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A 40.0-N force stretches a vertical spring 0.250 m. (a) What mass must be suspended from the spring so that the system will oscillate with a period of 1.00 s? (b) If the amplitude of the motion is 0.050 m and the period is that specified in part (a), where is the object and in what direction is it moving 0.35 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 m below the equilibrium position, moving upward?

Short Answer

Expert verified
(a) Mass: 4.05 kg. (b) 0.048 m below equilibrium, moving downward. (c) Spring exerts 4.8 N upward.

Step by step solution

01

Determine the Spring Constant

The spring constant \( k \) can be determined using Hooke's Law, which states that \( F = kx \), where \( F \) is the force applied and \( x \) is the displacement. Here, \( F = 40.0 \) N and \( x = 0.250 \) m. \[ k = \frac{F}{x} = \frac{40.0 \text{ N}}{0.250 \text{ m}} = 160 \text{ N/m} \]
02

Calculate the Mass for a Period of 1.00 s

The period \( T \) of a mass-spring system is given by \( T = 2\pi \sqrt{\frac{m}{k}} \). To find \( m \), rearrange this formula: \[ m = \frac{T^2 k}{4\pi^2} \]Substituting \( T = 1.00 \) s and \( k = 160 \text{ N/m} \), we get:\[ m = \frac{(1.00)^2 \times 160}{4\pi^2} \approx 4.05 \text{ kg} \]
03

Analyze the Position and Direction

To find the position and direction at \( t = 0.35 \) s, use the formula for position in simple harmonic motion, \( x(t) = A \cos(\omega t + \phi) \), where \( \omega = \frac{2\pi}{T} \), and \( \phi \) is the phase constant. Since it passes equilibrium moving downward at \( t = 0 \), \( x(0) = 0 \) and starting case is \( \sin \) shift:\[ x(t) = A \sin(\omega t) \]Calculate \( \omega = \frac{2\pi}{1.00} = 2\pi \) rad/s. Substitute:\[ x(0.35) = 0.05 \sin(2\pi \times 0.35) \approx -0.048 \text{ m} \] (moving downward)
04

Calculate the Spring Force at 0.030 m

The force exerted by the spring is given by \( F = kx \), where \( x \) is the displacement from the equilibrium position. Here, \( x = 0.030 \) m and \( k = 160 \text{ N/m} \):\[ F = 160 \times 0.030 = 4.8 \text{ N} \]Since the object is below equilibrium and moving upward, the spring force is upwards.

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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 represented by the letter \( k \), is an essential part of understanding how springs work in various physical systems. It tells us how stiff a spring is.
When you apply a force to a spring, it either compresses or stretches, depending on the direction of the force. The spring constant measures how resistant the spring is to this change in length.
In simple words, the larger the spring constant, the stiffer the spring, and the harder you need to push or pull to change its length. In the formula
  • \( F = kx \)
\( F \) represents the force applied to the spring, \( x \) is the displacement from the spring's original length, and \( k \) is the spring constant.
By understanding the spring constant, we can also predict how much a spring will stretch or compress when a force is applied. It's a key component to solving many physics problems involving springs.
Hooke's Law
Hooke's Law defines the relationship between the force exerted on a spring and the displacement it causes. The law is expressed as:
  • \( F = kx \)
where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the change in length of the spring.
Essentially, Hooke's Law tells us that the force required to compress or extend a spring is proportional to the distance it is stretched or compressed, so long as the spring is not stretched beyond its elastic limit.
This principle is key in tasks like calculating how much weight a spring can hold before it deforms permanently.
In practical terms, Hooke's Law is so useful because it simplifies complex physical dynamics into an easy equation, helping us design everything from mattresses to vehicle suspension systems.
Mass-Spring System
A mass-spring system is a common model in physics used to study oscillatory and vibratory systems. It consists of a mass attached to a spring that can move back and forth.
When the mass is displaced from its equilibrium position, the spring exerts a force to return it to equilibrium, causing the system to oscillate. The motion is a classic example of simple harmonic motion, where the force is proportional to the displacement and directed towards the equilibrium position.
In this type of system, two factors are critical:
  • the mass attached to the spring
  • the stiffness of the spring, or the spring constant \( k \).
These determine how the system oscillates, including the frequency and the period of the oscillation. As such, the mass-spring system is an ideal model for studying physical vibrations and resonates with many real-world applications.
Period of Oscillation
The period of oscillation, designated as \( T \), is the time it takes for a complete cycle of motion in a simple harmonic oscillator, such as a mass-spring system.
  • In the context of a mass-spring system, the period \( T \) is given by the formula
  • \( T = 2\pi \sqrt{\frac{m}{k}} \)
Here, \( m \) represents the mass, and \( k \) is the spring constant.
This formula shows us that the period of oscillation depends on both the mass and the stiffness of the spring:
  • A heavier mass takes longer to complete an oscillation, increasing the period.
  • A stiffer spring (with a higher spring constant) results in a shorter period, as the system oscillates faster.
Understanding the period of oscillation helps us design systems that must operate at particular frequencies, such as clocks, musical instruments, and various types of sensors.

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Most popular questions from this chapter

A 175-g glider on a horizontal, frictionless air track is attached to a fixed ideal spring with force constant 155 N/m. At the instant you make measurements on the glider, it is moving at 0.815 m/s and is 3.00 cm from its equilibrium point. Use \({energy\ conservation}\) to find (a) the amplitude of the motion and (b) the maximum speed of the glider. (c) What is the angular frequency of the oscillations?

An object is undergoing \(\textbf{SHM}\) with period 0.900 s and amplitude 0.320 m. At \(t =\) 0 the object is at \(x =\) 0.320 m and is instantaneously at rest. Calculate the time it takes the object to go (a) from \(x =\) 0.320 m to \(x =\) 0.160 m and (b) from \(x =\) 0.160 m to \(x =\) 0.

The balance wheel of a watch vibrates with an angular amplitude \(\Theta\), angular frequency \(\omega\), and phase angle \(\phi =\) 0. (a) Find expressions for the angular velocity \(d\theta/dt\) and angular acceleration \(d^2\theta/dt^2\) as functions of time. (b) Find the balance wheel's angular velocity and angular acceleration when its angular displacement is \(\Theta\), and when its angular displacement is \(\Theta\)/2 and \(\theta\) is decreasing. (\(Hint\): Sketch a graph of \(\theta\) versus \(t\).)

You want to find the moment of inertia of a complicated machine part about an axis through its center of mass. You suspend it from a wire along this axis. The wire has a torsion constant of 0.450 N \(\cdot\) m/rad. You twist the part a small amount about this axis and let it go, timing 165 oscillations in 265 s. What is its moment of inertia?

An object with height \(h\), mass \(M\), and a uniform cross-sectional area \(A\) floats upright in a liquid with density \(\rho\). (a) Calculate the vertical distance from the surface of the liquid to the bottom of the floating object at equilibrium. (b) A downward force with magnitude \(F\) is applied to the top of the object. At the new equilibrium position, how much farther below the surface of the liquid is the bottom of the object than it was in part (a)? (Assume that some of the object remains above the surface of the liquid.) (c) Your result in part (b) shows that if the force is suddenly removed, the object will oscillate up and down in SHM. Calculate the period of this motion in terms of the density \(\rho\) of the liquid, the mass \(M\), and the cross- sectional area A of the object. You can ignore the damping due to fluid friction (see Section 14.7).

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