Question

A proud deep-sea fisherman hangs a 65.0-kg fish from an ideal spring having negligible mass. The fish stretches the spring 0.180 m. (a) Find the force constant of the spring. The fish is now pulled down 5.00 cm and released. (b) What is the period of oscillation of the fish? (c) What is the maximum speed it will reach?

Short answer

(a) 3538.89 N/m (b) 0.54 s (c) 0.369 m/s

Step-by-step solution

Identify the Known Values

We are given the mass of the fish, the distance the spring is stretched initially, and the additional stretch for oscillation. Specifically, the mass of the fish is 65.0 kg, the initial stretch is 0.180 m, and the additional stretch is 5.00 cm or 0.05 m.

Calculate the Force Constant of the Spring

Using Hooke's Law, we calculate the spring constant. Hooke's Law is given by \( F = kx \), where \( F \) is the force, \( k \) is the spring constant, and \( x \) is the displacement of the spring.The force exerted by the fish is its weight, \( F = mg = 65.0 \text{ kg} \times 9.8 \text{ m/s}^2 = 637 \text{ N} \).Solving for \( k \):\[ k = \frac{F}{x} = \frac{637 \text{ N}}{0.180 \text{ m}} = 3538.89 \text{ N/m}\]

Determine the Period of Oscillation

The period of oscillation for a mass-spring system is given by \( T = 2\pi \sqrt{\frac{m}{k}} \).Substituting the known values:\[ T = 2\pi \sqrt{\frac{65.0 \text{ kg}}{3538.89 \text{ N/m}}} \approx 0.54 \text{ s}\]

Calculate the Maximum Speed of the Fish

The maximum speed in simple harmonic motion is given by \( v_{max} = \omega A \), where \( \omega \) is the angular frequency \( \omega = \sqrt{\frac{k}{m}} \) and \( A \) is the amplitude.First, calculate \( \omega \):\[ \omega = \sqrt{\frac{3538.89}{65.0}} \approx 7.383 \text{ rad/s}\]Then, calculate \( v_{max} \):\[ v_{max} = 7.383 \text{ rad/s} \times 0.05 \text{ m} \approx 0.369 \text{ m/s}\]

Key concepts

Hooke's Law

Hooke's Law is a fundamental principle that describes the behavior of springs and other elastic materials when they are compressed or stretched. It states that the force exerted by a spring is directly proportional to its displacement. In mathematical terms, it's expressed as:- \( F = kx \)where:

• \( F \) is the force applied on the spring (in newtons, N)
• \( k \) is the spring constant (in newtons per meter, N/m)
• \( x \) is the displacement of the spring from its equilibrium position (in meters, m)

Hooke's Law is applicable until the spring reaches its elastic limit. For the problem, we calculate the force on the spring due to the fish's weight (\( F = mg \)) and then use that force to find the spring constant \( k \). This relationship is crucial for understanding how springs behave under various loads.

Spring Constant

The spring constant, denoted as \( k \), is a measure of a spring's stiffness. A larger \( k \) value means the spring is stiffer and more resistant to deformation. - To determine \( k \), you can rearrange Hooke's Law: \( k = \frac{F}{x} \)In our exercise, we have:1. Force exerted by the fish: \( 637 \text{ N} \)2. Displacement of the spring with the fish hanging: \( 0.180 \text{ m} \)Substitute these values to find \( k \):- \( k = \frac{637 \text{ N}}{0.180 \text{ m}} \approx 3538.89 \text{ N/m} \)This calculation provides the spring constant, helping us understand how much force the spring can handle for a given stretch. Understanding the spring constant allows us to further study the spring's oscillations and other properties.

Oscillation Period

The oscillation period \( T \) is the time it takes for an object to complete one full cycle of motion in a simple harmonic oscillator, such as a mass-spring system. The formula to determine the period is:- \( T = 2\pi \sqrt{\frac{m}{k}} \)For our problem:

• Mass \( m = 65.0 \text{ kg} \)
• Spring constant \( k = 3538.89 \text{ N/m} \)

Substituting, we have:- \( T = 2\pi \sqrt{\frac{65.0}{3538.89}} \approx 0.54 \text{ s} \)This means the fish will complete one full oscillation cycle, from starting point back to starting point, in approximately 0.54 seconds. This concept is essential for predicting the behavior of oscillating objects in simple harmonic motion environments.

Maximum Speed

Maximum speed in simple harmonic motion refers to the highest velocity an oscillating object reaches as it moves between extremes of its path. This speed occurs as it passes through its equilibrium position. To find the maximum speed \( v_{max} \), use the equation:- \( v_{max} = \omega A \)where:

• \( \omega \) is the angular frequency, \( \omega = \sqrt{\frac{k}{m}} \)
• \( A \) is the amplitude or maximum displacement from equilibrium

Given:- Calculated \( \omega = 7.383 \text{ rad/s} \)- Amplitude \( A = 0.05 \text{ m} \)We find:- \( v_{max} = 7.383 \text{ rad/s} \times 0.05 \text{ m} \approx 0.369 \text{ m/s} \)Understanding maximum speed helps in predicting how fast an oscillating object moves at its midpoint. It also indicates kinetic energy peaks, crucial for energy-based calculations.