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Chapter 14: Problem 57

An unhappy \(0.300 \mathrm{~kg}\) rodent, moving on the end of a spring with force constant \(k=2.50 \mathrm{~N} / \mathrm{m},\) is acted on by a damping force \(F_{x}=-b v_{x}\). (a) If the constant \(b\) has the value \(0.900 \mathrm{~kg} / \mathrm{s},\) what is the frequency of oscillation of the rodent? (b) For what value of the constant \(b\) will the motion be critically damped?

Short Answer

Expert verified
The frequency of the oscillation of the rodent is given by \(\omega '\) calculated in step 2. The damping constant for the critical damping is \(b_c\) calculated in step 3.

Step by step solution

01

Calculate Frequency of Oscillation

The frequency \(\omega\) of the oscillation can be calculated using the formula \(\omega = [\sqrt{k/m}]\), where \(k = 2.50 N/m\) is the spring constant and \(m = 0.300 kg\) is mass of the rodent. Substituting \(k\) and \(m\) into the frequency formula yields \(\omega = \sqrt{2.50/0.300}\).
02

Calculate Final Frequency

We now adjust the frequency due to the damping force by subtracting \(b^2 / (4m^2)\) from \(\omega^2\) which gives the actual oscillation frequency \(\omega '\). Hence, \(\omega ' = \sqrt {\omega^2 - (b^2 / 4m^2)}\), where \(b = 0.90 kg/s\). We obtain \(\omega '\) by substituting the values of \(\omega\), \(b\), and \(m\) into the above equation.
03

Calculate Damping Constant for Critical Damping

For a system to be critically damped, the damping constant \(b_c\) has to be equal to \(2\sqrt{mk}\). Substituting the given values \(k = 2.50 N/m\) and \(m=0.300 kg\), we find the damping constant \(b_c\) when the system is critically damped.

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Key Concepts

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

Spring-Mass System
A spring-mass system is a fundamental concept often encountered in physics and mechanical engineering. It consists of a mass attached to a spring, allowing it to move back and forth when displaced from its equilibrium position. Imagine a small block connected to a spring. If you pull on the block and release it, the block will oscillate around its original position. This system is governed by Hooke's Law, which tells us that the force required to change the spring's length is proportional to the distance it is stretched or compressed. The force constant, known as the spring constant \( k \), describes how stiff the spring is. In this exercise, the spring constant is \( k = 2.50 \text{ N/m} \). The rodent acts as the mass \( m = 0.300 \text{ kg} \) in this system. Understanding how this system behaves is crucial for solving many real-world problems, such as vehicle suspension systems and even some biological processes.
  • Hooke's Law: \( F = -kx \)
  • Equation of Motion: \( m\frac{d^2 x}{dt^2} = -kx - b\frac{dx}{dt} \) where \( b \) is a damping factor.
Critical Damping
Critical damping is an important concept in the study of oscillatory systems, such as the spring-mass system. It refers to a damping setting that allows the system to return to equilibrium as quickly as possible, without oscillating. Imagine pushing a swinging door open and having it return to rest without wobbling back and forth. When a system is critically damped, it moves as rapidly back to its resting state as feasible without overshooting and oscillating around the equilibrium.In our spring-mass system example, to achieve critical damping, the damping constant \( b_c \) needs to equal \( 2\sqrt{mk} \). This relationship ensures that the resistive forces perfectly balance the system's inherent tendency to oscillate, preventing additional unwanted motion.
  • Critical Damping Condition: \( b = 2\sqrt{mk} \)
  • Equilibrium Return: Fastest return to equilibrium without oscillation.
Frequency of Oscillation
The frequency of oscillation is a measure of how many cycles of back-and-forth movement occur per unit time in an oscillating system, such as our spring-mass system. It is closely related to the natural tendency of the system to oscillate once displaced from equilibrium. For an undamped spring-mass system, the angular frequency \( \omega \) is determined by the spring constant \( k \) and the mass \( m \) using the formula \( \omega = \sqrt{\frac{k}{m}} \). However, when damping is present, this frequency is adjusted to account for the resistive damping forces. The new damped frequency \( \omega' \) is slightly lower and is calculated as \( \omega' = \sqrt{\omega^2 - \frac{b^2}{4m^2}} \).In our given problem, we determine this damped frequency using the damping constant \( b = 0.90 \text{ kg/s} \), the spring constant \( k = 2.50 \text{ N/m} \), and the mass \( m = 0.300 \text{ kg} \).
  • Frequency Formula (Undamped): \( \omega = \sqrt{\frac{k}{m}} \)
  • Frequency Formula (Damped): \( \omega' = \sqrt{\omega^2 - \frac{b^2}{4m^2}} \)

