t)

For the pebble to remain in contact with the block, the acceleration of the spring system plus block with pebble must be less than the acceleration of gravity in the descending oscillation

Let's look for system acceleration

         a = d²x / dt²

         dx / dt = - A w sin (wt + Ф)

         d²x / dt² = - A w² cos (wt+Ф)

To find the maximum value cos (wt) = ±1

         g = A w²

         A = g / w²

         A = 9.8 / 6.67²

         A = 0.22 m

When the amplitude of the oscillation exceeds this value the pebble is delayed with respect to the block