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Chapter 5: Problem 2

When a body is suspended from a fixed point by a certain linear spring, the angular frequency of its vertical oscillations is found to be \(\Omega_{1}\). When a different linear spring is used, the oscillations have angular frequency \(\Omega_{2}\). Find the angular frequency of vertical oscillations when the two springs are used together (i) in parallel, and (ii) in series. Show that the first of these frequencies is at least twice the second.

Short Answer

Expert verified
The angular frequencies of the vertical oscillations when two springs are used together in parallel and in series are \(\sqrt{\Omega_{1}^{2}+ \Omega_{2}^{2}}\) and \(\sqrt{\frac{\Omega_{1}^{2}\Omega_{2}^{2}}{\Omega_{1}^{2}+ \Omega_{2}^{2}}\) respectively. It has been shown that \(\Omega_{par}\) is at least twice as large as \(\Omega_{series}\).

Step by step solution

01

Calculate the Angular Frequency for Springs in Parallel

In a parallel arrangement, the combined spring constant \(k_{kom}\) becomes the summation of the individual spring constants \(k_{1}\) and \(k_{2}\). \(k_{1}\) and \(k_{2}\) have angular frequencies \(\Omega_{1}\) and \(\Omega_{2}\) respectively, thus \(k_{1} = m \Omega_{1}^{2}\) and \(k_{2} = m \Omega_{2}^{2}\). Hence the combined spring constant \(k_{kom} = k_{1} + k_{2} = m \Omega_{1}^{2} + m \Omega_{2}^{2} = m (\Omega_{1}^{2}+\Omega_{2}^{2})\). The angular frequency \(\Omega_{par}\) for the parallel scenario can then be found by using the angular frequency formula \(\Omega = \sqrt{\frac{k}{m}}\). Substituting the combined spring constant yields \(\Omega_{par} = \sqrt{\frac{k_{kom}}{m}} = \sqrt{\Omega_{1}^{2}+ \Omega_{2}^{2}}\).
02

Calculate the Angular Frequency for Springs in Series

In a series arrangement, the combined spring constant \(k_{kom}\) is the inverse of the sum of the reciprocals of the individual spring constants \(k_{1}\) and \(k_{2}\). \(k_{1} = m \Omega_{1}^{2}\) and \(k_{2} = m \Omega_{2}^{2}\), yielding \(k_{kom} = \frac{1}{\frac{1}{k_{1}} + \frac{1}{k_{2}}} = \frac{1}{\frac{1}{m \Omega_{1}^{2}} + \frac{1}{m \Omega_{2}^{2}}}= \frac{1}{\frac{\Omega_{1}^{2}+ \Omega_{2}^{2}}{m \Omega_{1}^{2}\Omega_{2}^{2}}}= \frac{m \Omega_{1}^{2}\Omega_{2}^{2}}{\Omega_{1}^{2}+ \Omega_{2}^{2}}\). The angular frequency \(\Omega_{series}\) for the series scenario can be found using \(\Omega = \sqrt{\frac{k}{m}}\), yielding \(\Omega_{series} = \sqrt{\frac{k_{kom}}{m}} = \sqrt{ \frac{\Omega_{1}^{2}\Omega_{2}^{2}}{\Omega_{1}^{2}+ \Omega_{2}^{2}}\).
03

Comparing the Angular Frequencies

From Step 1, \(\Omega_{par} = \sqrt{\Omega_{1}^{2}+ \Omega_{2}^{2}}\) and from Step 2, \(\Omega_{series} = \sqrt{ \frac{\Omega_{1}^{2}\Omega_{2}^{2}}{\Omega_{1}^{2}+ \Omega_{2}^{2}}\). Squaring both sides in both equations, we get \(\Omega_{par}^{2} = \Omega_{1}^{2}+ \Omega_{2}^{2}\) and \(\Omega_{series}^{2} = \frac{\Omega_{1}^{2}\Omega_{2}^{2}}{\Omega_{1}^{2}+ \Omega_{2}^{2}}\). Comparing these two equations by taking the ratio, we find that \(\frac{\Omega_{par}^{2}}{\Omega_{series}^{2}} = \frac{\Omega_{1}^{2}+ \Omega_{2}^{2}}{\frac{\Omega_{1}^{2}\Omega_{2}^{2}}{\Omega_{1}^{2}+ \Omega_{2}^{2}}} = \Omega_{1}^{2}+ \Omega_{2}^{2}\). Since \(\Omega_{1}^{2}+ \Omega_{2}^{2}\) is always greater than or equal to \(2\Omega_{1}\Omega_{2}\), it implies that \(\Omega_{par}\) is at least twice as large as \(\Omega_{series}\).

