183_notes:examples:energy_in_a_spring-mass_system [Projects & Practices in Physics]

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===== Example: Energy in a Spring-Mass System =====

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?


=== Facts ===


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)


=== Lacking ===


Amplitude (maximum stretch) of the oscillation

Maximum speed of the block


=== Approximations & Assumptions ===


Friction is negligible.


=== Representations ===


System: Mass and spring

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

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

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

$U_{spring} = \dfrac{1}{2}k_{s}s^2$


=== Solution ===


a: 

From the Energy Principle:

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

The initial energy and the final energy can be expressed as the addition of kinetic and potential parts. 

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

We know that no work is done by the surroundings because the point where the wall force is applied does not move.

We also know that when the stretch is maximum, the block is momentarily at rest (the turning point).

With this information we know that $K_{f}$ and $W$ are zero.

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

The amplitude corresponds to the stretch $s_{f}$. We want to rearrange the equations to isolate $s_{f}$.

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

We can further simplify the equation by substituting values in for the variables.

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

Solve for $s_{f}$.

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

$s_{f} = 0.07 m$


b:

$v_{max}$ is maximum when its final state is when $s_{f} = 0$ which is when the block has only kinetic energy.

The total initial energy when there is only kinetic energy is equal to the total final energy when there is only potential energy.

$K_f + 0 = 0.03 J$

Substitute in $\dfrac{1}{2}mv^2$  for kinetic energy.

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

Isolate $v_{max}$ to solve for the maximum velocity.

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