Problem 8 - University Physics with Modern Physics

Question

When a 0.750 -kg mass oscillates on an ideal spring, the frequency is 1.33 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.

Step-by-Step Solution

Answer

The new frequency is 1.10 Hz with added mass and 1.71 Hz with subtracted mass.

Step 1: Understanding the Relation Between Mass and Frequency

The frequency of a mass-spring system is related to the mass and the spring constant by the formula \( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \). Since we are not given the spring constant \( k \), and it is the same spring, the ratio of the frequencies for different masses can be used: \( \frac{f_1}{f_2} = \sqrt{\frac{m_2}{m_1}} \).

Step 2: Calculating the New Frequency With Added Mass

For part (a), add 0.220 kg to the original mass, making the new mass \( m_2 = 0.750 + 0.220 = 0.970 \) kg. Using the ratio formula \( \frac{f_1}{f_2} = \sqrt{\frac{m_2}{m_1}} \) and solving for \( f_2 \), you get \( f_2 = \frac{f_1}{\sqrt{\frac{m_2}{m_1}}} = \frac{1.33}{\sqrt{\frac{0.970}{0.750}}} \). Calculate this to find \( f_2 \).

Step 3: Computing Frequency With Decreased Mass

For part (b), subtract 0.220 kg from the original mass, so the new mass \( m_2 = 0.750 - 0.220 = 0.530 \) kg. Again, use the relation \( f_2 = \frac{f_1}{\sqrt{\frac{m_2}{m_1}}} = \frac{1.33}{\sqrt{\frac{0.530}{0.750}}} \). Calculate this to obtain \( f_2 \).

Step 4: Final Calculation and Verification

For both cases (a) and (b), plug the calculated values into the respective formulas and simplify the square root to determine the new frequencies. Verify the calculations by ensuring the logical consistency that increased mass should decrease frequency, and decreased mass should increase frequency.

Key Concepts

Mass-spring system

In physics, a mass-spring system is a simple model used to describe oscillatory motions, where a mass is attached to a spring. The mass moves back and forth under the influence of the spring's restoring force and its own inertia. This system is a classic example of simple harmonic motion, which is a type of periodic motion where the restoring force is directly proportional to the displacement.

• The spring constant, often denoted as \( k \), is a measure of the spring's stiffness.
• When the mass is displaced from its equilibrium position, the spring exerts a force that tries to bring the mass back to equilibrium.
• This back-and-forth motion continues indefinitely in an ideal mass-spring system with no external forces like friction or air resistance.

Understanding how this system works helps in grasping more complex systems in mechanics. The interaction between the mass and the spring's properties determines how the system behaves during its oscillations.

Frequency calculation

The frequency of oscillation in a mass-spring system refers to how often the mass completes one full cycle of motion per second, and it is measured in Hertz (Hz). The fundamental frequency formula for a mass-spring system is given by:\[f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}\]Where:

• \( f \) is the frequency in Hz,
• \( k \) is the spring constant in Newtons per meter,
• \( m \) is the mass in kilograms.

This formula shows that the frequency depends on both the spring constant and the mass. To find the frequency when the mass changes without needing to re-calculate the spring constant, the ratio of frequencies for different masses can be used. This allows us to solve problems efficiently by comparing the change in mass and directly calculating the change in frequency.

• As the mass increases, the frequency decreases because the mass takes longer to oscillate.
• Conversely, when the mass decreases, the frequency increases.

This understanding is key in predicting how a mass-spring system reacts to changes in mass.

Relation between mass and frequency

The relation between mass and frequency in a mass-spring system can be easily understood through the frequency formula. As illustrated before, the frequency \( f \) is inversely proportional to the square root of the mass \( m \). This means that any increase in mass will result in a decrease in frequency, and similarly, any decrease in mass will cause the frequency to increase.

• This inverse relationship can be expressed as: \( f \propto \frac{1}{\sqrt{m}} \).
• If you add mass, you effectively slow down the system's ability to oscillate quickly.
• On the other hand, subtracting mass makes it easier for the system to complete cycles more rapidly.

In practice, this helps in understanding phenomena like why adding cargo to a spring-loaded vehicle suspension makes it oscillate more slowly. By manipulating the mass attached to a spring, one can precisely control the frequency of oscillation, which is essential in designing systems that rely on predictable oscillatory behavior.