Question

Consider a harmonic oscillator (mass on a spring without friction). Taking the mass alone to be the system, how much work is done on the system as the spring of stiffness${\mathbf{K}}_{\mathbf{S}}$ contracts from its maximum stretch A to its relaxed length? What is the change in kinetic energy of the system during this motion? For what choice of system does energy remain constant during this motion?

Short answer

The amount of work done on the system is $W=\frac{1}{2}{k}_s{A}^2$ and the change in kinetic energy during the motion is $\nabla K=\frac{1}{2}{k}_s{A}^2$ . The energy remains constant for the system spring plus mass.

Step-by-step solution

Define the kinetic energy

The energy that an item has as a result of its motion is known as kinetic energy.

We must apply force to an item if we want it to accelerate. Applying force necessitates effort. After the job is completed, energy is transferred to the item, which then moves at a new constant speed. Kinetic energy is the amount of energy transmitted and is determined by the mass and speed achieved.

Calculation for amount of work done

The formula to find work is $W={\int }_0^{A}Fdx$where $F$is force and $x$is the displacement of the block.

For the spring force is $F=kx$. The stretch of the spring as a function of the displacement will be $A-x$.

Substitute $F=k_s\left(A-x\right)$ into the formula of work.

role="math" localid="1657788251253" $$\begin{gathered} W={\int }_0^A{k}_s\left(A-x\right)dx \\ =k_s{\overline{)\left(Ax-\frac{x^2}{2}\right)}}_0^A \\ =k_s\left(A^2-\frac{A^2}{2}\right) \\ =\frac{1}{2}{k}_s{A}^2 \end{gathered}$$

Therefore, the amount of work done is obtained as$W=\frac{1}{2}{k}_s{A}^2$ .

Change in kinetic energy

No non-conservative forces act on the system and so, energy is conserved between the initial and final states that is$E_i=E_f$.

In the initial state the kinetic energy is zero, therefore the initial energy is equal to the elastic potential energy of the spring.

$$\begin{gathered} E_i=\frac{1}{2}{k}_s{A}^2 \\ As{E}_f=K_{f}so,K_f=\frac{1}{2}{k}_s{A}^2 \end{gathered}$$

Similarly, when the spring is in its relaxed position, there is only kinetic energy.

$$\begin{gathered} \nabla K=K_f\cancel{K_i} \\ =\frac{1}{2}{k}_s{A}^2 \end{gathered}$$

The energy would remain constant for the system spring plus mass, because no non-conservative forces act on the system, so energy would only be shared between elastic potential energy (in the spring) and kinetic energy (in the mass). However, the total energy is always the same.

Therefore, the change in kinetic energy is obtained as $\nabla K=\frac{1}{2}{k}_s{A}^2$.