/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 Two point masses of \(3 \mathrm{... [FREE SOLUTION] | 91Ó°ÊÓ

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Two point masses of \(3 \mathrm{~kg}\) and \(1 \mathrm{~kg}\) are attached to opposite ends of a horizontal spring whose spring constant is \(300 \mathrm{~N} / \mathrm{m}\) as shown in figure. The natural vibrational frequency of the system is of order : (a) \(4 \mathrm{~Hz}\) (b) \(3 \mathrm{~Hz}\) (c) \(2 \mathrm{~Hz}\) (d) \(1 \mathrm{~Hz}\)

Short Answer

Expert verified
The natural vibrational frequency is approximately \(3 \mathrm{~Hz}\), so option (b) is correct.

Step by step solution

01

Identify System Type

This is a two-mass oscillator problem where two point masses are connected by a spring. The masses at each end oscillate due to the spring force.
02

Calculate Reduced Mass

The reduced mass \(\mu\) for this two-mass system is calculated using the formula \(\mu = \frac{m_1 m_2}{m_1 + m_2}\), where \(m_1 = 3 \mathrm{~kg}\) and \(m_2 = 1 \mathrm{~kg}\). Plugging in the values, we get \(\mu = \frac{3 \times 1}{3 + 1} = 0.75 \mathrm{~kg}\).
03

Apply Natural Frequency Formula

For a system composed of two masses connected by a spring, the natural frequency \(f\) is given by the formula \(f = \frac{1}{2\pi} \sqrt{\frac{k}{\mu}}\), where \(k = 300 \mathrm{~N/m}\) and \(\mu = 0.75 \mathrm{~kg}\).
04

Calculate the Frequency

Substitute the known values into the formula: \(f = \frac{1}{2\pi} \sqrt{\frac{300}{0.75}} = \frac{1}{2\pi} \sqrt{400} = \frac{1}{2\pi} \times 20\).
05

Approximating the Frequency Value

Evaluate \(\frac{20}{2\pi}\). Since \(\pi \approx 3.14\), this simplifies to \(\frac{20}{6.28} \approx 3.18\).
06

Identify Closest Option

The available answer choices are in whole Hertz values. Since our calculated value is approximately \(3.18\mathrm{~Hz}\), the frequency closest to this value is \(3 \mathrm{~Hz}\). Therefore, option (b) \(3 \mathrm{~Hz}\) is the correct answer.

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Key Concepts

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

Natural Frequency
In harmonic oscillators, such as a two-mass system connected by a spring, **natural frequency** is the rate at which the system oscillates when it is not disturbed by outside forces. This concept is crucial because it determines how fast the system vibrates. Naturally occurring oscillations depend on the system's physical characteristics, like mass and spring stiffness.

In the context of a two-mass system, the natural frequency is influenced by the spring constant and the combined effect of the masses, calculated using their reduced mass. This frequency is determined using the formula: \[ f = \frac{1}{2\pi} \sqrt{\frac{k}{\mu}} \]where \( f \) is the natural frequency, \( k \) is the spring constant, and \( \mu \) is the reduced mass. This equation shows that a stiffer spring or a smaller reduced mass results in a higher natural frequency, meaning faster oscillations.
Reduced Mass
In a two-mass system connected by a spring, the **reduced mass** plays a vital role in understanding how the system behaves. It is not simply the average of the two masses but a specific calculation that allows us to treat the system as if it were a single mass. This reduced mass captures the inertia of both masses relative to each other.

The reduced mass \( \mu \) can be calculated using the formula: \[ \mu = \frac{m_1 m_2}{m_1 + m_2} \]This equation takes into account both masses, \( m_1 \) and \( m_2 \), and allows us to simplify the problem of two masses into one easier system. It is especially important in calculating the natural frequency, ensuring that we accurately factor in the dynamics of both masses. For example, with masses of 3 kg and 1 kg, the reduced mass is calculated as 0.75 kg.
Spring Constant
The **spring constant**, denoted as \( k \), is a measure of a spring's stiffness. In a physical system like a two-mass oscillator, the spring constant indicates how much force is needed to stretch or compress the spring by a unit length.

