Question

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

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

Step-by-step solution

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. \[ k = \frac{F}{x} = \frac{40.0 \text{ N}}{0.250 \text{ m}} = 160 \text{ N/m} \]

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:\[ m = \frac{(1.00)^2 \times 160}{4\pi^2} \approx 4.05 \text{ kg} \]

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)

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.

Key concepts

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.