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\(\bullet\bullet\) (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?

Short Answer

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

Step by step solution

01

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.
02

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.
03

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.
04

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, \( T_{\text{new}} = 2 \times \sqrt{3} \).
05

Calculate the numeric value for the new period

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

\(\bullet\bullet\) What is the maximum elastic potential energy of a simple horizontal mass-spring oscillator whose equation of motion is given by \(x=(0.350\mathrm{~m}) \sin [(7 \mathrm{rad} / \mathrm{s}) t] ?\) The mass on the end of the spring is \(0.900 \mathrm{~kg}\).

\(\bullet\bullet\bullet\) A clock uses a pendulum that is \(75 \mathrm{~cm}\) long. The clock is accidentally broken, and when it is repaired, the length of the pendulum is shortened by \(2.0 \mathrm{~mm} .\) Consider the pendulum to be a simple pendulum. (a) Will the repaired clock gain or lose time? (b) By how much will the time indicated by the repaired clock differ from the correct time (taken to be the time determined by the original pendulum in \(24 \mathrm{~h}\) )? (c) If the pendulum rod were metal, would the surrounding temperature make a difference in the timekeeping of the clock? Explain.

\(\bullet\bullet\) A mass-spring system is in SHM in the horizontal direction. If the mass is \(0.25 \mathrm{~kg}\), the spring constant is \(12 \mathrm{~N} / \mathrm{m},\) and the amplitude is \(15 \mathrm{~cm}\) (a) what is the maximum speed of the mass, and (b) where does this occur? (c) What is the speed at a half-amplitude position?

A piece of steel string is under tension. (a) If the tension doubles, the transverse wave speed (1) doubles, (2) halves, (3) increases by \(\sqrt{2},\) (4) decreases by \(\sqrt{2}\). Why? (b) If the linear mass density of a 10.0 -m length of string is \(0.125 \mathrm{~kg} / \mathrm{m}\) and it is under a tension of \(9.00 \mathrm{~N}\), what is the transverse wave speed in the string? (c) What are its waves' natural frequencies?

\(\bullet\) An object of mass \(0.50 \mathrm{~kg}\) is attached to a spring with spring constant \(10 \mathrm{~N} / \mathrm{m}\). If the object is pulled down \(0.050 \mathrm{~m}\) from the equilibrium position and released, what is its maximum speed?

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