Chapter 11, Problem 12P

To determine

To Find: The time required by the spring.

Answer to Problem 12P

The spring will take $\text{0}\text{.282 s}$ .

Explanation of Solution

Given info:

Mass $m=2.62\text{ kg}$ .

Distance $x=0.315\text{ m}$ .

Formula used:

Spring constant that can be found from the stretch distance corresponding to the weight suspended from spring.

Expression of force :

  $\begin{array}{l}F=k\times x\\ k=\frac{F}{x}=\frac{m\times g}{x}\end{array}$

Here,

  $F$ is force,

  $x$ is the displacement of spring,

  $k$ is spring constant,

  $m$ is mass on spring.

After being stretched further & released the maximum oscillate it takes one quarter of a period for a mass to move from a maximum displacement to the equilibrium position.

  $\begin{array}{l}t=\frac{1}{4}\times T\\ T=2\pi \times \sqrt{\frac{m}{k}}\end{array}$

Here,

  $T$ is the time period of oscillation.

  $m$ is mass on spring.

  $k$ is spring constant.

Calculation:

Substitute the given values in the above equation,

  $\begin{array}{l}k=\frac{2.62\times 9.8}{0.315}=81.5\text{ N/m }\\ t=\frac{1}{4}\times 2\pi \times \sqrt{\frac{2.62}{81.5}}\\ t=0.282\text{ s}\end{array}$

Conclusion:

Thus, the new equilibrium position will be reach by spring in $0.282\text{ s}$ .