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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 guitar string vibrates at a frequency of 440 Hz. A point at its center moves in SHM with an amplitude of 3.0 mm and a phase angle of zero. (a) Write an equation for the position of the center of the string as a function of time. (b) What are the maximum values of the magnitudes of the velocity and acceleration of the center of the string? (c) The derivative of the acceleration with respect to time is a quantity called the \(jerk\). Write an equation for the jerk of the center of the string as a function of time, and find the maximum value of the magnitude of the jerk.

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. The amplitude of the motion is 0.250 m and the period is 3.20 s. What are the speed and acceleration of the block when \(x =\) 0.160 m?

A 1.50-kg, horizontal, uniform tray is attached to a vertical ideal spring of force constant 185 N/m and a 275-g metal ball is in the tray. The spring is below the tray, so it can oscillate up and down. The tray is then pushed down to point \(A\), which is 15.0 cm below the equilibrium point, and released from rest. (a) How high above point \(A\) will the tray be when the metal ball leaves the tray? (\(Hint\): This does \(not\) occur when the ball and tray reach their maximum speeds.) (b) How much time elapses between releasing the system at point \(A\) and the ball leaving the tray? (c) How fast is the ball moving just as it leaves the tray?

A small block is attached to an ideal spring and is moving in SHM on a horizontal frictionless surface. The amplitude of the motion is 0.165 m. The maximum speed of the block is 3.90 m/s. What is the maximum magnitude of the acceleration of the block?

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the block is at \(x =\) 0.280 m, the acceleration of the block is \(-\)5.30 m/s\(^2\). What is the frequency of the motion?

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