Chapter 11, Problem 7P

(a)

To determine

To find: The spring constant value.

Answer to Problem 7P

The value of spring constant is $k=0.158\text{ N/m}$ .

Explanation of Solution

Given:

Mass of spider, $\text{M = 0}\text{.25}\times {\text{10}}^{-3}\text{ kg}$ .

Frequency of vibration, $v\text{ = 4 Hz}$ .

Formula used:

Write the expression for time period.

  $\begin{array}{l}T=2\pi \sqrt{\frac{m}{k}}\\ \nu =\text{}\frac{1}{T}\end{array}$

Here,

  $m$ is mass on spring,

  $k$ is spring constant,

  $\nu$ is frequency of oscillation,

  $T$ is time period of spring.

Calculation:

Substituting the given values

  $v\text{=}\frac{1}{2\pi }\sqrt{\frac{k}{m}}$

  $4\text{=}\frac{1}{2\pi }\sqrt{\frac{k}{0.25\times {10}^{-3}}}$

  $k=0.158\text{ N/m}$

Conclusion:

Thus, the value of spring constant is $k=0.158\text{ N/m}$

(b)

To determine

To find: The value of frequency.

Answer to Problem 7P

The frequency of vibrations is $2.83\text{Hz}$ .

Explanation of Solution

Given:

Mass of insect $m=0.5\times {10}^{-3}\text{ kg}$

Formula used:

Time period of oscillation of spring.

  $\begin{array}{l}T=2\pi \sqrt{\frac{m}{k}}\\ \nu =\text{}\frac{1}{T}\end{array}$

Here,

  $m$ is mass on spring,

  $k$ is spring constant,

  $\nu$ is frequency of oscillation,

  $T$ is time period of spring.

Calculation:

Substituting the given values

  $v\text{=}\frac{1}{2\pi }\sqrt{\frac{k}{m}}$

  $\begin{array}{l}v\text{=}\frac{1}{2\pi }\sqrt{\frac{0.158}{0.50\times {10}^{-3}}}\\ =2.83\text{Hz}\end{array}$

Conclusion:

Thus, the frequency of vibrations is $2.83\text{Hz}$ .