Problem 19 - University Physics with Modern Physics

Question

A 1.50 -kg mass on a spring has displacement as a function of time given by the equation $$x(t)=(7.40 \mathrm{cm}) \cos \left[\left(4.16 \mathrm{s}^{-1}\right) t-2.42\right]$$ Find (a) the time for one complete vibration; (b) the force constant of the spring; (c) the maximum speed of the mass; (d) the maximum force on the mass; (e) the position, speed, and acceleration of the mass at \(t=1.00 \mathrm{s} ;\) (f) the force on the mass at that time.

Step-by-Step Solution

Answer

(a) 1.51 s; (b) 25.97 N/m; (c) 0.308 m/s; (d) 1.922 N; (e) x ≈ 0.0185 m, v ≈ -0.240 m/s, a ≈ -0.321 m/s²; (f) F ≈ -0.481 N.

Step 1: Determine the Time for One Complete Vibration

The time for one complete vibration is the period of the oscillation. The angular frequency \(\omega\) is given in the cosine function as \(4.16 \, \text{s}^{-1}\). The period \(T\) is found using the formula \(T = \frac{2\pi}{\omega}\).\[T = \frac{2\pi}{4.16 \, \text{s}^{-1}} \approx 1.51 \, \text{s}\]

Step 2: Calculate the Force Constant of the Spring

The force constant \(k\) is related to the mass \(m\) and angular frequency \(\omega\) by the formula \(\omega = \sqrt{\frac{k}{m}}\). Rearranging gives:\[k = m\omega^2 = (1.50 \, \text{kg}) \times (4.16 \, \text{s}^{-1})^2 = 25.97 \, \text{N/m}\]

Step 3: Find the Maximum Speed of the Mass

The maximum speed \(v_{max}\) for a simple harmonic oscillator occurs when the displacement is zero and is given by \(v_{max} = \omega A\), where \(A\) is the amplitude:\[v_{max} = 4.16 \, \text{s}^{-1} \times 0.074 \, \text{m} = 0.308 \, \text{m/s}\]

Step 4: Determine the Maximum Force on the Mass

The maximum force \(F_{max}\) is calculated using \(F_{max} = kA\):\[F_{max} = 25.97 \, \text{N/m} \times 0.074 \, \text{m} = 1.922 \, \text{N}\]

Step 5: Evaluate the Position at t = 1.00 s

Substitute \(t = 1.00 \, \text{s}\) into the displacement function to find position \(x(1.00)\):\[x(t) = 0.074 \, \text{m} \cos(4.16 \, \text{s}^{-1} \times 1.00 \, \text{s} - 2.42)\]Computing gives:\[x(1.00) = 0.074 \, \text{m} \cos(1.74) \approx 0.0185 \, \text{m}\]

Step 6: Compute the Speed at t = 1.00 s

The speed \(v(t)\) is obtained by differentiating the displacement function with respect to time and evaluating at \(t = 1.00\) s:\[v(t) = -0.074 \, \text{m} \cdot 4.16 \, \text{s}^{-1} \sin(4.16 \, \text{s}^{-1} \times 1.00 \, \text{s} - 2.42)\]Evaluating gives:\[v(1.00) \approx -0.240 \, \text{m/s}\]

Step 7: Calculate the Acceleration at t = 1.00 s

To find acceleration \(a(t)\), differentiate the velocity function with respect to time and evaluate at \(t = 1.00\) s:\[a(t) = -0.074 \, \text{m} \cdot (4.16 \, \text{s}^{-1})^2 \cos(4.16 \, \text{s}^{-1} \times 1.00 \, \text{s} - 2.42)\]Evaluating:\[a(1.00) \approx -0.321 \, \text{m/s}^2\]

Step 8: Calculate the Force on the Mass at t = 1.00 s

The force \(F(t)\) is given by Hooke's Law \(F(t) = -kx(t)\), where \(x(t)\) is the position at \(t\):\[F(1.00) = -25.97 \, \text{N/m} \times 0.0185 \, \text{m} \approx -0.481 \, \text{N}\]

Key Concepts

Spring Constant

The spring constant, often denoted as the letter \(k\), is a fundamental concept when dealing with harmonic motion. It represents the stiffness of a spring. A higher spring constant indicates a stiffer spring, while a lower spring constant suggests a more flexible spring. The formula for calculating the spring constant is:

• \(k = \frac{F}{x}\)

where \(F\) is the force applied and \(x\) is the displacement of the spring from its equilibrium position. This formula shows us how much force is needed to extend or compress the spring by a unit length.
In our problem, we use the relationship from harmonic motion, \(\omega = \sqrt{\frac{k}{m}}\), to find \(k\). By rearranging this formula, we can write:

• \(k = m\omega^2\)

This is useful when you know the mass \(m\) and the angular frequency \(\omega\). Understanding the spring constant helps us comprehend how forces influence the motion of objects attached to springs.

Angular Frequency

Angular frequency, symbolized by \(\omega\), is a measure of how quickly an object undergoes oscillations in simple harmonic motion. It is related to the frequency and the period of the oscillations. The formula for angular frequency is:

• \(\omega = 2\pi f\)

or, alternatively:

• \(\omega = \frac{2\pi}{T}\)

where \(f\) is the frequency and \(T\) is the period of the motion.
Angular frequency is measured in radians per second and gives us a rotational perspective on oscillation rates. It tells us how many radians (a full circle is \(2\pi\) radians) the system covers per second. By knowing \(\omega\), we can further calculate the period \(T\) of the mass on a spring using:

• \(T = \frac{2\pi}{\omega}\)

This helps to determine how long it takes for the system to complete one full cycle of motion.

Maximum Speed

In simple harmonic motion, the maximum speed of an object is an important quantity, representing the highest velocity the object reaches as it moves back and forth. This maximum speed occurs when the object passes through its equilibrium position, where potential energy is minimal, and kinetic energy is maximal.
The maximum speed \(v_{max}\) of an object in harmonic motion is calculated through:

• \(v_{max} = \omega A\)

Here, \(\omega\) denotes the angular frequency, and \(A\) is the amplitude, which is the maximum displacement from the equilibrium position.
This formula shows us that the maximum speed is directly proportional to both the angular frequency of the oscillation and the amplitude of the motion. Understanding this concept helps us predict how fast the object will move at its fastest point in motion.

Hooke's Law

Hooke's Law is a principle of physics that describes the behavior of springs and is crucial in understanding harmonic motion. It states that the force \(F\) needed to extend or compress a spring is proportional to the distance \(x\) it is stretched or compressed from its rest position:

• \(F = -kx\)

In this equation, \(k\) is the spring constant, and the negative sign indicates that the force exerted by the spring is in the opposite direction of the displacement.
This law is fundamental for analyzing the forces in systems undergoing simple harmonic motion. It helps provide a basis for calculating how an object will behave when attached to a spring. By using Hooke's Law, one can determine how a spring will respond to external forces, further predicting motion dynamics for systems like that of a mass attached to a spring. This law allows us to calculate the force at any point in time, such as at \(t=1.00 \, \text{s}\) in our exercise, by knowing the spring constant and displacement.