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Key Concepts

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

Angular Frequency
Angular frequency is a key concept within oscillatory systems, particularly in the study of harmonic motion. To understand it better, think of angular frequency as the rate at which an object moves through its oscillatory cycle. It is denoted by the symbol \( \Omega \) and closely related to the period and frequency of oscillation.\(
\)
Angular frequency is measured in radians per second. This represents how quickly or rapidly an object oscillates around its equilibrium point. The formula \( \Omega = 2\pi f \), where \( f \) is the frequency in hertz, helps to convert between frequency and angular frequency.\(
\)
In physics problems such as those involving springs, the angular frequency tells us how fast a spring-mass system vibrates back and forth. Understanding the angular frequency is essential when you want to analyze the behavior of different spring combinations, such as when springs are arranged in parallel or series.
Spring Constant
The spring constant is an important factor in determining how a spring behaves in dynamic systems. Represented by the symbol \( k \), the spring constant measures the stiffness of a spring. A higher spring constant indicates a stiffer spring, while a lower value implies a more flexible spring.\(
\)
The spring constant is calculated using Hooke鈥檚 Law which is expressed as \( F = kx \), where \( F \) is the force applied to the spring and \( x \) is the displacement of the spring from its equilibrium position. This means that the more force is applied, the more the spring will stretch, assuming a linear relationship.\(
\)
When dealing with oscillations, the spring constant helps us determine the system鈥檚 natural frequency, which influences the angular frequency. In any scenario involving dynamic systems, knowing the spring constant aids in predicting how an object will respond to various forces.
Parallel and Series Springs
Understanding how springs behave together, whether in parallel or series, is crucial for solving problems involving combined spring systems.\(
\)
- **Parallel Springs**: When springs are used together in parallel, they work together to support a load. In this setup, the total spring constant \( k_{kom} \) is the sum of the individual spring constants: \( k_{kom} = k_1 + k_2 \). This results in a system that is stiffer and has a higher angular frequency compared to the individual springs.\(
\)
- **Series Springs**: In contrast, when springs are arranged in a series, they share the load. The combined spring constant \( k_{kom} \) is calculated using the reciprocals of the individual spring constants, resulting in: \( k_{kom} = \frac{1}{\frac{1}{k_1} + \frac{1}{k_2}} \). This results in a system that is less stiff compared to individual springs.\(
\)
These configurations significantly affect the stiffness of the system and the resulting angular frequency of oscillations, making them a vital concept in both theoretical calculations and practical applications.

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Most popular questions from this chapter

A block of mass \(2 \mathrm{~kg}\) is suspended from a fixed support by a spring of strength \(2000 \mathrm{Nm}^{-1}\). The block is subject to the vertical driving force \(36 \cos p t \mathrm{~N}\). Given that the spring will yield if its extension exceeds \(4 \mathrm{~cm}\), find the range of frequencies that can safely be applied.

An overdamped harmonic oscillator satisfies the equation $$ \ddot{x}+10 \dot{x}+16 x=0 $$ At time \(t=0\) the particle is projected from the point \(x=1\) towards the origin with speed \(u\). Find \(x\) in the subsequent motion. Show that the particle will reach the origin at some later time \(t\) if $$ \frac{u-2}{u-8}=e^{6 t} $$ How large must \(u\) be so that the particle will pass through the origin?

The oscillations of a galvanometer satisfy the equation $$ \ddot{x}+2 K \dot{x}+\Omega^{2} x=0 $$ The galvanometer is released from rest with \(x=a\) and we wish to bring the reading permanently within the interval \(-\epsilon a \leq x \leq \epsilon a\) as quickly as possible, where \(\epsilon\) is a small positive constant. What value of \(K\) should be chosen? One possibility is to choose a sub-critical value of \(K\) such that the first minimum point of \(x(t)\) occurs when \(x=-\epsilon a\). [Sketch the graph of \(x(t)\) in this case.] Show that this can be acheived by setting the value of \(K\) to be $$ K=\Omega\left[1+\left(\frac{\pi}{\ln (1 / \epsilon)}\right)^{2}\right]^{-1 / 2} $$ If \(K\) has this value, show that the time taken for \(x\) to reach its first minimum is approximately \(\Omega^{-1} \ln (1 / \epsilon)\) when \(\epsilon\) is small.

Two particles \(P\) and \(Q\), each of mass \(m\), are secured at the points of trisection of a light string that is stretched to tension \(T_{0}\) between two fixed supports a distance \(3 a\) apart. The particles undergo small transverse oscillations perpendicular to the equlilibrium line of the string. Find the normal frequencies, the forms of the normal modes, and the general motion of this system. [Note that the forms of the modes could have been deduced from the symmetry of the system.] Is the general motion periodic?

A block of mass \(M\) is connected to a second block of mass \(m\) by a linear spring of natural length \(8 a\). When the system is in equilibrium with the first block on the floor, and with the spring and second block vertically above it, the length of the spring is \(7 a\). The upper block is then pressed down until the spring has half its natural length and is then resleased from rest. Show that the lower block will leave the floor if \(M<2 m\). For the case in which \(M=3 \mathrm{~m} / 2\), find when the lower block leaves the floor.
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