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Chapter 7: Problem 114

Springs of spring cosnstant \(K, 3 K, 9 K, 27 K, \cdots, \infty\) are connected in series. Equivalent spring constant of the combination is a. \(\frac{3 K}{2}\) b. \(\frac{K}{2}\) c. \(\frac{2 K}{3}\) d. \(\infty\)

Short Answer

Expert verified
The equivalent spring constant is \( \frac{2K}{3} \) (Option c).

Step by step solution

01

Understand the Series Combination Formula

When springs are connected in series, their equivalent spring constant \( k_{eq} \) can be found using the formula: \[ \frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2} + \frac{1}{k_3} + \cdots \] where \( k_1, k_2, k_3, \ldots \) are the spring constants of the individual springs in series.
02

Identify the Pattern of Spring Constants

The series follows the pattern \( K, 3K, 9K, 27K, \ldots \). This is a geometric series where each spring constant is three times the previous one. The general term for the \( n \)-th spring is \( 3^{n-1}K \).
03

Express the Series Using the Formula

Substitute the spring constants into the series formula: \[ \frac{1}{k_{eq}} = \frac{1}{K} + \frac{1}{3K} + \frac{1}{9K} + \frac{1}{27K} + \cdots \]
04

Simplify the Equation

Factor out \( \frac{1}{K} \) from each term: \[ \frac{1}{k_{eq}} = \frac{1}{K} \left( 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots \right) \] The remaining series inside the parentheses is an infinite geometric series with first term \( a = 1 \) and common ratio \( r = \frac{1}{3} \).
05

Calculate the Sum of the Geometric Series

The sum \( S \) of an infinite geometric series is given by \( S = \frac{a}{1-r} \). For our series: \[ S = \frac{1}{1 - \frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2} \]
06

Solve for the Equivalent Spring Constant

Substitute the sum back into the equation from Step 4: \[ \frac{1}{k_{eq}} = \frac{1}{K} \cdot \frac{3}{2} = \frac{3}{2K} \] Thus, \[ k_{eq} = \frac{2K}{3} \].

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Key Concepts

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

Spring Series Combination
When springs are arranged in a series combination, it essentially means the springs are connected end-to-end. Each spring individually contributes to the overall extension or compression when a force is applied. This type of configuration results in an equivalent spring constant that can be calculated using the reciprocal sum of the individual spring constants.
In mathematical terms, if you have multiple springs with constants \(k_1, k_2, k_3, \ldots\), the equivalent spring constant \(k_{eq}\) is determined by:
\[ \frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2} + \frac{1}{k_3} + \cdots \]
This means in a series, the equivalent spring constant is always less than the smallest individual spring constant. Springs connected in series will behave like a single, more flexible spring.
Geometric Series in Physics
A geometric series is a sequence of numbers where each term is a constant multiple of the previous one. In this exercise, the spring constants follow a pattern where each subsequent spring has a constant, in this case, being multiplied by 3 as you progress through the series (\(K, 3K, 9K, 27K, \ldots\)).
This sequence is a geometric series with a first term \(a\) of \(K\) and a common ratio \(r\) of 3. The infinite series representing this spring problem can be expressed in terms of its equivalent spring constant by the formula for geometric series:
\[ S = \frac{a}{1 - r} \]
Notably, geometric series are widely used in physics to simplify problems involving repeated multiplicative patterns, making complex systems much more manageable in analysis.
Mechanical Systems Analysis
Understanding mechanical systems often involves breaking down the components and analyzing their interactions. Series and parallel springs are fundamental concepts in mechanical systems.
By examining these configurations, you can predict how the system will respond to external forces, like tension or compression. This knowledge allows engineers to design systems that are reliable, strong, and efficient. In the given problem, analyzing springs connected in series means that you must focus on accumulative elongation, considering the reduced stiffness that such an arrangement entails.
With a sound understanding of these principles, you can accurately determine how a system will perform and address potential weaknesses early in the design phase. This dismisses trial and error, saving both time and resources.

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Most popular questions from this chapter

A particles is moving in \(x-y\) plane. At certain instant.of time, the components of its velocity and acceleration are as follows: \(v_{x}=3 \mathrm{~ms}^{-1}, v_{y}=4 \mathrm{~ms}^{-1}, a_{x}=2 \mathrm{~ms}^{-2}\), and \(a_{y}=1 \mathrm{~ms}^{-2}\). The rateof change of speed at this moment is a. \(\sqrt{10} \mathrm{~ms}^{-2}\) b. \(4 \mathrm{~ms}^{-2}\) c. \(\sqrt{5} \mathrm{~ms}^{-2}\) d. \(2 \mathrm{~ms}^{-2}\)

A plank is held at an angle \(\alpha\) to the horizontal (Fig. 7.442) on two fixed supports \(A\) and \(B\). The plank can slide against the supports (without friction) because of the weight \(M g\). Acieration and direction in which a man of mass \(m\) should move so that the plank does not move. a. \(g \sin \alpha\left(1+\frac{m}{M}\right)\) down the incline b. \(g \sin \alpha\left(1+\frac{M}{m}\right)\) down the incline c. \(g \sin \alpha\left(1+\frac{m}{M}\right)\) up the incline d. \(g \sin \alpha\left(1+\frac{M}{m}\right)\) up the incline

A block of metal weighing \(2 \mathrm{~kg}\) is resting on a frictionless plane. It is struck by a jet releasing water at a rate of \(1 \mathrm{~kg} / \mathrm{s}\) and at a speed of \(5 \mathrm{~m} / \mathrm{s}\). The initial acceleration of the block is a. \(\frac{5}{3} \mathrm{~m} / \mathrm{s}^{2}\) b. \(\frac{25}{4} \mathrm{~m} / \mathrm{s}^{2}\) c. \(\frac{25}{6} \mathrm{~m} / \mathrm{s}^{2}\) d. \(\frac{5}{2} \mathrm{~m} / \mathrm{s}^{2}\)

An inclined plane makes an angle \(30^{\circ}\) with the horizontal. \(A\) groove \((O A)\) of length \(5 \mathrm{~m}\) cut, in the plane makes an angle \(30^{\circ}\) with \(O X\). A short smooth cylinder is free to slide down the influence of gravity. The time taken by the cylinder to reach from \(A\) to \(O\) is \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) a. \(4 \mathrm{~s}\) b. \(2 \mathrm{~s}\) c. \(2 \mathrm{~s}\) d. \(1 \mathrm{~s}\)

A particle of mass \(2 \mathrm{~kg}\) moves with an initial velocity of \(v=(4 \hat{i}+4 \hat{j}) \mathrm{m} / \mathrm{s}\). A constant force of \(F=-20 \hat{j} \mathrm{~N}\) is applied on the particle. Initially, the particle was at (0, 0). The \(x\) -coordinate of the particle when its \(y\) -coordinate again becomes zero is given by a. \(1.2 \mathrm{~m}\) b. \(4.8 \mathrm{~m}\) c. \(6.0 \mathrm{~m}\) d. \(3.2 \mathrm{~m}\)
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