Chapter 11, Problem 25P

To determine

To Find: The period of the two spring.

Answer to Problem 25P

The period of the two spring with different spring constants is $T=2\pi \times \sqrt{\frac{m}{k_1+k_2}}$ .

Explanation of Solution

Given info:

Mass of object = $m\text{kg}$ .

Spring first constant = $k_1\text{}\text{N/m}$ .

Spring second constant = $k_2\text{}\text{N/m}$ .

Formula used:

  $F=-k\times x$ .

Net force $F=F_1+F_2$ .

Period of motion $T=2\pi \omega$ .

Angular speed $\omega =\sqrt{\frac{m}{k}}$ .

Here,

  $m$ is mass on spring.

  $k$ is spring constant.

  $x$ is spring displacement.

Calculation:

Spring first force $F_1=-k_1\times x$

Spring second force $F_2=-k_2\times x$

Net force is calculated :

  $\begin{array}{l}F=F_1+F_2\\ F=-k_{1}x-k_{2}x\\ F=-x(k_1+k_2)\end{array}$

Effective spring constant $k=k_1+k_2$ .

Period of motion is determined by:

Period of motion $T=2\times \pi \times \sqrt{\frac{m}{k}}$ .

Where $k=k_1+k_2$ .

  $T=2\times \pi \times \sqrt{\frac{m}{k_1+k_2}}$ .

Conclusion:

Thus, the period of the two spring with different spring constants is $T=2\pi \times \sqrt{\frac{m}{k_1+k_2}}$ .