Problem 35 - College Physics

Question

\(\bullet\bullet\) (a) If the spring constant in a mass-spring system is halved, the new period is \((1) 2,(2) \sqrt{2},(3) 1 / \sqrt{2},\) (4) \(1 / 2\) times the old period. Why? (b) If the initial period is \(2.0 \mathrm{~s}\) and the spring constant is reduced to \(1 / 3\) of its initial value, what is the new period?

Step-by-Step Solution

Answer

(a) New period is \(\sqrt{2}\) times the old; (b) New period is \(3.464\,\text{s}\).

Step 1: Understand the formula for the period of a mass-spring system

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

Step 2: Analyze how the period changes when the spring constant is halved

If the spring constant \( k \) is halved, replace \( k \) with \( \frac{k}{2} \). The new period becomes:\[ T_{\text{new}} = 2\pi \sqrt{\frac{m}{k/2}} = 2\pi \sqrt{\frac{2m}{k}} = \sqrt{2} \times 2\pi \sqrt{\frac{m}{k}} \]This means the new period is \( \sqrt{2} \) times the old period.

Step 3: Validate the new period when spring constant is halved

According to the calculation, when the spring constant is halved, the new period becomes \( \sqrt{2} \) times the old period. Therefore, option (2) is correct.

Step 4: Calculate the effect of reducing the spring constant to \( \frac{1}{3} \) of its value

With \( k \) replaced by \( \frac{k}{3} \), the new period \( T_{\text{new}} \) is:\[ T_{\text{new}} = 2\pi \sqrt{\frac{m}{k/3}} = 2\pi \sqrt{\frac{3m}{k}} = \sqrt{3} \times 2\pi \sqrt{\frac{m}{k}} \]And since the initial period \( T = 2 \) s, \( T_{\text{new}} = 2 \times \sqrt{3} \).

Step 5: Calculate the numeric value for the new period

Using the relation \( T_{\text{new}} = 2 \sqrt{3} \), calculate the new period:\[ T_{\text{new}} = 2 \times 1.732 = 3.464 \text{ s} \]

Key Concepts

Spring Constant

The spring constant, denoted by the symbol \( k \), is a fundamental parameter in mechanics that describes how stiff or rigid a spring is. This constant is unique to each spring and determines how much force is needed to stretch or compress it by a unit length. In mathematical terms, the spring constant relates the force \( F \) applied to the spring and the displacement \( x \) that results from this force through the equation \( F = kx \).

• A high spring constant means the spring is stiff. More force is required to achieve deformation.
• A low spring constant indicates a more flexible spring. Less force is needed for deformation.

When you change the spring constant, it affects the behavior of the spring-mass system: - If you decrease the spring constant, the system becomes less stiff, and vice versa.- Variations in \( k \) can drastically alter the dynamic responses of the system, such as the oscillation period.

Oscillation Period

The oscillation period \( T \) in a mass-spring system is the time it takes for the system to complete one full cycle of motion. This period is central to the study of oscillations, as it tells us about the timing of the system's movements.

According to the formula \( T = 2\pi \sqrt{\frac{m}{k}} \):

• \( m \) is the mass attached to the spring. A larger mass results in a longer period, meaning the system oscillates more slowly.
• \( k \) is the spring constant. A larger spring constant results in a shorter period, meaning the system oscillates more quickly.

Thus, the period is affected by both mass and spring constant. For example:- Halving the spring constant results in a period of \( \sqrt{2} \) times the original period, reflecting a slower oscillation because the spring is less stiff.- Reducing the spring constant to a third extends the period to \( \sqrt{3} \times T \), signaling even slower oscillations due to increased flexibility.

Hooke's Law

Hooke's Law is a principle of physics that relates the force on a spring to the displacement it causes, expressed as \( F = kx \), where:

• \( F \) is the force applied to the spring.
• \( k \) is the spring constant, indicating the spring's stiffness.
• \( x \) is the displacement from the equilibrium position.

This law assumes that the spring behaves linearly within its elastic limit, meaning that its deformation is proportional to the force applied. It’s important to note that Hooke's Law simplifies the calculations and provides a fundamental understanding of spring dynamics.

Applications of Hooke's Law:- Designing spring systems in mechanical devices and structures.- Understanding elastic materials in various engineering applications.In the context of a mass-spring system, Hooke's Law is vital for calculating not only the force but also predicting the motion behavior over time, such as determining oscillation periods.