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Chapter 9: Problem 2

The root mean square value of a signal \(x(t), x_{\mathrm{rms}},\) is defined as $$ x_{\mathrm{rms}}=\left\\{\begin{array}{l} \left.\lim _{T \rightarrow \infty} \frac{1}{T} \int_{0}^{T} x^{2}(t) d t\right\\}^{1 / 2} \end{array}\right. $$ Using this definition, find the root mean square values of the displacement \(\left(x_{\mathrm{rms}}\right),\) velocity \(\left(\dot{x}_{\mathrm{rms}}\right),\) and acceleration \(\left(\ddot{x}_{\mathrm{rms}}\right)\) corresponding to \(x(t)=X \cos \omega t\).

Short Answer

Expert verified
The root mean square values for the displacement, velocity, and acceleration of the given signal \(x(t) = X \cos \omega t\) are as follows: Displacement rms value: \(x_{\mathrm{rms}} = \frac{X}{\sqrt{2}}\) Velocity rms value: \(\dot{x}_{\mathrm{rms}} = \frac{X\omega}{\sqrt{2}}\) Acceleration rms value: \(\ddot{x}_{\mathrm{rms}} = \frac{X{\omega}^2}{\sqrt{2}}\)

Step by step solution

01

1. Compute velocity and acceleration functions

First, we need to find the velocity and acceleration functions by differentiating the displacement function $$x(t) = X \cos \omega t$$ with respect to time t. \\ Velocity function: \(\dot{x}(t) = \frac{dx(t)}{dt} = -X\omega \sin \omega t\) Acceleration function: \(\ddot{x}(t) = \frac{d^2x(t)}{dt^2} = -X{\omega}^2 \cos \omega t\) Now that we have the functions for velocity and acceleration, we can use the given formula to calculate the root mean square values.
02

2. Substitute functions into the formula for rms value and evaluate

Displacement rms value: \(x_{\mathrm{rms}} = \left\lbrace\left.\lim _{T \rightarrow \infty} \frac{1}{T} \int_{0}^{T} (X \cos \omega t)^2 dt \right \rbrace^{1/2}\) Velocity rms value: \(\dot{x}_{\mathrm{rms}} = \left\lbrace\left.\lim _{T \rightarrow \infty} \frac{1}{T} \int_{0}^{T} (-X\omega \sin \omega t)^2 dt\right \rbrace^{1/2}\) Acceleration rms value: \(\ddot{x}_{\mathrm{rms}} = \left\lbrace\left.\lim _{T \rightarrow \infty} \frac{1}{T} \int_{0}^{T} (-X{\omega}^2 \cos \omega t)^2 dt\right \rbrace^{1/2}\) Now evaluate the integrals and compute the rms values for each: Displacement rms value: \(x_{\mathrm{rms}} = \left\lbrace\frac{1}{T} \int_{0}^{T} X^2 \cos ^2 \omega t dt\right\rbrace^{1/2}\) By using the trigonometric identity: \( \cos^2 \omega t = \frac{1+\cos 2\omega t }{2}\) \(x_{\mathrm{rms}} = \left\lbrace\frac{1}{T} \int_{0}^{T} \frac{X^2}{2}+\frac{X^2}{2}\cos 2\omega t \ dt\right\rbrace^{1/2}\) Taking limit and evaluating, \(x_{\mathrm{rms}} = (\frac{X^2}{2})^{1/2} = \frac{X}{\sqrt{2}}\) Velocity rms value: \(\dot{x}_{\mathrm{rms}} = \left\lbrace\frac{1}{T} \int_{0}^{T} X^2\omega^2 \sin ^2 \omega t dt\right \rbrace^{1/2}\) By using the trigonometric identity: \(\sin^2 \omega t = \frac{1-\cos 2\omega t}{2}\) \(\dot{x}_{\mathrm{rms}} = \left\lbrace\frac{1}{T} \int_{0}^{T} \frac{X^2\omega^2}{2}-\frac{X^2\omega^2}{2}\cos 2\omega t \ dt \right\rbrace^{1/2}\) Taking limit and evaluating, \(\dot{x}_{\mathrm{rms}} = (\frac{X^2\omega^2}{2})^{1/2} = \frac{X\omega}{\sqrt{2}}\) Acceleration rms value: \(\ddot{x}_{\mathrm{rms}} = \left\lbrace\frac{1}{T} \int_{0}^{T} X^2{\omega}^4 \cos ^2 \omega t dt\right \rbrace^{1/2}\) Using the trigonometric identity: \( \cos^2 \omega t = \frac{1+\cos 2\omega t }{2}\) \(\ddot{x}_{\mathrm{rms}} = \left\lbrace\frac{1}{T} \int_{0}^{T} \frac{X^2{\omega}^4}{2}+\frac{X^2{\omega}^4}{2}\cos 2\omega t \ dt\right\rbrace^{1/2}\) Taking limit and evaluating, \(\ddot{x}_{\mathrm{rms}} = (\frac{X^2{\omega}^4}{2})^{1/2} = \frac{X{\omega}^2}{\sqrt{2}}\) The root mean square values are: Displacement rms value: \(x_{\mathrm{rms}} = \frac{X}{\sqrt{2}}\) Velocity rms value: \(\dot{x}_{\mathrm{rms}} = \frac{X\omega}{\sqrt{2}}\) Acceleration rms value: \(\ddot{x}_{\mathrm{rms}} = \frac{X{\omega}^2}{\sqrt{2}}\)

