Question

When a 0.750-kg mass oscillates on an ideal spring, the frequency is 1.75 Hz. What will the frequency be if 0.220 kg are (a) added to the original mass and (b) subtracted from the original mass? Try to solve this problem \(without\) finding the force constant of the spring.

Short answer

New frequency is 1.55 Hz when mass is added, and 2.10 Hz when mass is subtracted.

Step-by-step solution

Understanding the problem

We have a 0.750 kg mass attached to an ideal spring with an oscillation frequency of 1.75 Hz. We need to calculate the new frequencies when 0.220 kg is added and when 0.220 kg is subtracted from the original mass.

Recall the formula for frequency of oscillation

The frequency of a mass-spring system is given by the formula: \[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]where \( f \) is the frequency, \( k \) is the spring constant, and \( m \) is the mass.

Express the initial frequency

Given the initial mass \( m_1 = 0.750 \) kg and frequency \( f_1 = 1.75 \) Hz, we use: \[ f_1 = \frac{1}{2\pi} \sqrt{\frac{k}{m_1}} \]Solving for the spring constant, we don't need its value, but note how it scales: \[ f_1^2 = \frac{k}{4\pi^2 m_1} \]

Calculate new frequency when additional mass is added

New mass \( m_2 = 0.750 + 0.220 = 0.970 \) kg. The new frequency \( f_2 \) is:\[ f_2 = \frac{1}{2\pi} \sqrt{\frac{k}{m_2}} \]Using the relationship, \( f_2^2 = \frac{k}{4\pi^2 m_2} \). Therefore,\[ \frac{f_2}{f_1} = \sqrt{\frac{m_1}{m_2}} \]\[ f_2 = f_1 \times \sqrt{\frac{m_1}{m_2}} \]\[ f_2 = 1.75 \times \sqrt{\frac{0.750}{0.970}} \approx 1.55 \text{ Hz} \]

Calculate new frequency when mass is subtracted

New mass \( m_3 = 0.750 - 0.220 = 0.530 \) kg. The new frequency \( f_3 \) is: \[ f_3 = \frac{1}{2\pi} \sqrt{\frac{k}{m_3}} \]Using our relationship, \( f_3^2 = \frac{k}{4\pi^2 m_3} \). Therefore,\[ \frac{f_3}{f_1} = \sqrt{\frac{m_1}{m_3}} \]\[ f_3 = f_1 \times \sqrt{\frac{m_1}{m_3}} \]\[ f_3 = 1.75 \times \sqrt{\frac{0.750}{0.530}} \approx 2.10 \text{ Hz} \]

Key concepts

Frequency of Oscillation

Let's dig into what frequency of oscillation means, especially in a mass-spring system. Frequency refers to how many times an object vibrates back and forth in one second. It is measured in Hertz (Hz), where 1 Hz equals one complete oscillation or cycle per second. In the context of a mass-spring system, the frequency is determined by both the mass attached to the spring and the spring's stiffness, or spring constant.The essential formula for frequency in such a system is:\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]In this formula:

• \( f \) stands for frequency.
• \( k \) is the spring constant, representing the spring's stiffness.
• \( m \) represents the mass attached to the spring.

Understanding the effect of mass on frequency is crucial. As the mass increases, the frequency decreases. This is because a heavier mass makes the system oscillate slower. Conversely, when the mass decreases, the frequency increases, making the system oscillate faster.So, when additional mass is added to or removed from the system, recalculating the frequency can determine how the oscillation behavior changes.

Mass-Spring System

A mass-spring system is a fundamental concept in physics where a mass (an object with weight) is attached to a spring. This system is significant because it demonstrates simple harmonic motion, a type of periodic motion where the restoring force is directly proportional to the displacement. The force that pulls or pushes the mass back toward its equilibrium position can be described by Hooke's Law:\[ F = -kx \]Where:

• \( F \) is the force exerted by the spring.
• \( k \) is the spring constant (a measure of the spring's stiffness).
• \( x \) is the displacement from the equilibrium position.

This system oscillates when set in motion, and its behavior can be predicted using certain formulas, especially the frequency formula discussed previously. A critical aspect is that the system's motion is purely dependent on the mass and the spring constant, not on factors such as amplitude, provided the amplitude remains small. Such systems are found everywhere in the world around us, from car suspensions to the function of clocks. Understanding these systems helps in designing various practical applications that rely on controlled oscillations and vibrations.

Physics Problem Solving

Tackling physics problems, like understanding the frequency of a mass-spring system, requires breaking down the problem into manageable parts. Here's a general approach for solving such problems related to oscillations:

• **Understand the Problem**: Clearly define what is given and what needs to be found. In the original exercise, the frequency of a spring system needed to be calculated with updated mass values.
• **Recall Relevant Formulas**: Identify which physics formulas apply to the situation. For oscillation problems, the frequency formula \( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \) is key.
• **Manipulate the Math**: Rearrange equations as needed to solve for the unknown. In this case, you manipulate the relationship between frequency and mass to find the new frequencies without needing the spring constant explicitly.
• **Calculate Carefully**: When performing calculations, ensure accuracy, especially when dealing with square roots and ratios.
• **Analyze Results**: Ensure results make sense logically. Added mass leads to lower frequency, while reduced mass gives higher frequency. Cross-check your calculations by considering this relationship.

This structured approach not only helps solve specific physics problems but also strengthens overall problem-solving skills for diverse scientific inquiries.