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Chapter 19: Problem 32

The force-deflection equation for a nonlinear spring fixed at one end is \(F=1.5 x^{112}\) where \(F\) is the force, expressed in newtons, applied at the other end and \(x\) is the deflection expressed in meters. (a) Determine the deflection \(x_{0}\) if a 4 -oz block is suspended from the spring and is at rest. (b) Assuming that the slope of the force-deflection curve at the point corresponding to this loading can be used as an equivalent spring constant, determine the frequency of vibration of the block if it is given a very small downward displacement from its equilibrium position and released.

Short Answer

Expert verified
Deflection: \( x_0 \approx 0.999 \, \text{m} \); Frequency of vibration: \( 4.8 \, \text{Hz} \).

Step by step solution

01

Convert Block Weight to Newtons

First, convert the weight of the block from ounces to newtons (N). Given that 1 oz is approximately 0.0283495 kg, and using the acceleration due to gravity as \(9.81 \, \text{m/s}^2\), a 4-oz block has a weight \( W = 4 \, \text{oz} \times 0.0283495 \, \text{kg/oz} \times 9.81 \, \text{m/s}^2 = 1.112 \, \text{N} \).
02

Solve for Deflection x鈧�

Since the block is at rest, the force from the block's weight (1.112 N) equals the spring force. Thus, solve the equation \( F = 1.5 x^{112} = 1.112 \) to find \( x_0 \). This results in \( x_0 = \left( \frac{1.112}{1.5} \right)^{\frac{1}{112}} \). Calculate this to find \( x_0 \approx 0.999\,\text{m} \).
03

Find Spring Constant Equivalent

The spring constant equivalent, \( k_{eq} \), is the derivative of the force with respect to deflection evaluated at \( x_0 \): \( k_{eq} = \frac{dF}{dx} = 1.5 \times 112 \times x^{111} \). Evaluate this at \( x = x_0 \): \( k_{eq} = 1.5 \times 112 \times (0.999)^{111} \approx 167.54 \, \text{N/m} \).
04

Solve for Natural Frequency

The frequency of vibration \( f \) can be found using the formula \( f = \frac{1}{2\pi} \sqrt{\frac{k_{eq}}{m}} \), where \( m \) is the mass of the block in kg (from Step 1). So, \( f = \frac{1}{2\pi} \sqrt{\frac{167.54}{0.113398}} \approx 4.8 \, \text{Hz} \).

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Key Concepts

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

Force-Deflection Equation
In the world of springs, the force-deflection equation helps us understand how much a spring stretches or compresses when a force is applied. Specifically, for a nonlinear spring, the relationship between the force and the deflection is not straightforward like in linear springs. Instead, the force is related to the deflection raised to a power.

For instance, the equation given is
  • \( F = 1.5 x^{112} \)
where \( F \) is the force (in newtons) and \( x \) is the deflection (in meters). This indicates that the force grows very rapidly with even a slight increase in deflection due to the high exponent.

For practical purposes, if a known weight is suspended and the spring is at rest, we can set this force equal to the weight's force to find the deflection \( x \) at equilibrium.
Natural Frequency
Natural frequency is an essential concept in understanding how systems vibrate. When an object like a block attached to a spring is displaced slightly from its resting position and then released, it tends to "bounce" back and forth at a certain rate, known as the natural frequency.

The vibration frequency depends on the mass of the object and the stiffness of the spring. For a system like our nonlinear spring example, if the spring has a high equivalent stiffness and the mass is low, the natural frequency will be high, meaning it vibrates quickly.

The formula to calculate the natural frequency \( f \) of the system is:
  • \[ f = \frac{1}{2\pi} \sqrt{\frac{k_{eq}}{m}} \]
where \( m \) is the mass of the block and \( k_{eq} \) is the equivalent spring constant. This shows how both mass and spring stiffness directly affect the frequency.
Equivalent Spring Constant
An equivalent spring constant \( k_{eq} \) helps us approximate the behavior of a nonlinear spring through a linear spring model. It serves as a single parameter to describe how stiff the spring is at a certain deflection.

For our nonlinear spring, the equivalent spring constant is the derivative of the force-deflection equation at the specific deflection point. This gives us the rate at which the force increases with deflection and is calculated as:
  • \( k_{eq} = \frac{dF}{dx} = 1.5 \times 112 \times x^{111} \)
At equilibrium position \( x = x_0 \), evaluating this equation gives the equivalent spring constant value \( k_{eq} \), which can then be used to find the system's natural frequency or to estimate the system's response to small disturbances.

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Most popular questions from this chapter

A \(6-\) lb slender rod is suspended from a steel wire that is known to have a torsional spring constant \(K=1.5 \mathrm{ft}\) lb/rad. If the rod is rotated through \(180^{\circ}\) about the vertical and released, determine (a) the period of oscillation, (b) the maximum velocity of end \(A\) of the rod.

A \(10-\) -lb block \(A\) rests on a \(40-\) lb plate \(B\) that is attached to an unstretched spring with a constant of \(k=60\) Ib/ft. Plate \(B\) is slowly moved 2.4 in. to the left and released from rest. Assuming that block \(A\) does not slip on the plate, determine ( \(a\) ) the amplitude and frequency of the resulting motion, \((b)\) the corresponding smallest allowable value of the coefficient of static friction.

A simplified model of a washing machine is shown. A bundle of wet clothes forms a weight \(w_{b}\) of 20 lb in the machine and causes a rotating unbalance. The rotating weight is \(\left.40 \text { lb (including } w_{b}\right)\) and the radius of the washer basket \(e\) is 9 in. Knowing the washer has an equivalent spring constant \(k=70\) lb/t and damping ratio \(\zeta=c / c_{c}=0.05\) and during the spin cycle the drum rotates at 250 rpm, determine the amplitude of the motion and the magnitude of the force transmitted to the sides of the washing machine.

A \(36-\) bb motor is bolted to a light horizontal beam that has a static deflection of 0.075 in. due to the weight of the motor. Knowing that the unbalance of the rotor is equivalent to a weight of 0.64 oz located 6.25 in. from the axis of rotation, determine the amplitude of the vibration of the motor at a speed of 900 rpm, assuming \((a)\) that no damping is present, \((b)\) that the damping factor \(c / c_{c}\) is equal to \(0.055 .\)

A particle moves in simple harmonic motion. Knowing that the maximum velocity is \(200 \mathrm{mm} / \mathrm{s}\) and the maximum acceleration is \(4 \mathrm{m} / \mathrm{s}^{2},\) determine the amplitude and frequency of the motion.
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