Problem 34 - College Physics

Question

\(\bullet\bullet\) (a) If the mass 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 \(3.0 \mathrm{~s}\) and the mass is reduced to \(1 / 3\) of its initial value, what is the new period?

Step-by-Step Solution

Answer

(a) \( \frac{1}{\sqrt{2}} \); (b) ~1.73 s

Step 1: Understanding the Period of a Mass-Spring System

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

Step 2: Effect of Halving the Mass on the Period

If the mass \( m \) is halved, the new period \( T' \) becomes \( T' = 2\pi \sqrt{\frac{m/2}{k}} = 2\pi \sqrt{\frac{m}{2k}} = 2\pi \frac{1}{\sqrt{2}} \sqrt{\frac{m}{k}} = \frac{1}{\sqrt{2}} T \). This means the new period is \( \frac{1}{\sqrt{2}} \) times the old period.

Step 3: Selecting the Correct Option

From the given options, \((3) \frac{1}{\sqrt{2}} \) times the old period is the correct one when the mass is halved.

Step 4: Calculating the New Period with Reduced Mass

For part (b), if the initial period \( T = 3.0 \mathrm{~s} \) and the mass is reduced to \( \frac{1}{3} \) of its initial value, the new period \( T' \) is given by:\[T' = 2\pi \sqrt{\frac{m/3}{k}} = 2\pi \sqrt{\frac{m}{3k}} = 2\pi \frac{1}{\sqrt{3}} \sqrt{\frac{m}{k}} = \frac{1}{\sqrt{3}} T = \frac{1}{\sqrt{3}} \times 3.0 \mathrm{~s} \approx 1.73 \mathrm{~s}\]

Step 5: Conclusion

For part (a), the period is \( \frac{1}{\sqrt{2}} \) times the old period when the mass is halved. For part (b), the new period is approximately \( 1.73 \mathrm{~s} \) when the mass is reduced to \( \frac{1}{3} \) of its initial value.

Key Concepts

Simple Harmonic Motion

In the context of a mass-spring system, simple harmonic motion refers to the oscillatory movement of an object where the restoring force is proportional to the displacement. This motion is predictable and repeats in a regular cycle. The primary feature of simple harmonic motion is that it is characterized by a sine or cosine function when graphed over time.
This type of motion can be observed in everyday phenomena, such as playground swings, vibrating guitar strings, and of course, the mass-spring systems discussed here.
The key components involved in this motion are:

• The mass attached to the spring, which influences how quickly the system oscillates.
• The spring constant, defining the stiffness of the spring and affecting the speed of oscillation.

Observing simple harmonic motion helps to understand the dynamics of oscillatory systems and provides foundational knowledge for studying more complex mechanical systems.

Period of Oscillation

For a mass-spring system, the period of oscillation is the time it takes for the system to complete one full cycle of motion. This period is a crucial characteristic of simple harmonic motion, as it remains constant under consistent conditions.
The formula for the period of a mass-spring system is given by: \[ T = 2\pi \sqrt{\frac{m}{k}} \] where:

• \(T\) is the period of oscillation.
• \(m\) is the mass of the object.
• \(k\) is the spring constant.

The formula indicates that the period is dependent on both the mass and the spring constant. When the mass is changed, the period will adjust accordingly. For example:

• If the mass is halved, the new period is reduced by a factor of \(1/\sqrt{2}\).
• If the mass is reduced to one-third, the period becomes \(1/\sqrt{3}\) of its original value.

Understanding this relationship is essential for predicting how changes in the system affect its periodic motion.

Spring Constant

The spring constant, denoted by \(k\), is a core parameter in describing a spring's stiffness. A larger spring constant means a stiffer spring, while a smaller one indicates a more flexible spring. This constant plays a significant role in determining the period of oscillation in a mass-spring system.
The value of the spring constant comes from Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement experienced by the spring. Mathematically, it's expressed as: \[ F = -kx \] where:

• \(F\) is the force exerted by the spring.
• \(k\) is the spring constant.
• \(x\) is the displacement from the equilibrium position.

The negative sign indicates that the force exerted by the spring is always directed towards the equilibrium position, acting as a restoring force.
In the context of the period of oscillation formula, the spring constant is part of the equation \(T = 2\pi \sqrt{\frac{m}{k}} \). Here, you can see how increasing the spring constant will decrease the period, making the oscillation faster, while a lower spring constant will have the opposite effect, resulting in a longer period and slower motion.