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Chapter 1: Problem 99

If the displacement of a machine is described as \(x(t)=0.4 \sin 4 t+5.0 \cos 4 t,\) where \(x\) is in centimetres and \(t\) is in seconds, find the expressions for the velocity and acceleration of the machine. Also find the amplitudes of displacement, velocity, and acceleration of the machine.

Short Answer

Expert verified
The expressions for the velocity and acceleration of the machine are \(v(t) = 1.6\cos(4t) - 20\sin(4t)\) and \(a(t) = -6.4\sin(4t) - 80\cos(4t)\), respectively. The amplitudes of displacement, velocity, and acceleration are approximately 5.02 cm, 20.06 cm/s, and 80.26 cm/s虏, respectively.

Step by step solution

01

Derive the velocity function

To find the velocity of the machine, we need to take the first derivative of the displacement function with respect to time. The given displacement function is: \[x(t) = 0.4\sin(4t) + 5.0\cos(4t)\] The first derivative of the sine function is the cosine function, and the first derivative of the cosine function is the negative sine function. So, we can derive the given function to find the velocity function: \[v(t) = \frac{d}{dt}(0.4\sin(4t) + 5.0\cos(4t))\] Using the chain rule, we get: \[v(t) = 0.4 \cdot 4 \cos(4t) - 5.0 \cdot 4\sin(4t)\] Simplifying this expression, we get: \[v(t) = 1.6\cos(4t) - 20\sin(4t)\]
02

Derive the acceleration function

To find the acceleration of the machine, we need to take the second derivative of the displacement function (or the first derivative of the velocity function) with respect to time. The velocity function is: \[v(t) = 1.6\cos(4t) - 20\sin(4t)\] Taking the derivative of the velocity function with respect to time, we get: \[a(t) = \frac{d}{dt}(1.6\cos(4t) - 20\sin(4t))\] Using the chain rule again, we get: \[a(t) = -1.6 \cdot 4 \sin(4t) - 20 \cdot 4\cos(4t)\] Simplifying this expression, we get: \[a(t) = -6.4\sin(4t) - 80\cos(4t)\]
03

Find amplitudes of displacement, velocity, and acceleration

Now, we have the displacement, velocity, and acceleration functions: \[x(t) = 0.4\sin(4t) + 5.0\cos(4t)\] \[v(t) = 1.6\cos(4t) - 20\sin(4t)\] \[a(t) = -6.4\sin(4t) - 80\cos(4t)\] To find the amplitudes of these functions, we need to determine the maximum values of these functions. We can use the formula for the amplitude of a sinusoidal function: Amplitude \(= \sqrt{A^2 + B^2}\), where A is the coefficient of the sine term and B is the coefficient of the cosine term. For displacement, A = 0.4 and B = 5.0. So, the amplitude of displacement is: Amplitude of displacement \(= \sqrt{0.4^2 + 5.0^2} = \sqrt{0.16 + 25} = \sqrt{25.16} \approx 5.02 \) cm For velocity, A = 20 and B = 1.6. So, the amplitude of velocity is: Amplitude of velocity \(= \sqrt{20^2 + 1.6^2} = \sqrt{400 + 2.56} = \sqrt{402.56} \approx 20.06 \) cm/s For acceleration, A = 6.4 and B = 80. So, the amplitude of acceleration is: Amplitude of acceleration \(= \sqrt{6.4^2 + 80^2} = \sqrt{40.96 + 6400} = \sqrt{6440.96} \approx 80.26 \) cm/s虏 Thus, the amplitudes of displacement, velocity, and acceleration are approximately 5.02 cm, 20.06 cm/s, and 80.26 cm/s虏, respectively.

