Chapter 11, Problem 4P

(a)

To determine

To find: The spring constant value.

Answer to Problem 4P

The value of spring constant is $750\text{N/m}$ .

Explanation of Solution

Given:

Mass of fish, $\text{M = 2}\text{.7 kg}$ .

Scale reading, $\text{x = 0}\text{.036 m}$ .

Formula used:

Restoring force of spring

  $F=-kx$

Here,

  $k$ is spring constant,

  $F$ is spring restoring force,

  $x$ is displacement of spring.

Calculation:

Substituting the given values

  $F=kx$

  $2.7\times 10=k\times 0.036$

  $k=750\text{ N/m}$

Conclusion:

Thus, the value of spring constant is $750\text{N/m}$ .

(b)

To determine

To find: The value of amplitude and frequency.

Answer to Problem 4P

The amplitude and frequency of vibrations are $\text{A = 0}\text{.025 m}$ and $2.652\text{Hz}$ .

Explanation of Solution

Given:

Fish pulled down $\text{0}\text{.025 m}$ more, which is nothing but Amplitude of vibration, $\text{A = 0}\text{.025 m}$ .

Formula used:

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

Here,

  $m$ is mass on spring,

  $k$ is spring constant,

  $T$ is time period of spring,

  $\nu$ is frequency of oscillation.

Calculation:

Substituting the given values

  $\text{T =}2\pi \sqrt{\frac{2.7}{750}}=0.377\text{s}$

  $v=\text{}\frac{1}{0.377}=2.652\text{ Hz}$ .

Conclusion:

Thus, the amplitude and frequency of vibrations are $\text{A = 0}\text{.025 m}$ and $2.652\text{Hz}$ .