Problem 6 - University Physics with Modern Physics

Question

A spring is hanging from the ceiling with a mass attached to it. The mass is pulled downward, causing it to oscillate vertically with simple harmonic motion. Which of the following will increase the frequency of oscillation? a) adding a second, identical spring with one end attached to the mass and the other to the ceiling b) adding a second, identical spring with one end attached to the mass and the other to the floor c) increasing the mass d) adding both springs, as described in (a) and (b)

Step-by-Step Solution

Answer

A) Adding a second, identical spring with one end attached to the mass and the other to the ceiling B) Adding a second, identical spring with one end attached to the mass and the other to the floor C) Increasing the mass D) Adding both springs, as described in (A) and (B) Answer: A) Adding a second, identical spring with one end attached to the mass and the other to the ceiling

Step 1: Option A: Adding a second, identical spring with one end attached to the mass and the other to the ceiling

When a second, identical spring is added to the mass and attached to the ceiling in parallel, the effective spring constant of the system becomes \(k_\mathrm{effective} = k_1 + k_2\), where \(k_1\) and \(k_2\) are the spring constants of the two springs. Since the springs are identical, \(k_\mathrm{effective} = 2k\). The frequency will now depend on the effective spring constant and can be represented as \(f_A = \frac{1}{2\pi}\sqrt{\frac{2k}{m}}\). Comparing this to the initial frequency, we can see that the frequency has increased.

Step 2: Option B: Adding a second, identical spring with one end attached to the mass and the other to the floor

When a second, identical spring is attached to the mass and the floor, the second spring will be in series with the first one . The effective spring constant of the system can be determined using the formula: \(\frac{1}{k_\mathrm{effective}} = \frac{1}{k_1} + \frac{1}{k_2}\). As the springs are identical, we have \(k_\mathrm{effective} = \frac{k}{2}\). The frequency is now given by \(f_B = \frac{1}{2\pi}\sqrt{\frac{k/2}{m}}\). Comparing it with the initial frequency, we see that the frequency will decrease in this case.

Step 3: Option C: Increasing the mass

As we can see from the frequency formula \(f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\), increasing the mass will result in a decrease in the frequency of oscillation.

Step 4: Option D: Adding both springs, as described in (A) and (B)

In this situation, we have a parallel setup with the ceiling (Option A) and a series setup with the floor (Option B). The result of Option A was to increase the frequency, while the result of Option B was to decrease the frequency. When we add both springs, their effects will counteract each other, and the overall effect on the frequency of oscillation will not be an increase. In conclusion, only Option A (adding a second, identical spring with one end attached to the mass and the other to the ceiling) will increase the frequency of oscillation.

Key Concepts

Spring Constant

In simple harmonic motion, the spring constant, often denoted as \(k\), plays a crucial role in determining how a spring system behaves. Think of the spring constant as a measure of stiffness. A higher spring constant indicates a stiffer spring, meaning it requires more force to stretch or compress it. This concept is essential in understanding how springs affect oscillation frequency. When two identical springs are connected in parallel, their spring constants add together. For example:

• For two springs in parallel: \(k_\text{effective} = k_1 + k_2\)
• If both springs have the same spring constant \(k\), then \(k_\text{effective} = 2k\)

Increasing \(k_\text{effective}\) leads to a higher oscillation frequency, meaning the system will oscillate faster. This is because the system exerts more force against displacement, causing quicker movements back to equilibrium.
Understanding the connection between spring constants and oscillation helps in analyzing real-world systems such as vehicle suspensions or even heartbeats, where elasticity and regular cycles are important.

Oscillation Frequency

Oscillation frequency refers to how often an oscillating system completes a full cycle of motion in one second. In the context of springs in simple harmonic motion, frequency can be determined using:

• Frequency formula: \(f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\)

Where \(k\) is the spring constant and \(m\) is the mass attached to the spring. This formula reveals a few critical points:

• A stronger spring (higher \(k\)) results in a higher frequency.
• A larger mass (higher \(m\)) leads to a lower frequency.

Option A from the exercise, where a second spring is added in parallel, demonstrates how increasing the effective spring constant boosts the frequency. The increased stiffness accelerates the system’s return to its equilibrium position. Conversely, more mass means the system is harder to move, thus slowing oscillation and reducing frequency.
This principle is vital in engineering and science, where controlling motion frequency can lead to systems that are more efficient, more stable, or produce desired effects in manufacturing processes.

Parallel and Series Springs

Understanding how springs behave in parallel and series combinations extends their utility in various mechanical systems.

Parallel Springs:

• Much like adding resistors in parallel, adding springs in parallel increases the system's overall stiffness.
• The formula \(k_{\text{parallel}} = k_1 + k_2\) applies, leading to an increased effective spring constant.
• This setup causes the system to oscillate at a higher frequency if all other factors remain constant.

Series Springs:

• When springs are linked in series, the system as a whole becomes more flexible.
• The reciprocal of the effective spring constant is the sum of the reciprocals of each spring: \(\frac{1}{k_{\text{series}}} = \frac{1}{k_1} + \frac{1}{k_2}\).
• This setup generally reduces the system’s effective spring constant and lowers the oscillation frequency.

In the exercise, in scenario B, when a second spring is added in series, it demonstrates the decrease in effective stiffness, hence lowering the frequency of oscillation.
Understanding these principles is fundamental in designing devices that require precise control over motion, such as in clock mechanisms or vibration isolators.