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A proud deep-sea fisherman hangs a \(65.0-\mathrm{kg}\) fish from an ideal spring having negligible mass. The fish stretches the spring 0.120 \(\mathrm{m}\) m. Find the force constant of the spring. The fish is now pulled down 5.00 \(\mathrm{cm}\) and released. (b) What is the period of oscillation of the fish? (c) What is the maximum speed it will reach?

Short Answer

Expert verified
(a) \(k = 5313.75 \text{ N/m}\); (b) \(T = 0.694 \text{ s}\); (c) \(v_{max} = 0.452 \text{ m/s}\).

Step by step solution

01

Define the Problem for Part (a)

We need to find the force constant, or the spring constant \(k\), of the spring. According to Hooke's Law, the force \(F\) exerted by the spring is directly proportional to the displacement \(x\) from its equilibrium position: \[ F = kx \] The force exerted by the fish on the spring is equal to its weight, which can be calculated using the formula: \( F = mg \), where \( m = 65.0 \text{ kg} \) and gravitational acceleration \( g = 9.81 \text{ m/s}^2 \). The displacement \(x\) is \(0.120 \text{ m}\).
02

Calculate the Force Constant (Part a)

The weight of the fish (force) is: \[ F = mg = 65.0 \times 9.81 = 637.65 \text{ N} \] Substitute into Hooke's Law to solve for \(k\): \[ 637.65 = k \times 0.120 \] \[ k = \frac{637.65}{0.120} \approx 5313.75 \text{ N/m} \]
03

Define the Problem for Part (b)

We need to find the period \(T\) of oscillation of the fish. The period of a simple harmonic oscillator like this spring-mass system can be calculated using the formula: \[ T = 2\pi \sqrt{\frac{m}{k}} \] where \(m = 65.0 \, \mathrm{kg}\) is the mass of the fish and \(k = 5313.75 \, \mathrm{N/m}\) is the force constant we calculated.
04

Calculate the Period of Oscillation (Part b)

Substitute the known values into the period formula: \[ T = 2\pi \sqrt{\frac{65.0}{5313.75}} \] \[ T \approx 2\pi \sqrt{0.01223} \] \[ T \approx 2\pi \times 0.1106 \] \[ T \approx 0.694 \text{ seconds} \]
05

Define the Problem for Part (c)

To find the maximum speed of the fish, we first note that the maximum speed occurs at the equilibrium position where kinetic energy is at its maximum. The maximum speed \(v_{max}\) can be found from the maximum potential energy stored in the spring being converted to kinetic energy. Use \( v_{max} = \omega A \), where \( \omega = \sqrt{\frac{k}{m}} \) is the angular frequency and \( A = 0.050 \text{ m} \) is the amplitude of displacement.
06

Calculate the Maximum Speed (Part c)

Calculate the angular frequency: \[ \omega = \sqrt{\frac{5313.75}{65.0}} \approx 9.04 \text{ rad/s} \] Then find the maximum speed: \[ v_{max} = \omega A = 9.04 \times 0.050 = 0.452 \text{ m/s} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Spring Constant
The spring constant, often denoted as \( k \), is a measure of a spring's stiffness. According to Hooke's Law, the force \( F \) exerted by a spring is directly proportional to the displacement \( x \) it experiences from its equilibrium position. This relationship is expressed by the equation \( F = kx \). If the spring is not very stiff, \( k \) will be smaller, indicating that a given force will cause a larger displacement.
The spring constant is crucial in determining how a spring behaves when a force is applied to it. In the context of the exercise, we calculated \( k \) by considering the weight of the fish, which acts as the force stretching the spring. By rearranging Hooke's Law, we found that \( k = \frac{F}{x} \), where \( F \) is the gravitational force (\( mg \)) and \( x \) is the displacement caused by this force. This gives us information about how much force is required to compress or extend a spring by a specific distance. In this example, the spring constant was determined to be approximately \( 5313.75 \text{ N/m} \), indicating a fairly stiff spring suitable for measuring larger weights.
Simple Harmonic Motion
Simple Harmonic Motion (SHM) describes the oscillatory motion that repeats itself in a regular cycle. This type of motion is characterized by the restoring force being directly proportional to the displacement and acting in the opposite direction.
Consider a mass-spring system, like in the exercise. When the fish is pulled slightly from its equilibrium position and released, it oscillates up and down. The springs have this tendency because their force is always directed towards the equilibrium position, making the motion periodic. SHM can be described mathematically with the equation for the period \( T \), the time for one complete cycle. It is given by \[ T = 2\pi \sqrt{\frac{m}{k}} \] where \( m \) is the mass and \( k \) is the spring constant.
In our exercise, we found the period of the oscillation to be approximately \( 0.694 \) seconds. This tells us how long it takes for one full up-and-down oscillation of the fish, which is due to the natural properties of the mass-spring system.
Angular Frequency
Angular frequency, often symbolized by \( \omega \), relates to how rapidly something oscillates in simple harmonic motion. It is measured in radians per second and offers a way to understand the rate of motion in a circular or oscillatory system.
The angular frequency is related to the spring constant \( k \) and mass \( m \) by the equation \( \omega = \sqrt{\frac{k}{m}} \). It essentially captures the frequency of oscillations in terms of radians, rather than cycles, which is particularly useful in various physics and engineering contexts.
In the given problem, the fish's oscillation frequency helps predict how often the maximum speed occurs at the equilibrium point. We calculated \( \omega \) to be about \( 9.04 \text{ rad/s} \). Knowing this allows us to find the maximum speed of the fish using \( v_{max} = \omega A \), where \( A \) is the amplitude of motion. Understanding \( \omega \) gives insights into not just how fast but also how the motion is orchestrated in periodic systems like springs.

