Problem 40 - College Physics

Question

An object of unknown mass is attached to an ideal spring with force constant \(120 \mathrm{~N} / \mathrm{m}\) and is found to vibrate with a frequency of \(6.00 \mathrm{~Hz}\). Find (a) the period, (b) the angular frequency, and (c) the mass of this object.

Step-by-Step Solution

Answer

(a) 0.167 seconds, (b) 37.70 rad/s, (c) 0.084 kg.

Step 1: Calculate the Period

The frequency of vibration is given as \(6.00\ \text{Hz}\). The period \(T\) is the reciprocal of the frequency. Use the formula: \[ T = \frac{1}{f} \] Substituting the given frequency: \[ T = \frac{1}{6.00} \approx 0.167\ \text{seconds} \]

Step 2: Calculate the Angular Frequency

The angular frequency \(\omega\) is related to the frequency by the formula: \[ \omega = 2\pi f \] Substituting the frequency: \[ \omega = 2\pi \times 6.00\ \approx 37.70\ \text{rad/s} \]

Step 3: Calculate the Mass of the Object

The formula relating angular frequency \(\omega\), spring constant \(k\), and mass \(m\) is: \[ \omega = \sqrt{\frac{k}{m}} \] Rearrange to solve for \(m\): \[ m = \frac{k}{\omega^2} \] Substitute the given spring constant and calculated angular frequency: \[ m = \frac{120}{(37.70)^2} \approx 0.084\ \text{kg} \]

Key Concepts

Spring Constant

The spring constant, often denoted by the symbol \( k \), is a measure of a spring's stiffness. It describes how much force is needed to stretch or compress the spring by a certain distance. For example, in our exercise, the spring constant is given as \( 120 \, \text{N/m} \). This means that 120 newtons of force are needed to extend or compress the spring by one meter. Understanding the spring constant is crucial because it determines how strong the spring's restoring force is when displaced from its equilibrium position. The larger the value of \( k \), the stiffer the spring.

• The unit of the spring constant is newtons per meter (\( \text{N/m} \)).
• It plays a central role in Hooke's Law, which states that the force exerted by a spring is proportional to its extension or compression: \( F = -kx \).
• The negative sign signifies that the spring force is a restoring force, acting in the opposite direction of displacement.

Understanding how the spring constant relates to simple harmonic motion allows us to analyze various oscillating systems effectively.

Frequency of Vibration

The frequency of vibration, denoted by \( f \), is a fundamental concept in simple harmonic motion. It tells us how many complete cycles of vibration occur in one second. The unit for frequency is hertz (Hz), where 1 Hz equals one cycle per second. In our problem, the frequency is specified as \( 6.00 \, \text{Hz} \), indicating that the object completes six oscillations every second.
In the context of simple harmonic motion, such as with a spring-mass system, the frequency is determined by both the mass attached to the spring and the spring constant \( k \). These two factors influence how quickly the spring can return to its equilibrium position after being disturbed.

• A higher frequency means more cycles are completed in a given time, indicating fast oscillation.
• Frequency is inversely related to the period (\( T \)), the time for one complete cycle: \( f = \frac{1}{T} \).
• Frequency can be used to derive the angular frequency, which is another vital parameter of simple harmonic motion.

Understanding frequency is critical in applications ranging from engineering to music, where the concepts of pitch and tone are directly related to vibration frequency.

Angular Frequency

Angular frequency, denoted by \( \omega \), is closely related to the frequency of vibration but provides a measure of how quickly an object moves through its cycle in terms of radians per second. It is calculated using the formula \( \omega = 2\pi f \), where \( f \) is the frequency. In our case, substituting the given frequency of \( 6.00 \, \text{Hz} \) into the formula provides an angular frequency of approximately \( 37.70 \, \text{rad/s} \).
Unlike regular frequency, which counts cycles per second, angular frequency reveals how much angle an oscillating object covers per unit of time.

• Angular frequency is measured in radians per second (rad/s).
• This parameter is especially useful in the analysis of rotational systems and circuits.
• It allows us to relate linear displacement to angular motion through equations involving sinusoids typical of oscillating systems.

Mastering the concept of angular frequency helps in understanding not only simple harmonic motion but also more complex waves and oscillations.

Mass of an Object

The mass of an object in a spring-mass system plays a crucial role in determining the characteristics of its oscillation. In our exercise, we determine the mass by using the relationship between angular frequency \( \omega \), spring constant \( k \), and mass \( m \) described by the equation \( \omega = \sqrt{\frac{k}{m}} \). By rearranging this equation, we solve for mass: \( m = \frac{k}{\omega^2} \). Substituting the known values, we calculate the mass to be approximately \( 0.084 \, \text{kg} \).
Mass determines how much inertia an object has; the more massive it is, the slower it will accelerate under the same force, impacting the oscillation frequency.

• The mass is inversely related to the frequency of oscillation in simple harmonic motion; more mass typically leads to slower vibrations.
• Analyzing this parameter helps in predicting how changes in mass will alter the dynamics of oscillating systems.

Understanding the role of mass in such systems is vital because it affects both the system's stability and responsiveness to external forces.