Chapter 13: Problem 34
\(\bullet\bullet\) (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?
Short Answer
Expert verified
(a) \( \frac{1}{\sqrt{2}} \); (b) ~1.73 s
Step by step solution
01
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.
02
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.
03
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.
04
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:\[T' = 2\pi \sqrt{\frac{m/3}{k}} = 2\pi \sqrt{\frac{m}{3k}} = 2\pi \frac{1}{\sqrt{3}} \sqrt{\frac{m}{k}} = \frac{1}{\sqrt{3}} T = \frac{1}{\sqrt{3}} \times 3.0 \mathrm{~s} \approx 1.73 \mathrm{~s}\]
05
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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.
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:
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.
- 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.
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:
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.
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.
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.