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Key Concepts

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

Displacement
The concept of displacement in the context of root mean square (RMS) vibrations refers to the continuous movement of an object from its original position due to oscillations. In a vibrating system, displacement signifies how far an object moves from its equilibrium point. The formula given to calculate the RMS value of displacement for the function \(x(t) = X \cos \omega t\) illustrates this. Here, \(X\) is the amplitude, indicating the maximum range of movement from the central position, and \(\omega\) is the angular frequency. To determine the RMS value, integrate the square of the displacement function over time and then take its average. This involves using the trigonometric identity \(\cos^2 \omega t = \frac{1 + \cos 2\omega t}{2}\) to simplify the computation. By evaluating, we find the RMS of displacement as \(x_{\mathrm{rms}} = \frac{X}{\sqrt{2}}\). This value represents a type of "average" displacement over time, allowing for easier comparison and understanding of vibration magnitudes. It's a crucial parameter for analyzing the behavior of vibrating systems.
Velocity
In the study of vibrations, velocity is defined as the rate of change of displacement with respect to time. It's a pivotal aspect that illustrates how fast the position of a point changes within a given oscillatory wave. Starting with the displacement function \(x(t) = X \cos \omega t\), velocity can be derived as \(\dot{x}(t) = -X\omega \sin \omega t\) by differentiating the displacement function with respect to time.Calculating the RMS value of velocity involves integrating the square of this velocity function throughout time, ensuring to simplify using the trigonometric identity \(\sin^2 \omega t = \frac{1 - \cos 2\omega t}{2}\). Once simplified and integrated, the expression shows that \(\dot{x}_{\mathrm{rms}} = \frac{X\omega}{\sqrt{2}}\). This RMS value of velocity is crucial because it provides a measure of how rapidly the vibrating system travels back and forth over time, accounting for both the magnitude of the velocity and its distribution over the oscillation cycle. Recognizing this value helps us to predict forces acting on structures and components due to vibrations.
Acceleration
Acceleration in the realm of vibrations highlights the rate at which velocity changes over time. It serves as an indicator of how significantly the force impacts an object during oscillation. Using the original displacement function \(x(t) = X \cos \omega t\), acceleration can be derived as \(\ddot{x}(t) = -X{\omega}^2 \cos \omega t\) by differentiating the velocity function with respect to time.To find the RMS value of acceleration, the square of the acceleration function is integrated over time, employing the trigonometric identity for simplification where \( \cos^2 \omega t = \frac{1 + \cos 2\omega t}{2}\). After evaluating the integral, we find the RMS acceleration as \(\ddot{x}_{\mathrm{rms}} = \frac{X{\omega}^2}{\sqrt{2}}\). This RMS value provides insight into the vigor of the forces involved in the vibrating motion. Understanding these forces is vital in disciplines like mechanical engineering and physics, where vibrations could significantly affect the performance and safety of structures and machinery.

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Most popular questions from this chapter

During operation, the compressor unit of a refrigerator, with mass \(75 \mathrm{~kg}\) and rotational speed 900 rpm, experiences a dynamic force of \(200 \mathrm{~N}\). The compressor unit is supported on four identical springs, each with a stiffness of \(k\) and negligible damping. Find the value of \(k\) if only 15 percent of the dynamic force is to be transmitted to the support or base. Also, find the clearance space to be provided to the compressor unit.

Derive an expression for the displacement transmissibility of a damped single- degree-offreedom system whose base is subjected to a general periodic displacement.

The force transmissibility of a damped single-degree-of-freedom system with base motion is given by Eq. (9.106): $$ T_{f}=\frac{F_{t}}{k Y}=r^{2}\left\\{\frac{1+(2 \zeta r)^{2}}{\left(1-r^{2}\right)^{2}+(2 \zeta r)^{2}}\right\\}^{\frac{1}{2}} $$ where \(F_{t}\) is the magnitude of the force transmitted to the mass. Determine the frequency ratios (r) at which the force transmissibility attains maximum and minimum values. Discuss your results.

An electronic instrument, of mass \(20 \mathrm{~kg}\), is to be isolated to achieve a natural frequency of \(15 \mathrm{rad} / \mathrm{s}\) and a damping ratio of 0.95 . The available dashpots can produce a damping constant (c) in the range \(10 \mathrm{~N}-\mathrm{s} / \mathrm{m}\) to \(80 \mathrm{~N}-\mathrm{s} / \mathrm{m}\). Determine whether the desired damping ratio can be achieved using a passive system. If a passive system cannot be used, design a suitable active control system to achieve the desired damping ratio.

The armature of a variable-speed electric motor, of mass \(200 \mathrm{~kg}\), has an unbalance due to manufacturing errors. The motor is mounted on an isolator having a stiffness of \(10 \mathrm{kN} / \mathrm{m}\) and a dashpot having a damping ratio of 0.15 . (a) Find the speed range over which the amplitude of the fluctuating force transmitted to the foundation will be larger than the exciting force. (b) Find the speed range over which the transmitted force amplitude will be less than \(10 \%\) of the exciting force amplitude.
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