Question

A 5-kilogram block is fastened to an ideal vertical spring that has an unknown spring constant. A 3-kilogram block rests on top of the 5-kilogram block, as shown.

 

a. When the blocks are at rest, the spring is compressed to its equilibrium position a distance of ∆x₁ = 20 cm, from its original length. Determine the spring constant of the spring The 3kg block is then raised 50 cm above the 5 kg block and dropped onto it. 

b. Determine the speed of the combined blocks after the collision 

c. Setup, plug in known values, but do not solve an equation to determine the amplitude ∆x₂ 

d. Determine the resulting frequency of this oscillation. of the resulting oscillation 

e. Where will the block attain its maximum speed, explain. 

f. Is this motion simple harmonic.

Answer 1

(a) Equilibrium so F_net = 0, 

F_sp = mg 

k∆x = mg 

k(0.20) = (8)(9.8) 

k = 392N/m

(b) First determine the speed of the 3 kg block prior to impact using energy conservation 

U = K 

mgh = ½ mv² 

 (9.8)(0.50) = ½ v² 

 v = 3.13m/s

Then solve perfect inelastic collision. 

p_before = p_after 

m₁v_1i = (m₁ + m₂)v_f 

(3)(3.13)=(8)v_f 

v_f = 1.17 m/s

(c) Since we do not know the speed at equilibrium nor do we know the amplitude ∆x₂ the turn horizontal trick would not work initially. If you first solve for the speed at equilibrium as was done in you could then use the turn horizontal trick. However, since this question is simply looking for an equation to be solved, we will use energy conservation from the top position to the lowest position where the max amplitude is reached. For these two positions, the total distance traveled is equal to the distance traveled to equilibrium + the distance traveled to the max compression (∆x₁ + ∆x₂) = (0.20 + ∆x₂) which will serve as both the initial height as well as the total compression distance. We separate it this way because the distance traveled to the maximum compression from equilibrium is the resulting amplitude ∆x₂ that the question is asking for.

Apply energy conservation 

 U_top + K_top = U_sp(max-comp) 

mgh + ½ mv² = ½ k ∆x²₂ 

(8)(9.8)(0.20 + ∆x²) + ½(8)(1.17)² = ½ (392)(0.20 + ∆x₂)² 

The solution of this quadratic would lead to the answer for ∆x₂ which is the amplitude.

(e) The maximum speed will occur at equilibrium because the net force is zero here and the blocks stop accelerating in the direction of motion momentarily. Past this point, an upwards net force begins to exist which will slow the blocks down as they approach maximum compressions and begin to oscillate.

(f) This motion is simple harmonic because the force acting on the masses is given by F = k∆x and is therefore directly proportional to the displacement meeting the definition of simple harmonic motion.