Question:

A mass of 0.2 kg is attached to a spring with a spring constant of 200 N/m. The mass is pulled down 0.1 meters from its equilibrium position and released. Calculate the maximum speed attained by the mass during its oscillation.

Answer:

To find the maximum speed attained by the mass, we will apply Hooke's Law and the principles of mechanical energy.

First, we can determine the maximum potential energy stored in the spring when it is stretched 0.1 meters from its equilibrium position. This can be found using the formula for potential energy in a spring: [ PE_{max} = \frac{1}{2}kx^2 ] Where: $k = 200 \, \text{N/m}$ (spring constant) $x = 0.1 \, \text{m}$ (displacement from equilibrium)

At the equilibrium position, the potential energy is zero, so the total mechanical energy of the mass-spring system is equal to the maximum potential energy: [ E_{mech} = PE_{max} ] [ E_{mech} = 1 , \text{J} ]

The maximum speed is attained at the equilibrium position, where all of the potential energy has been converted to kinetic energy. Using the equation for kinetic energy: [ KE_{max} = \frac{1}{2}mv^2 ] Where: $m = 0.2 \, \text{kg}$ (mass) $v$ = maximum speed (to be determined)

Setting the mechanical energy equal to the kinetic energy at the maximal position: [ KE_{max} = E_{mech} ] [ \frac{1}{2}mv^2 = 1 , \text{J} ]

Solving for $v$: [ v^2 = \frac{2 \times 1 , \text{J}}{0.2 , \text{kg}} ] [ v = \sqrt{10} , \text{m/s} ] [ v ≈ 3.16 , \text{m/s} ]

Therefore, the maximum speed attained by the mass during its oscillation is approximately $3.16 \, \text{m/s}$.