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A proud deep-sea fisherman hangs a 65.0 -kg fish from an ideal spring having negligible mass. The fish stretches the spring 0.120 \(\mathrm{m}\) . (a) 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
Force constant: 5313.75 N/m; Period: 0.697 s; Max speed: 0.451 m/s.

Step by step solution

01

Define Variables and Given Data

1. The mass of the fish, \( m = 65.0 \) kg.2. The initial displacement of the spring, \( x = 0.120 \) m.3. The additional displacement, \( x_0 = 0.050 \) m (since 5.00 cm = 0.050 m).
02

Calculate the Force Constant of the Spring (Part a)

The force exerted by the fish due to its weight is given by \( F = mg \), where \( g = 9.81 \) m/s².\[F = 65.0 \times 9.81 = 637.65 \text{ N}\]According to Hooke's Law, \( F = kx \), where \( k \) is the force constant of the spring.Rearranging the formula gives \( k = \frac{F}{x} \).\[k = \frac{637.65}{0.120} = 5313.75 \text{ N/m}\]
03

Find the Period of Oscillation (Part b)

The period of oscillation for a mass-spring system is given by the formula \( T = 2\pi \sqrt{\frac{m}{k}} \).Substitute the values:\[T = 2\pi \sqrt{\frac{65.0}{5313.75}} \approx 2\pi \sqrt{0.01223} \approx 0.697 \text{ seconds}\]
04

Determine the Maximum Speed (Part c)

The maximum speed \( v_{\text{max}} \) in simple harmonic motion is given by \( v_{\text{max}} = \omega A \), where \( \omega = \sqrt{\frac{k}{m}} \) is the angular frequency and \( A \) is the amplitude.Calculate \( \omega \):\[\omega = \sqrt{\frac{5313.75}{65.0}} \approx 9.022 \text{ rad/s}\]The amplitude \( A = 0.050 \text{ m} \).\[v_{\text{max}} = 9.022 \times 0.050 \approx 0.451 \text{ m/s}\]
05

Compile the Solutions

The force constant of the spring is approximately 5313.75 N/m. The period of oscillation of the fish is approximately 0.697 seconds. The maximum speed the fish reaches is approximately 0.451 m/s.

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Key Concepts

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

Hooke's Law
Hooke's Law describes how springs work in terms of force and displacement. It's crucial for understanding simple harmonic motion. In this scenario, the fisherman hangs a fish, which causes the spring to stretch.
- The key relationship in Hooke's Law is:
\[ F = kx \]
Here, \( F \) is the force applied by the fish (due to gravity), \( k \) is the spring's force constant (also known as the spring constant), and \( x \) is the displacement from the spring's original position.
- How to calculate the force constant? You can rearrange the formula to solve for \( k \):
\[ k = \frac{F}{x} \]
For the fish, \( F \) is determined by multiplying the mass of the fish (65.0 kg) by gravitational acceleration (9.81 m/s²), resulting in 637.65 N. With a displacement \( x = 0.120 \) m, the force constant \( k \) comes out to be approximately 5313.75 N/m.
This tells us how stiff the spring is. A larger \( k \) means a stiffer spring, which requires more force to stretch by the same distance.
Oscillation Period
The period of oscillation, \( T \), is the time it takes to complete one full cycle of motion. For a spring-mass system, you find \( T \) using the following formula:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
This means that the period depends on both the mass (\( m \)) and the spring constant (\( k \)).
- Heavier objects or softer springs (lower \( k \)) will result in a longer period of oscillation.
- Conversely, lighter objects or stiffer springs (higher \( k \)) will lead to a shorter period.
In our fish scenario, after substituting the mass (65.0 kg) and the calculated spring constant (5313.75 N/m) into the formula, the oscillation period comes out to approximately 0.697 seconds.
This means every 0.697 seconds, the fish completes one full up-and-down motion in the spring. The idea is that this system repeats its motion predictably, governed by the spring's properties and the object's mass.
Angular Frequency
Angular frequency, \( \omega \), is a measure of how fast the object oscillates in simple harmonic motion. It's defined as:
\[ \omega = \sqrt{\frac{k}{m}} \]
- It offers an alternative way to think about how quickly the system oscillates, compared to the period.
- In rad/s, it describes how many radians the system moves through per second.
For the oscillating fish and spring, with a spring constant \( k \) of 5313.75 N/m and a fish mass \( m \) of 65.0 kg, the angular frequency calculates to approximately 9.022 rad/s.
This value helps further in determining dynamic aspects of the motion, such as the maximum speed, where:
\[ v_{\text{max}} = \omega A \]
Here, \( A \) is the amplitude of the oscillation. For our fish, pulled an additional 0.050 m down, the maximum speed comes out to approximately 0.451 m/s. Angular frequency gives insights into the pace and liveliness of oscillations, linking directly to energy and velocity in the system.

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Most popular questions from this chapter

In a physics lab, you attach a \(0.200-\mathrm{kg}\) air-track glider to the end of an ideal spring of negligible mass and start it oscillating. The elapsed time from when the glider first moves through the equilibrium point to the second time it moves through that point is 2.60 s. Find the spring's force constant.

CP Two uniform solid spheres, each with mass \(M=0.800 \mathrm{kg}\) and radius \(R=0.0800 \mathrm{m},\) are connected by a short, light rod that is along a diameter of each sphere and are at rest on a horizontal tabletop. A spring with force constant \(k=160 \mathrm{N} / \mathrm{m}\) has one end attached to the wall and the other end attached to a frictionless ring that passes over the rod at the center of mass of the spheres, which is midway between the centers of the two spheres. The spheres are each pulled the same distance from the wall, stretching the spring, and released. There is sufficient friction between the tabletop and the spheres for the spheres to roll without slipping as they move back and forth on the end of the spring. Show that the motion of the center of mass of the spheres is simple harmonic and calculate the period.

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the amplitude of the motion is \(0.090 \mathrm{m},\) it takes the block 2.70 s to travel from \(x=0.090 \mathrm{m}\) to \(x=-0.090 \mathrm{m} .\) If the amplitude is doubled, to \(0.180 \mathrm{m},\) how long does it take the block to travel (a) from \(x=0.180 \mathrm{m}\) to \(x=-0.180 \mathrm{m}\) and (b) from \(x=0.090 \mathrm{m}\) to \(x=-0.090 \mathrm{m} ?\)

A 2.50 -kg rock is attached at the end of a thin, very light rope 1.45 \(\mathrm{m}\) long. You start it swinging by releasing it when the rope makes an \(11^{\circ}\) angle with the vertical. You record the observation that it rises only to an angle of \(4.5^{\circ}\) with the vertical after 10\(\frac{1}{2}\) swings. (a) How much energy has this system lost during that time? (b) What happened to the "lost" energy? Explain how it could have been "lost."

An apple weighs 1.00 \(\mathrm{N}\) . When you hang it from the end of a long spring of force constant 1.50 \(\mathrm{N} / \mathrm{m}\) and negligible mass, it bounces up and down in SHM. If you stop the bouncing and let the apple swing from side to side through a small angle, the frequency of this simple pendulum is half the bounce frequency. (Because the angle is small, the back- and-forth swings do not cause any appreciable change in the length of the spring.) What is the unstretched length of the spring (with the apple removed?

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