Problem 88 - University Physics with Modern Physics

Question

A 40.0 -N force stretches a vertical spring 0.250 \(\mathrm{m}\) . (a) What mass must be suspended from the spring so that the system will oscillare with a period of 1.00 \(\mathrm{s} ?\) (b) 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 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?

Step-by-Step Solution

Answer

(a) 4.05 kg; (b) Moving down, x ≈ -0.034 m; (c) 4.8 N upwards.

Step 1: Determine Spring Constant using Hooke's Law

The spring constant (k) can be found using Hooke's law, which states that the force (F) is proportional to the extension (x) of the spring: \[ F = kx \]Given: F = 40.0 \text{ N} x = 0.250 \text{ m}So, \[ k = \frac{F}{x} = \frac{40.0}{0.250} = 160 \text{ N/m} \]

Step 2: Calculate Mass for Desired Period

The period (T) of oscillation for a spring-mass system is given by: \[ T = 2\pi \sqrt{\frac{m}{k}} \]We rearrange to find the mass (m): \[ m = \frac{T^2 k}{4\pi^2} \]Given: T = 1.00 \text{ s} k = 160 \text{ N/m}Substitute these values:\[ m = \frac{(1.00)^2 \times 160}{4\times (\pi)^2} = \frac{160}{39.478} = 4.05 \text{ kg} \]

Step 3: Analyze Position and Velocity for Motion

The position as a function of time is given by:\[ x(t) = A\cos\left(\frac{2\pi t}{T}\right) \]The velocity is the derivative of position with respect to time, \[ v(t) = -A \frac{2\pi}{T} \sin\left(\frac{2\pi t}{T}\right) \]Given: A = 0.050 \text{ m} T = 1.00 \text{ s} t = 0.35 \text{ s}\[ x(0.35) = 0.050 \cos\left(\frac{2\pi \times 0.35}{1.00}\right) = 0.050 \cos(2.2) \approx -0.034 \text{ m} \]\[ v(0.35) = -0.050 \times 2\pi \times \sin(2.2) \approx -0.314 \text{ m/s}\]Since velocity is negative, the object is moving downward.

Step 4: Calculate Force Exerted by the Spring

The spring force when the object is below equilibrium is given by Hooke’s law: \[ F = kx \]Given that the object is 0.030 m below equilibrium: \[ F = 160 \times (-0.030) = -4.8 \text{ N} \]Since the object is moving upward, by Newton’s third law, the spring exerts an equal and opposite force, 4.8 N upwards.

Key Concepts

Hooke's Law

Hooke's Law describes how springs behave when forces are applied to them. It states that the force needed to extend or compress a spring by a distance is proportional to that distance. Mathematically, this can be written as:

• \[ F = kx \]

Here,

• F is the force applied to the spring,
• k is the spring constant, which measures the stiffness of the spring,
• x is the displacement or change in length of the spring from its rest position.

Hooke's Law holds true only within the elastic limit of a spring. If you stretch the spring too much, it may permanently deform. Thus, this law is pivotal in problems involving springs, as it allows us to calculate how much force a spring will exert when displaced by a specific amount.

Spring Constant Calculation

The spring constant, denoted as \( k \), is a measure of a spring's stiffness. A higher spring constant means a stiffer spring. Calculating the spring constant is fundamental for understanding how a spring behaves under force.
To calculate the spring constant, we use Hooke’s Law. For example, given a force \( F = 40.0 \, \text{N} \) and a displacement \( x = 0.250 \, \text{m} \), the spring constant is calculated as:

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

This means that for every meter the spring is stretched, it will exert a 160 N force. Calculating \( k \) is a foundational step, essential for solving further motion-related aspects of spring systems.

Period of Oscillation

The period of oscillation of a mass-spring system is the time it takes to complete one full back-and-forth cycle of motion. It is a crucial concept when examining how these systems behave over time. The formula to determine the period \( T \) is:

• \[ T = 2\pi \sqrt{\frac{m}{k}} \]

Where

• m is the mass attached to the spring, and
• k is the spring constant.

This formula shows that the period of oscillation depends on the mass and the spring constant. For instance, a larger mass will result in a longer period, indicating that it takes more time for a heavier object to complete one cycle of oscillation.

Solving for a system with \( T = 1.00 \, \text{s} \) and \( k = 160 \, \text{N/m} \), we calculated the mass \( m \) to be approximately 4.05 kg. This shows that altering either the spring constant or the mass affects the oscillation period.

Mass-Spring System

A mass-spring system is a foundational model in physics that showcases simple harmonic motion (SHM). It consists of a mass attached to a spring, which when displaced, brings about oscillatory motion. Such systems are characterized by their back-and-forth motion around an equilibrium position.

Key aspects of a mass-spring system include:

• Equilibrium Position: Where the spring is at rest, and no net force acts on the mass.
• Amplitude: Maximum displacement from the equilibrium position. In our case, the amplitude \( A \) is 0.050 m.
• Velocity and Position: They change sinusoidally over time, affecting how the system oscillates.
• Forces: The spring exerts a restoring force that pulls the mass toward the equilibrium position.

Using functions like \( x(t) = A \cos\left(\frac{2\pi t}{T}\right) \) and its derivative for velocity, the behavior of this system over time can be analyzed extensively to predict the movement and forces at different positions during the oscillation.