Question

A horizontal spring—mass system has low friction, spring stiffness $\mathbf{200}\mathbf{}\mathit{N}\mathbf{/}\mathit{m}$and mass$\mathbf{0}\mathbf{.}\mathbf{4}\mathbf{}\mathit{k}\mathit{g}$. The system is released with an initial compression of the spring of$\mathbf{10}\mathbf{}\mathit{c}\mathit{m}$ and an initial speed of the mass of$\mathbf{3}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$.

(a) What is the maximum stretch during the motion?

(b) What is the maximum speed during the motion?

(c) Now suppose that there is energy dissipation of 0.01 J per cycle of the spring—mass system. What is the average power input in watts required to maintain a steady oscillation?

Short answer

1. The maximum stretchis $0.165m$.
2. The maximum speed is,$3.74m/s$ .
3. The average input energy per second is,$0.030watts$ .

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The stiffness of the spring is,$k=200N/m$ .
• The mass of the spring is,role="math" localid="1668488385102" $m=0.4kg$.
• The initial compression of the spring is,$x_c=10 \text{cm}$.
• The speed of the mass is,$v_i=3m/s$.

Significance of spring Stiffness.

A body's stiffness is a measurement of how resistant an elastic body is to deforming. In general, the force necessary to generate unit deformation for a spring is determined by the spring stiffness.

(a)  Determination of the maximum stretch during the motion.

The expression for the conservation of energy is expressed as,

$\frac{1}{2}k{x}_c^2+\frac{1}{2}m{v}_i^2=\frac{1}{2}k{x}_m^2+\frac{1}{2}m{v}_f^2$

Here, $k$is the spring constant,$x_c$ is the initial compression,$m$ is the mass,$x_m$ is the maximum stretch,$v_i$ is the initial speed and$v_f$ is the final speed.

Substitute all the value in the above equation.

$$\begin{gathered} \frac{1}{2}\left(200N/m\right)\times {\left(0.1m\right)}^2+\frac{1}{2}\left(0.4kg\right)\times {\left(3m/s\right)}^2=\frac{1}{2}\left(200N/m\right)x_m^2+\frac{1}{2}\left(0.4kg\right)\times 0 \\ {x}_m=0.167m \end{gathered}$$

Hence the maximum stretch is $0.167 \text{m}$.

(b) Determination of the maximum speed during the motion

The conservation of energy between the initial and final states is expressed as,

$\frac{1}{2}k{x}_c^2+\frac{1}{2}m{v}_i^2=\frac{1}{2}m{v}_m^2$

Substitute all the value in the above equation.

$$\begin{gathered} \frac{1}{2}\left(200N/m\right)\times {\left(0.1m\right)}^2+\frac{1}{2}\left(0.4kg\right)\times {\left(3m/s\right)}^2=\frac{1}{2}\left(0.4kg\right)\times v_m^2 \\ {v}_m=3.74m/s \end{gathered}$$

Hencethe maximum speed is,$3.74m/s$.

(c) Determination of the average power input

Consider each cycle has energy dissipation is, $0.01 \text{J}$

The energy must be supply steady state oscillation is $0.01 \text{J}$per cycle.

The time period per cycle of per cycle is,

$$\begin{gathered} T=\frac{2\pi }{\omega } \\ T=2\pi \sqrt{\frac{m}{k}} \end{gathered}$$

Substitute all the value in the above equation.

$$\begin{gathered} T=2\mathrm{\pi }\sqrt{\frac{0.4\mathrm{kg}}{200\mathrm{N}/\mathrm{m}}} \\ \mathrm{T}=0.28\mathrm{s} \end{gathered}$$

Average input energy per second is,

$$\begin{gathered} \frac{\Delta E}{T}=\frac{0.01 \text{J}}{0.28 \text{s}} \\ =0.036J/s \\ =0.036watts \end{gathered}$$

Hence the average input energy per second is, $0.036watts$.