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The instrument shown has a mass of \(43 \mathrm{kg}\) and is spring-mounted to the horizontal base. If the amplitude of vertical vibration of the base is \(0.10 \mathrm{mm}\) calculate the range of frequencies \(f_{n}\) of the base vibration which must be prohibited if the amplitude of vertical vibration of the instrument is not to exceed \(0.15 \mathrm{mm}\). Each of the four identical springs has a stiffness of \(7.2 \mathrm{kN} / \mathrm{m}\)

Short Answer

Expert verified
The base vibration frequencies between 2.38 Hz and 5.31 Hz must be prohibited.

Step by step solution

01

Calculate the total stiffness of springs

Each of the four springs arranged in parallel has a stiffness of \(7.2 \text{kN/m}\). The total stiffness \(k\) can be calculated as:\[k = 4 \times 7.2 \times 10^3 \text{ N/m} = 28,800 \text{ N/m}\]
02

Determine the natural frequency

The natural frequency \(f_n\) for the undamped system can be found using the formula:\[f_n = \frac{1}{2\pi} \sqrt{\frac{k}{m}}\]Substitute \(k = 28800 \text{ N/m}\) and \(m = 43 \text{ kg}\):\[f_n = \frac{1}{2\pi} \sqrt{\frac{28800}{43}} \ \approx 4.115 \text{ Hz}\]
03

Identify the frequency range to be prohibited

Since the amplitude \(A\) of vibration of the instrument is not to exceed 0.15 mm, apply the relationship of maximum amplitude \(A_m\): \[A_m = A_b \times \frac{1}{|1-(f/f_n)^2|}\]where \(A_b = 0.10 \text{ mm}\) is the amplitude of the base. Solving for \(f\) such that:\[0.15 = 0.10 \times \frac{1}{|1-(f/f_n)^2|}\]\[\frac{1}{|1-(f/f_n)^2|}=1.5\] \[|1-(f/f_n)^2| = \frac{1}{1.5} \approx 0.6667\]This implies:\[(f/f_n)^2 = 0.3333 \quad \text{or} \quad 1.6667\]Taking the square root:\[f = f_n \times (0.5774 \text{ or } 1.2909)\]Substituting \(f_n\), the restricted frequencies are found:\[f = 2.376 \text{ Hz or } 5.31 \text{ Hz}\]
04

Conclude the prohibited frequency range

Therefore, the prohibited frequency range for the base vibration is between \(2.38 \text{ Hz and } 5.31 \text{ Hz}\). This is the range where the resonance would cause the instrument's amplitude to exceed 0.15 mm.

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Key Concepts

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

Natural Frequency Calculation
Understanding the natural frequency of a system is key in vibration analysis. It represents the frequency at which a system oscillates when not subjected to external forces except its own dynamics.
To calculate the natural frequency \(f_n\) of a spring-mass system, we make use of the following formula:
  • \(f_n = \frac{1}{2\pi} \sqrt{\frac{k}{m}}\)
Here, \(k\) is the total stiffness of the springs in the system, and \(m\) is the mass of the object. In our example, the total stiffness was calculated by summing the stiffness of all springs in parallel, resulting in \(28,800\; \text{N/m}\).
Substituting values, the natural frequency was determined to be approximately \(4.115\; \text{Hz}\). This frequency plays a crucial role in understanding how systems react to external vibrations.
Spring-mass System
A spring-mass system is foundational in understanding mechanical vibrations. It consists of a mass attached to one or more springs. This arrangement allows us to study how objects move under the influence of that spring force.
In this system, the springs create a restoring force proportional to the displacement from an equilibrium position. In this example, a mass of \(43\; \text{kg}\) was mounted on four identical springs each with a stiffness of \(7.2\; \text{kN/m}\).
Spring-mass systems are important for modeling oscillatory motion, making them invaluable in engineering and physics fields.
Vibration Analysis
Vibration analysis involves studying the oscillations of a system to predict its response to different disturbances. It is an essential process in avoiding damage caused by resonance.
Resonance happens when the frequency of an external force matches the natural frequency of a system. This can lead to predominantly larger amplitude vibrations.
In our example, prohibiting frequencies that could cause excessive amplitudes ensured the instrument's vibrations did not exceed \(0.15\; \text{mm}\). By analyzing vibrations, we determine frequencies to avoid, which shields structures from resonance-related damage.
Amplitude of Vibration
Amplitude in vibration analysis represents the maximum displacement of the system from its rest position. Larger amplitudes indicate more considerable energy in the oscillation.
The task at hand was ensuring the instrument's amplitude did not exceed \(0.15\; \text{mm}\). This constraint is crucial for maintaining its operational effectiveness and preventing damage.
The formula connecting base and system amplitudes was utilized:
  • \(A_m = A_b \times \frac{1}{|1-(f/f_n)^2|}\)
Here, \(A_b\) is the base amplitude, and \(f\) is the frequency of operation. Managing the amplitude avoids problems like system fatigue and failure.

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Most popular questions from this chapter

A single-cylinder four-stroke gasoline engine with a mass of \(90 \mathrm{kg}\) is mounted on four stiff spring pads, each with a stiffness of \(30\left(10^{3}\right) \mathrm{kN} / \mathrm{m},\) and is designed to run at 3600 rev/min. The mounting system is equipped with viscous dampers which have a large enough combined viscous damping coefficient \(c\) so that the system is critically damped when it is given a vertical displacement and then released while not running. When the engine is running, it fires every other revolution, causing a periodic vertical displacement modeled by \(1.2 \cos \omega t \mathrm{mm}\) with \(t\) in seconds. Determine the magnification factor \(M\) and the overall damping coefficient \(c\)

The block of weight \(W=100\) lb is suspended by two springs each of stiffness \(k=200 \mathrm{lb} / \mathrm{ft}\) and is acted upon by the force \(F=75 \cos 15 t\) lb where \(t\) is the time in seconds. Determine the amplitude \(X\) of the steady-state motion if the viscous damping coefficient \(c\) is \((a) 0\) and (b) 60 lb-sec/ft. Compare these amplitudes to the static spring deflection \(\delta_{\mathrm{st}}\)

The addition of damping to an undamped springmass system causes its period to increase by 25 percent. Determine the damping ratio \(\zeta\)

A 200 -kg machine rests on four floor mounts, each of which has an effective spring constant \(k=250\) \(\mathrm{kN} / \mathrm{m}\) and an effective viscous damping coefficient \(c=1000 \mathrm{N} \cdot \mathrm{s} / \mathrm{m} .\) The floor is known to vibrate vertically with a frequency of \(24 \mathrm{Hz}\). What would be the effect on the amplitude of the absolute machine oscillation if the mounts were replaced with new ones which have the same effective spring constant but twice the effective damping coefficient?

During the design of the spring-support system for the 4000 -kg weighing platform, it is decided that the frequency of free vertical vibration in the unloaded condition shall not exceed 3 cycles per second. (a) Determine the maximum acceptable spring constant \(k\) for each of the three identical springs. (b) For this spring constant, what would be the natural frequency \(f_{n}\) of vertical vibration of the platform loaded by the 40 -Mg truck?

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