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Most popular questions from this chapter

If an object on a horizontal, frictionless surface is attached to a spring, displaced, and then released, it will oscillate. If it is displaced 0.120 \(\mathrm{m}\) from its equilibrium position and released with zero initial speed, then after 0.800 \(\mathrm{s}\) its displacement is found to be 0.120 \(\mathrm{m}\) on the opposite side, and it has passed the equilibrium position once during this interval. Find (a) the amplitude; (b) the period; (c) the frequency.

On a horizontal, frictionless table, an open-topped \(5.20-\mathrm{kg}\) box is attached to an ideal horizontal spring having force constant 375 \(\mathrm{N} / \mathrm{m}\) . Inside the box is a 3.44 \(\mathrm{kg}\) stone. The system is oscillaring with an amplitude of 7.50 \(\mathrm{cm} .\) When the box has reached its maximum speed, the stone is suddenly plucked vertically out of the box without touching the box. Find (a) the period and (b) the amplitude of the resulting motion of the box. (c) Without doing any calculations, is the new period greater or smaller than the original period? How do you know?

A \(40.0 .\) N force strethes a vertical spring 0.250 \(\mathrm{m}\) . (a) What mass must be suspended from the spring so that the system will oscillate with a period of 1.00 \(\mathrm{s} ?(6)\) If the amplitude of the motion is 0.050 \(\mathrm{m}\) and the period is that specified in part (a), where is the object and in what direction is it moving 0.35 \(\mathrm{s}\) after it has passed the equilibrium position, moving downward? (c) What force (magnitude and direction) does the spring exert on the object when it is 0.030 \(\mathrm{m}\) below the equilibrium position, moving upward?

An \(85.0-\mathrm{kg}\) mountain climber plans to swing down, starting from rest, from a ledge using a light rope 6.50 \(\mathrm{m}\) long. He holds one end of the rope, and the other end is tied higher up on a rock face. Since the ledge is not very far from the rock face, the rope makes a small angle with the vertical. At the lowest point of his swing, he plans to let go and drop a short distance to the ground. (a) How long after he begins his swing will the climber first reach his lowest point? (b) If he missed the first chance to drop off, how long after first beginning his swing will the climber reach his lowest point for the second time?

A 0.500 \(\mathrm{kg}\) glider, attached to the end of an ideal spring with force constant \(k=450 \mathrm{N} / \mathrm{m}\) , undergoes SHM with an amplitude of 0.040 \(\mathrm{m}\) . Compute (a) the maximum speed of the glider; (b) the speed of the glider when it is at \(x=-0.015 \mathrm{m} ;\) (c) the magnitude of the maximum acceleration of the glider; (d) the acceleration of the glider at \(x=-0.015 \mathrm{m} ;\) (e) the total mechanical energy of the glider at any point in its motion.

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