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Key Concepts

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

Displacement Function
In the world of mechanical vibrations, understanding the displacement function is crucial for analyzing how an object moves over time. For this problem, the displacement of a machine is given by the function: \[ x(t) = 0.4 \sin(4t) + 5.0 \cos(4t) \]Here, \(x(t)\) represents the displacement, with \(t\) being time in seconds. The components \(0.4\sin(4t)\) and \(5.0\cos(4t)\) are sinusoidal, indicating a harmonic motion.
- The term \(0.4\sin(4t)\) describes the oscillatory motion based on a sine wave with a specific frequency and phase shift.- The term \(5.0\cos(4t)\) describes another oscillatory motion, this time represented by a cosine wave.
Combined, these provide a comprehensive depiction of how the displacement changes over time, giving insights into the dynamic behavior of the machine.
Velocity Derivation
Velocity is the rate at which an object changes its position. To derive the velocity from the displacement function, we calculate the first derivative of the displacement with respect to time. The displacement function is given as:\[ x(t) = 0.4 \sin(4t) + 5.0 \cos(4t) \]The velocity function \(v(t)\) is derived by differentiating \(x(t)\):\[ v(t) = \frac{d}{dt}(0.4 \sin(4t) + 5.0 \cos(4t)) \]Utilizing the chain rule:- The derivative of \(\sin(4t)\) is \(4 \cos(4t)\), and multiplied by \(0.4\) results in \(1.6\cos(4t)\).- The derivative of \(\cos(4t)\) is \(-4 \sin(4t)\), and multiplied by \(5.0\) results in \(-20\sin(4t)\).
Thus, the velocity function is:\[ v(t) = 1.6 \cos(4t) - 20 \sin(4t) \]This function illustrates how quickly and in what manner the machine's position changes over time.
Acceleration Analysis
Acceleration indicates the rate of change of velocity. For finding the acceleration, we need the derivative of the velocity function. From the velocity function \(v(t)\):\[ v(t) = 1.6 \cos(4t) - 20 \sin(4t) \]We derive the acceleration function \(a(t)\) by differentiating \(v(t)\):\[ a(t) = \frac{d}{dt}(1.6 \cos(4t) - 20 \sin(4t)) \]Again using the chain rule:- The derivative of \(\cos(4t)\) becomes \(-4 \sin(4t)\), and multiplied by \(1.6\) results in \(-6.4\sin(4t)\).- The derivative of \(\sin(4t)\) becomes \(4 \cos(4t)\), and multiplied by \(-20\) results in \(-80\cos(4t)\).
Thus, the acceleration function is:\[ a(t) = -6.4 \sin(4t) - 80 \cos(4t) \]This expression shows how the velocity of the machine changes at each moment, capturing the essence of its dynamic acceleration.
Amplitude Calculation
Amplitude defines the maximum extent of vibration in sinusoidal waveforms, identifying how large the oscillations are. To calculate the amplitude for displacement, velocity, and acceleration, we use:Amplitude \( = \sqrt{A^2 + B^2} \)This formula applies where \(A\) is the coefficient of the \(\sin\) term and \(B\) is the coefficient of the \(\cos\) term.
  • **Displacement**: For \(x(t) = 0.4 \sin(4t) + 5.0 \cos(4t)\), \(A = 0.4\) and \(B = 5.0\). Amplitude = \(\sqrt{0.4^2 + 5.0^2} = \sqrt{25.16} \approx 5.02\) cm.
  • **Velocity**: For \(v(t) = 1.6 \cos(4t) - 20 \sin(4t)\), \(A = 20\) and \(B = 1.6\). Amplitude = \(\sqrt{20^2 + 1.6^2} = \sqrt{402.56} \approx 20.06\) cm/s.
  • **Acceleration**: For \(a(t) = -6.4 \sin(4t) - 80 \cos(4t)\), \(A = 6.4\) and \(B = 80\). Amplitude = \(\sqrt{6.4^2 + 80^2} = \sqrt{6440.96} \approx 80.26\) cm/s虏.
These amplitudes characterize the extent and limits of motion, velocity, and acceleration of the machine, reflecting its full dynamic range.

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Most popular questions from this chapter

The shell-and-tube exchanger shown in Fig. \(1.125(\) a) can be modeled as shown in Fig. \(1.125(\mathrm{~b})\) for a simplified vibration analysis. Find the cross-sectional area of the tubes so that the total stiffness of the heat exchanger exceeds a value of \(200 \times 10^{6} \mathrm{~N} / \mathrm{m}\) in the axial direction and \(20 \times 10^{6} \mathrm{~N}-\mathrm{m} / \mathrm{rad}\) in the tangential direction. Assume that the tubes have the same length and cross section and are spaced uniformly.

The amplitudes of displacement and acceleration of an unbalanced turbine rotor are found to be \(0.5 \mathrm{~mm}\) and \(0.5 \mathrm{~g},\) respectively. Find the rotational speed of the rotor using the value of \(g\) as \(9.81 \mathrm{~m} / \mathrm{s}^{2}\)

The Fourier series of a periodic function, \(x(t),\) is an infinite series given by $$ x(t)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left(a_{n} \cos n \omega t+b_{n} \sin n \omega t\right) $$ where $$ \begin{array}{l} a_{0}=\frac{\omega}{\pi} \int_{0}^{\frac{2 \pi}{\omega}} x(t) d t \\ a_{n}=\frac{\omega}{\pi} \int_{0}^{\frac{2 \pi}{\omega}} x(t) \cos n \omega t d t \\ b_{n}=\frac{\omega}{\pi} \int_{0}^{\frac{2 \pi}{\omega}} x(t) \sin n \omega t d t \end{array} $$ \(\omega\) is the circular frequency and \(2 \pi / \omega\) is the time period. Instead of including the infinite number of terms in Eq. (E.1), it is often truncated by retaining only \(k\) terms as $$ x(t) \approx \tilde{x}(t)=\frac{\widetilde{a}_{0}}{2}+\sum_{n=1}^{k}\left(\widetilde{a}_{n} \cos n \omega t+\tilde{b}_{n} \sin n \omega t\right) $$ so that the error, \(e(t),\) becomes $$ e(t)=x(t)-\tilde{x}(t) $$ Find the coefficients \(\widetilde{a}_{0}, \widetilde{a}_{n},\) and \(\tilde{b}_{n}\) which minimize the square of the error over a time period: $$ \int_{-\frac{\pi}{\omega}}^{\frac{\pi}{\omega}} e^{2}(t) d t $$ Compare the expressions of \(\tilde{a}_{0}, \widetilde{a}_{n}\), and \(\widetilde{b}_{n}\) with Eqs. (E.2)-(E.4) and state your observation(s).

Use MATLAB to plot the variation of spring stiffness \((k)\) with deformation \((x)\) given by the relations: a. \(k=1000 x-100 x^{2} ; 0 \leq x \leq 4\). b. \(k=500+500 x^{2} ; 0 \leq x \leq 4\).

Show that any linear combination of \(\sin \omega t\) and \(\cos \omega t\) such that \(x(t)=A_{1} \cos \omega t+A_{2}\) \(\sin \omega t\left(A_{1}, A_{2}=\right.\) constants ) represents a simple harmonic motion.
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