Mathematically, the spring constant is expressed via Hooke’s Law, \[ F = kx \]where \( F \) is the force applied, and \( x \) is the displacement of the spring from its equilibrium position. A higher spring constant means a stiffer spring, which requires more force to change its shape. In our example, the spring constant is 300 N/m, which tells us about the spring's stiffness and affects the system's natural frequency. Stiffer springs lead to higher natural frequencies due to their resistance to deformation.
Two-Mass System
A **two-mass system** connected by a spring is a fundamental model used in physics to understand harmonic oscillators. This model consists of two point masses at either end of a spring, and it helps us analyze the motion and interactions between the masses through the spring's force.

In such a system, each mass can move back and forth as the spring exerts force upon it, resulting in oscillations. The behavior of the system is characterized by the reduced mass and the spring constant. This example illustrates the importance of these parameters in determining the system's natural frequency.

The two-mass system is a simplification that allows us to apply mathematical models to understand more complex real-world systems, demonstrating fundamental oscillatory behavior found in numerous physical contexts, from molecular vibrations to mechanical systems.

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Most popular questions from this chapter

A block is performing S.H.M. along a vertical line with amplitude of \(40 \mathrm{~cm}\) on a horizontal plank. The block just lose the contact with plank when plank is momentarily at rest. Then : (Take \(g=10 \mathrm{~m} / \mathrm{s}^{-}\) ) (a) the period of its oscillation is \(\frac{2 \pi}{5} \mathrm{sec}\) (b) the period of its oscillation is \(\frac{2 \pi}{6} \mathrm{sec}\) (c) the period of its oscillation is \(\frac{\pi}{5} \mathrm{sec}\) (d) none of the above

A block of mass \(1 \mathrm{~kg}\) is connected with a massless spring of force constant 100 \(\mathrm{N} / \mathrm{m}\). At \(t=0\), a constant force \(F=10 \mathrm{~N}\) is applied on the block. The spring 8 is in its natural length at \(t=0\). The speed of particle at \(x=6 \mathrm{~cm}\) from mean position is : (a) \(4 \mathrm{~cm} / \mathrm{s}\) (b) \(10 \mathrm{~cm} / \mathrm{s}\) (c) \(80 \mathrm{~cm} / \mathrm{s}\) (d) \(50 \mathrm{~cm} / \mathrm{s}\)

The period of oscillation of a simple pendulum of length \(L\), suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination \(\alpha\), is given by : (a) \(2 \pi \sqrt{ \left.\frac{L}{\mid g \cos \alpha}\right)}\) (b) \(2 \pi \sqrt{\left(\frac{L}{g \sin \alpha}\right)}\) (c) \(2 \pi \sqrt{\left(\frac{L}{g}\right)}\) (d) \(2 \pi \sqrt{\left(\frac{L}{g \tan \alpha}\right)}\)

Ram say's, "The average value of displacement, velocity and acceleration for one time period in the case of S.H.M. is zero." Shyam say's "The acceleration of particle is maximum at extreme position." (a) Ram's statement is correct (b) Shyam's statement is correct (c) Both statements are wrong (d) Both (a) and (b) are correct

A body with speed \(v\) is moving along a straight line. At the same time it is at distance \(x\) from a fixed point on the line, the speed is given by \(v^{2}=144-9 x^{7}\). Then: (a) displacement of the body \(<\) distance moved by body (b) the magnitude of acceleration at a distance \(3 \mathrm{~m}\) from the fixed point is \(27 \mathrm{~m} / \mathrm{s}^{2}\) (c) the motion is S.H.M. with \(T=\frac{2 \pi}{3}\) unit (d) the maximum displacement from the fixed point is 4 unit (e) all the above

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