Question

A \(450-\mathrm{gm}\) mass is placed upon the lower end of a coil spring suspended from the ceiling. The mass comes to rest in its equilibrium position, thereby stretching the spring \(5 \mathrm{~cm}\). The mass is then pulled down \(3 \mathrm{~cm}\) below its equilibrium position and released at \(t=0\) with an initial velocity of \(2 \mathrm{ft} / \mathrm{sec}\) directed downward. Find the amplitude, period, and frequency of the resulting motion.

Short answer

The resulting motion of the mass-spring system has an amplitude of 0.62 m, a period of approximately 0.45 seconds, and a frequency of approximately 2.22 Hz.

Step-by-step solution

1: Convert units to the SI system

First, we need to convert all units to the SI system for consistency: - Mass: 450 gm = 0.45 kg - Initial spring stretch: 5 cm = 0.05 m - Initial displacement: 3 cm = 0.03 m - Initial velocity: 2 ft/sec ≈ 0.61 m/s (since 1 ft ≈ 0.3048 m) Now we have: Mass: 0.45 kg Initial spring stretch: 0.05 m Initial displacement: 0.03 m Initial velocity: 0.61 m/s

2: Use Hooke's law to find the spring constant

Hooke's law states that the force exerted by a spring is proportional to its displacement from its equilibrium position. In mathematical terms, the spring force F is given by: \[F = -kx\] where x is the displacement from equilibrium position, k is the spring constant, and F is the force exerted by the spring. Since the mass is in its equilibrium position at the initial spring stretch of 0.05 m, we know that gravitational force is balanced by the spring force. We can use this information to find the spring constant k: \[mg = kx_{0}\] Solving for the spring constant k: \[k = \frac{mg}{x_0} = \frac{(0.45)(9.81)}{0.05} = 88.47 \text{ N/m}\]

3: Find the angular frequency

We know that the angular frequency ω is given by the square root of the ratio of the spring constant k to the mass m: \[\omega = \sqrt{\frac{k}{m}}\] Plugging in the values we found for k and m: \[\omega = \sqrt{\frac{88.47}{0.45}} = 14.02 \text{ rad/s}\]

4: Write the general solution and solve for the constants

The general solution for the motion of a mass-spring system is given by: \[x(t) = A\cos(\omega t - \phi)\] At t=0, we have x(0)=0.03 m and v(0)=0.61 m/s. Plugging these into the position and velocity equations, we get: \[0.03 = A\cos(-\phi)\] \[0.61 = -14.02A\sin(-\phi)\] To find A and \(\phi\), we will solve this system of equations. Square both equations and add them together: \[(0.03)^2 + (0.61)^2 = A^2\cos^2(-\phi) + A^2\sin^2(-\phi)\] Using the identity \(\cos^2(x) + \sin^2(x) = 1\), we get: \[(0.03)^2 + (0.61)^2 = A^2\] Calculate the amplitude A: \[A = \sqrt{(0.03)^2 + (0.61)^2} = 0.62 \text{ m}\] Next, using the position equation, find the phase angle \(\phi\): \[\phi = \arccos\left(\frac{0.03}{0.62}\right) = -1.54 \text{ rad}\] Now we have the amplitude and phase angle, which are A = 0.62 m and \(\phi = -1.54\) rad.

5: Calculate the period and frequency

Finally, we need to find the period T and the frequency f. The period is given by the inverse of the angular frequency: \[T = \frac{2\pi}{\omega} = \frac{2\pi}{14.02} \approx 0.45 \text{ s}\] Then to calculate the frequency f, we take the reciprocal of the period T: \[f = \frac{1}{T} = \frac{1}{0.45} \approx 2.22 \text{ Hz}\] #Conclusion# The amplitude, period, and frequency of the resulting motion are: Amplitude: \(A = 0.62 \text{ m}\) Period: \(T \approx 0.45 \text{ s}\) Frequency: \(f \approx 2.22 \text{ Hz}\)

Key concepts

Hooke's Law

Hooke's Law is a fundamental principle in the study of mechanics, particularly within the context of mass-spring systems. It describes a linear relationship between the force exerted by a spring and the displacement caused by that force. The law is often articulated as:
\[ F = -kx \]
where \( F \) represents the force applied by the spring, \( k \) is the spring constant (a measure of the stiffness of the spring), and \( x \) is the displacement of the spring from its equilibrium position. The negative sign reflects that the force exerted by the spring is in the opposite direction of the displacement, embodying the restorative nature of spring force.
Understanding Hooke's Law is vital for determining the behavior of a mass when attached to a spring since it allows us to find the spring constant. This constant is crucial for predicting how the spring will act under varying conditions of force and displacement.

Spring Constant

The spring constant, denoted by \( k \), is a measure of the stiffness of a spring. It essentially determines how much force is required to stretch or compress the spring by a certain distance. Higher values of \( k \) indicate a stiffer spring that resists deformation, while lower values suggest a spring that’s easier to stretch or compress.
To calculate the spring constant, you can rearrange Hooke's Law:
\[ k = \frac{F}{x} \]
In practical situations, by measuring the force exerted at a known displacement, the spring constant can be determined. This characteristic of a spring is foundational to solving problems related to mass-spring oscillator systems, as it influences the system’s period and frequency.

Angular Frequency

Angular frequency, represented by \( \omega \), is a measure of how rapidly an oscillating system undergoes its cycles, expressed in radians per second. It is specifically important in the context of harmonic motion, such as that of a mass on a spring.
The formula for calculating angular frequency in a mass-spring system is:
\[ \omega = \sqrt{\frac{k}{m}} \]
Here, \( k \) is the spring constant and \( m \) is the mass attached to the spring. Angular frequency provides a way to determine other characteristics of an oscillator, such as its amplitude, period, and frequency, and is related to the energy within the system.

Amplitude

The amplitude of an oscillating system is the maximum displacement from its equilibrium position. It represents one-half of the total distance covered in one cycle of movement and is a measure of the energy in the system—the greater the amplitude, the more energy the system possesses.
In a mass-spring system after being displaced and released, the system will oscillate around the equilibrium position. The amplitude is not affected by the mass or spring constant but is determined by the initial conditions of the system, such as initial displacement and initial velocity. The amplitude remains constant as long as the system is isolated from external forces or damping.

Period

The period of an oscillating system, denoted by \( T \), is the duration it takes to complete one full cycle of motion. For a mass-spring system, the period reflects how long it takes for the mass to return to its starting point after being displaced.
You can find the period with the formula:
\[ T = \frac{2\pi}{\omega} \]
The period is inversely proportional to the angular frequency, meaning that a higher angular frequency implies a shorter period. The period is independent of the amplitude and depends on the mass and spring constant of the system. Knowing the period is essential for predicting the timing of oscillatory events and is part of the foundation for understanding wave phenomena.

Frequency

Frequency, symbolized by \( f \), is the number of oscillatory cycles that occur in one second, expressed in Hertz (Hz). It is the reciprocal of the period, thus given by the formula:
\[ f = \frac{1}{T} \]
This measurement is indicative of how fast an object is oscillating. In a mass-spring system, the frequency can be used to describe how often the mass passes a particular point in its path. It's related to the angular frequency by the equation:
\[ f = \frac{\omega}{2\pi} \]
Frequency plays a significant role in many areas of physics, from the analysis of circuits to the study of sound waves, making it a key concept in understanding oscillatory systems.