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Chapter 19: Problem 55

The 8 -kg uniform bar \(A B\) is hinged at \(C\) and is attached at \(A\) to a spring of constant \(k=500 \mathrm{N} / \mathrm{m}\). If end \(A\) is given a small displacement and released, determine \((a)\) the frequency of small oscillations, (b) the smallest value of the spring constant \(k\) for which oscillations will occur.

Short Answer

Expert verified
Frequency: Calculate using \( f = \frac{1}{2\pi} \sqrt{\frac{k L^2}{3m}} \). Minimum \( k \) ensures \( k L^2 > 0 \).

Step by step solution

01

Identify the governing equation

The frequency of small oscillations for a system involving a spring can be determined using the equation for the angular frequency of a simple harmonic oscillator, which is given by \( \omega = \sqrt{\frac{k}{m}} \), where \( k \) is the spring constant and \( m \) is the mass of the bar.
02

Calculate mass moment of inertia

Since the bar is hinged at point \( C \), we need to calculate the moment of inertia \( I_C \) about the pivot point. For a uniform bar of length \( L \), the moment of inertia about an axis through one end is \( I_C = \frac{1}{3}m L^2 \). The length \( L \) is given.
03

Relate angular frequency to moment of inertia

The frequency of oscillations depends on the angular frequency given by the equation \( \omega = \sqrt{\frac{k L^2}{3m}} \). Simplifying, we arrive at \( f = \frac{\omega}{2\pi} = \frac{1}{2\pi} \sqrt{\frac{k L^2}{3m}} \).
04

Calculate the frequency of oscillations

Substitute the values for \( k = 500 \text{ N/m} \), \( m = 8 \text{ kg}\), and the length \( L \) calculated in the previous step into the equation for \( f \) to find the frequency.
05

Determine the critical spring constant

To find the smallest value of the spring constant \( k \) for which oscillations will occur, recognize that this happens when \( \omega^2 > 0 \), so \( \frac{k L^2}{3m} > 0 \). The smallest \( k \) will thus ensure \( k L^2 > 0 \), solve for \( k \) in terms of \( L \) to find the minimum value ensuring oscillations. For simplicity, assume a standard length where \( L = 1 \).

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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 \( k \), is a measure of the stiffness of a spring. It describes how much force is needed to stretch or compress the spring by a unit length. A higher spring constant means a stiffer spring.

In the context of simple harmonic motion, the spring constant is crucial as it affects the angular frequency and, thus, the frequency of oscillations. In formulas for oscillating systems, the spring constant often appears as the numerator under a square root, indicating how stiffness impacts system dynamics.
  • The unit of the spring constant is Newtons per meter (\( \text{N/m} \)).
  • Its role is pivotal in determining both the angular frequency and the tendency of the object to return to equilibrium.
Mass Moment of Inertia
The mass moment of inertia, often symbolized as \( I_C \), is a measure of an object's resistance to rotational acceleration about an axis. It depends not only on mass but also on how that mass is distributed relative to the axis.

For a uniform bar hinged at one end, the formula used is \( I_C = \frac{1}{3} m L^2 \), where \( m \) is the mass and \( L \) is the length of the bar. Understanding this concept is essential for calculating how rotational dynamics come into play during oscillation.
  • The mass moment of inertia determines how much torque is needed for a desired angular acceleration.
  • It directly influences the angular frequency of oscillations in systems with rotational dynamics.
Angular Frequency
Angular frequency, denoted as \( \omega \), is an essential concept in simple harmonic motion. It represents how fast an object oscillates in radians per second around a circular path, even in linear cases like springs.

The formula \( \omega = \sqrt{\frac{k}{m}} \) describes it for a simple harmonic oscillator. However, in systems involving rotation, the adjusted formula might incorporate both the spring constant and the moment of inertia, as \( \omega = \sqrt{\frac{k L^2}{3m}} \).
  • Angular frequency offers insight into the periodic nature of oscillations, helping predict system behavior over time.
  • Understanding \( \omega \) gives clues on how quickly oscillations occur, influencing both practical applications and predictions.
Frequency of Oscillations
The frequency of oscillations refers to how often an oscillating object completes a full cycle per unit time. It is directly derived from the angular frequency, using the relationship \( f = \frac{\omega}{2\pi} \).

In practical terms, frequency helps us understand how quickly or slowly a system returns to its original position after displacement. For the given bar and spring system, substituting in the given values, including the spring constant \( k \) and the mass \( m \), one can determine the exact frequency.
  • Measured in Hertz (\( \text{Hz} \)), frequency tells us how the dynamics vary with different spring stiffness or mass amounts.
  • Finding the smallest value of \( k \) ensuring oscillations involves checking conditions using frequency formulas, ensuring positive oscillatory actions.

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Most popular questions from this chapter

A period of \(6.00 \mathrm{s}\) is observed for the angular oscillations of a \(4-\mathrm{oz}\) gyroscope rotor suspended from a wire as shown. Knowing that a period of \(3.80 \mathrm{s}\) is obtained when a 1.25 -in. -diameter steel sphere is suspended in the same fashion, determine the centroidal radius of gyration of the rotor. (Specific weight of steel \(=490 \mathrm{lb} / \mathrm{t}^{3}\) ).

A 4 -kg collar can slide on a frictionless horizontal rod and is attached to a spring with a constant of \(450 \mathrm{N} / \mathrm{m}\). It is acted upon by a periodic force with a magnitude of \(P=P_{m} \sin \omega_{f} t,\) where \(P_{m}=13 \mathrm{N}\). Determine the amplitude of the motion of the collar if \((a) \omega_{f}=5 \mathrm{rad} / \mathrm{s},\), \(\omega_{f}=10 \mathrm{rad} / \mathrm{s}\).

Three collars each have a mass \(m\) and are connected by pins to bars \(A C\) and \(B C,\) each having length \(l\) and negligible mass. Collars \(A\) and \(B\) can slide without friction on a horizontal rod and are connected by a spring of constant \(k\) Collar \(C\) can slide without friction on a vertical rod and the system is in equilibrium in the position shown. Knowing that collar \(C\) is given a small displacement and released, determine the frequency of the resulting motion of the system.

Two springs with constants \(k_{1}\) and \(k_{2}\) are connected in series to a block \(A\) that vibrates in simple harmonic motion with a period of 5 s. When the same two springs are connected in parallel to the same block, the block vibrates with a period of 2 s. Determine the ratio \(k_{1} / k_{2}\) of the two spring constants.

A 4 -kg collar can slide on a frictionless horizontal rod and is attached to a spring with constant \(k\). It is acted upon by a periodic force of magnitude \(P=P_{m} \sin \omega_{f} t,\) where \(P_{m}=9 \mathrm{N}\) and \(\omega_{f}=5 \mathrm{rad} / \mathrm{s}\). Determine the value of the spring constant \(k\) knowing that the motion of the collar has an amplitude of \(150 \mathrm{mm}\) and is \((a)\) in phase with the applied force, \((b)\) out of phase with the applied force.
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