Problem 7 - University Physics with Modern Physics

Question

When a body of unknown mass is attached to an idealspring with force constant \(120 \mathrm{N} / \mathrm{m},\) it is found to vibrate with a frequency of 6.00 \(\mathrm{Hz}\) . Find (a) the period of the motion; (b) the angular frequency; (c) the mass of the body.

Step-by-Step Solution

Answer

(a) 0.167 s, (b) 37.70 rad/s, (c) 0.0844 kg.

Step 1: Find the period of the motion

The period of oscillation \(T\) is the reciprocal of the frequency \(f\). Given the frequency \(f = 6.00\, \text{Hz}\), we can find the period using the formula: \(T = \frac{1}{f}\). Substituting the value, we get \(T = \frac{1}{6.00} \approx 0.167\, \text{s}\).

Step 2: Calculate the angular frequency

Angular frequency \(\omega\) is related to frequency by the formula \(\omega = 2\pi f\). Given \(f = 6.00\, \text{Hz}\), substituting in the values yields \(\omega = 2\pi \times 6.00 \approx 37.70\, \text{rad/s}\).

Step 3: Calculate the mass of the body

The angular frequency \(\omega\) is also related to the mass \(m\) and the spring constant \(k\) via the formula \(\omega = \sqrt{\frac{k}{m}}\). Rearranging for \(m\), we get \(m = \frac{k}{\omega^2}\). Given \(k = 120\, \text{N/m}\) and \(\omega \approx 37.70\, \text{rad/s}\), substituting in these values gives \(m = \frac{120}{(37.70)^2} \approx 0.0844\, \text{kg}\).

Key Concepts

Spring Constant

The spring constant, often symbolized as \( k \), is a measure of a spring's stiffness. It tells us how much force is needed to stretch or compress a spring by a certain amount. The higher the spring constant, the stiffer the spring is and the more force it takes to change its length.
To understand it better:

• Units: The unit of spring constant is Newton per meter (N/m).
• Hooke's Law: The relationship between force and spring constant is expressed by Hooke's Law \( F = kx \), where \( F \) is the force applied, and \( x \) is the displacement from its equilibrium position.
• Application: Knowing the spring constant helps in designing systems with desired elasticity and in predicting how a system will respond to an external force.

Oscillation Period

The oscillation period, termed as \( T \), is the time taken for one complete cycle of vibration to pass a given point. In this exercise, it is calculated as the reciprocal of the frequency of the vibrations.

• Calculation Formula: The period \( T \) is given by \( T = \frac{1}{f} \), where \( f \) is the frequency.
• Recurring Motion: In oscillatory motion, such as spring vibration, the period remains constant as long as the system is undisturbed.
• Units: The unit for the period is seconds (s).

By knowing how long one cycle takes, you can determine the pace of any periodic motion.

Angular Frequency

Angular frequency, denoted by \( \omega \), tells us how quickly an object oscillates in a circular path. Despite being called 'angular', it applies even in linear systems, like the vibrations of a spring.

• Relationship to Frequency: It's related to frequency \( f \) through the formula \( \omega = 2\pi f \). This shows the link between rotational and linear perspectives of motion.
• Units: Angular frequency is measured in radians per second (rad/s).
• Comprehensive Understanding: Knowing \( \omega \) helps in analyzing systems subject to harmonic motion. It is crucial for solutions involving trigonometric functions like sine and cosine.

Mass Calculation

Calculating the mass of an oscillating body in a spring-mass system can be intriguing. Given is the relationship between mass \( m \), spring constant \( k \), and angular frequency \( \omega \).

• Key Formula: The mass is derived from \( m = \frac{k}{\omega^2} \).
• Spring Influence: The stiffness of the spring affects how the system oscillates. Thus, \( k \) is a part of the mass calculation.
• Understanding Oscillation: Mass affects how quickly or slowly a system responds to forces—essential for things like shock absorbers or measuring devices.
• Units: Mass is measured in kilograms (kg).

This in-depth look into the calculations helps ensure that the components of any oscillating system are correctly designed for their purposes.