Problem 39 - University Physics with Modern Physics

Question

A 1.50-kg ball and a 2.00 -kg ball are glued together with the lighter one below the heavier one. The upper ball is attached to a vertical ideal spring of force constant \(165 \mathrm{N} / \mathrm{m},\) and the system is vibrating vertically with amplitude 15.0 \(\mathrm{cm} .\) The glue connecting the balls is old and weak, and it suddenly comes loose when the balls are at the lowest position in their motion. (a) Why is the glue more likely to fail at the lowest point than at any other point in the motion? (b) Find the amplitude and frequency of the vibrations after the lower ball has come loose.

Step-by-Step Solution

Answer

(a) The glue is more likely to fail because the restoring force is maximum at the lowest point. (b) The amplitude remains 15.0 cm, and the frequency is approximately 1.44 Hz.

Step 1: Understanding the Problem Context

When the system is at the lowest point in the oscillation, the spring is maximally stretched. This maximum stretch occurs because the system is at its maximum displacement downward from the equilibrium position. At this lowest point, the forces involved include the weight of both balls and the restoring force of the spring.

Step 2: Explaining Glue Failure at the Lowest Point

At the lowest point in the motion, the spring exerts its maximum force in the upward direction because it's stretched most at that point. Additionally, gravitational force acts downwards. Therefore, the net force on the glue is highest at this point, making it more likely to fail.

Step 3: Calculating New System Mass After Glue Failure

Once the glue fails, only the mass of the upper ball, 2.00 kg, interacts with the spring. This changes the dynamics of the oscillating system since the lower 1.50 kg ball is no longer part of the system.

Step 4: Determining New Amplitude

The amplitude of the mass-spring system remains unchanged, provided there is no energy loss due to the glue failure. Thus, the amplitude stays at 15.0 cm.

Step 5: Calculating New Frequency of Oscillations

The frequency of oscillations can be calculated using the formula: \( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \), where \(k\) is the spring constant \(165 \mathrm{N/m}\) and \(m\) is the mass of the single remaining ball (2.00 kg). Substituting the values, \( f = \frac{1}{2\pi} \sqrt{\frac{165}{2.00}} = \frac{1}{2\pi} \sqrt{82.5} \). Calculating further, the frequency \( f \approx 1.44 \) Hz.

Key Concepts

Spring-Mass System

A spring-mass system is a fundamental concept in physics, involving a mass attached to a spring. This setup can oscillate, meaning it moves back and forth, when disturbed from its equilibrium position. In the case of the given exercise, two balls glued together form the mass attached to the spring. The spring has a force constant, also known as the spring constant, which determines how stiff the spring is. For this system, the spring constant is given as 165 N/m.

In a spring-mass system, the mass can be stretched or compressed from its equilibrium position, creating potential energy within the spring. This energy converts oscillates between kinetic and potential forms as the system vibrates. The point where the spring returns naturally is called the equilibrium position, and when displaced, it follows Hooke's law: the force exerted by the spring is directly proportional and opposite to the displacement.

Oscillations

Oscillations are the repetitive variations, usually in time, of some measure about a central value. These can be periodic, occurring at regular intervals, as they do in a spring-mass system. In our exercise, the system is vibrating vertically.

Oscillations involve a few key points:

• The equilibrium position, where the net force is zero.
• The maximum displacement from this position, known as the amplitude.
• The time it takes to complete one cycle, called the period.

At different points in its motion, different forces act on the spring-mass system. At the lowest point, the spring is stretched the most, exerting a maximum upward restoring force. This is why the glue holding the two balls together is most likely to fail at this position, due to the high tension.

Frequency Calculation

To understand the motion of the spring-mass system, we can calculate its frequency, which indicates how often oscillations occur per second. Frequency is measured in Hertz (Hz).

The formula to calculate frequency is:
\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}\]
where:

• \( f \) is the frequency,
• \( k \) is the spring constant,
• \( m \) is the mass.

In the problem, after the glue breaks, we are left with a single 2.00 kg mass. Substituting the relevant values:\[f = \frac{1}{2\pi} \sqrt{\frac{165}{2.00}}\]This gives us a frequency of approximately 1.44 Hz.

Understanding frequency helps you know how quickly the spring-mass system oscillates, an essential aspect of simple harmonic motion.

Amplitude

Amplitude refers to the maximum extent of a vibration or oscillation, measured from the position of equilibrium. In the context of this spring-mass system, amplitude is the furthest distance from the equilibrium point that the mass reaches as it oscillates.

In our given problem, the amplitude is noted as 15.0 cm. This means that the mass moves 15.0 cm away from its central equilibrium position during the oscillation.

Importantly, the amplitude stays the same even when the system's mass changes after the glue breaks, assuming there is no loss of energy. The constancy of amplitude is key because it shows us that the energy of the system remains consistent, solely transitioning between kinetic and potential energy through the oscillation period.