A mass of 0.2 kg is attached to a horizontal spring whose stiffness is 12 N/m. Friction is negligible. At t = 0 the spring has a stretch of 3 cm and the mass has a speed of 0.5 m/s.

a: What is the amplitude (maximum stretch) of the oscillation?

b: What is the maximum speed of the block?

Mass of 0.2kg

Mass attached to horizontal spring

Horizontal spring stiffness = 12 N/m.

The initial states for both parts of the problem are separated into part a and part b.

a:

Initial State: $s_{i}$ = 3 cm, $v_{i}$ = 0.5 m/s

Final State: $v_{f}$ = 0 (maximum stretch)

b:

Initial State: $s_{i}$ = 3 cm, $v_{i}$ = 0.5 m/s

Final State: Maximum speed ($s_{f}$ = 0)

Amplitude (maximum stretch) of the oscillation

Maximum speed of the block

Friction is negligible.

System: Mass and spring

Surroundings: Earth, table, wall (neglect friction, air)

$E_{f} = E_{i} + W$ The energy principle.

$K = \dfrac{1}{2}mv^2$

a:

From the Energy Principle:

$E_{f} = E_{i} + W$

$K_{f} + U_{f} = K_{i} + U_{i} + W$

$K_{f}$ and $W$ cancel out.

$0 + \dfrac{1}{2}k_{s}s^2_{f} = \dfrac{1}{2}mv^2_{i} + \dfrac{1}{2}k_{s}s^2_{i}$

$\dfrac{1}{2}k_{s}s^2_{f} = \dfrac{1}{2}(0.2 kg)(0.5 m/s)^2 + \dfrac{1}{2}(12 N/m)(0.03 m)^2$

$\dfrac{1}{2}k_{s}s^2_{f} = 0.03 J$

$s_{f} = \sqrt {2(0.03 J)/(12 N/m)} = 0.07 m$

$s_{f} = 0.07 m$

b:

$K_f + 0 = 0.03 J$

$\dfrac{1}{2}mv^2_{f} = 0.03 J$

$v_{max} = \sqrt {2(0.03 J)/(0.2 kg)} = 0.55 m/s$