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Most popular questions from this chapter

A block of mass \(m\) is undergoing SHM on a horizontal, frictionless surface while attached to a light, horizontal spring. The spring has force constant \(k\), and the amplitude of the \(\mathrm{SHM}\) is \(A\). The block has \(v=0,\) and \(x=+A\) at \(t=0 .\) It first reaches \(x=0\) when \(t=T / 4\) where \(T\) is the period of the motion. (a) In terms of \(T,\) what is the time \(t\) when the block first reaches \(x=A / 2 ?\) (b) The block has its maximum speed when \(t=T / 4\). What is the value of \(t\) when the speed of the block first reaches the value \(v_{\max } / 2 ?\) (c) Does \(v=v_{\max } / 2\) when \(x=A / 2 ?\)

A slender rod of length \(80.0 \mathrm{~cm}\) and mass \(0.400 \mathrm{~kg}\) has its center of gravity at its geometrical center. But its density is not uniform; it increases by the same amount from the center of the rod out to either end. You want to determine the moment of inertia \(I_{\mathrm{cm}}\) of the rod for an axis perpendicular to the rod at its center, but you don't know its density as a function of distance along the rod, so you can't use an integration method to calculate \(I_{\mathrm{cm}}\). Therefore, you make the following measurements: You suspend the rod about an axis that is a distance \(d\) (measured in meters) above the center of the rod and measure the period \(T\) (measured in seconds) for small-amplitude oscillations about the axis. You repeat this for several values of \(d\). When you plot your data as \(T^{2}-4 \pi^{2} d / g\) versus \(1 / d\), the data lie close to a straight line that has slope \(0.320 \mathrm{~m} \cdot \mathrm{s}^{2} .\) What is the value of \(I_{\mathrm{cm}}\) for the rod?

The Silently Ringing Bell. A large, \(34.0 \mathrm{~kg}\) bell is hung from a wooden beam so it can swing back and forth with negligible friction. The bell's center of mass is \(0.60 \mathrm{~m}\) below the pivot. The bell's moment of inertia about an axis at the pivot is \(18.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\). The clapper is a small, \(1.8 \mathrm{~kg}\) mass attached to one end of a slender rod of length \(L\) and negligible mass. The other end of the rod is attached to the inside of the bell; the rod can swing freely about the same axis as the bell. What should be the length \(L\) of the clapper rod for the bell to ring silently - that is, for the period of oscillation for the bell to equal that of the clapper?

In a physics lab, you attach a \(0.200 \mathrm{~kg}\) air-track glider to the end of an ideal spring of negligible mass and start it oscillating. The elapsed time from when the glider first moves through the equilibrium point to the second time it moves through that point is \(2.60 \mathrm{~s}\). Find the spring's force constant.

A \(5.00 \mathrm{~kg}\) partridge is suspended from a pear tree by an ideal spring of negligible mass. When the partridge is pulled down \(0.100 \mathrm{~m}\) below its equilibrium position and released, it vibrates with a period of \(4.20 \mathrm{~s}\). (a) What is its speed as it passes through the equilibrium position? (b) What is its acceleration when it is \(0.050 \mathrm{~m}\) above the equilibrium position? (c) When it is moving upward, how much time is required for it to move from a point \(0.050 \mathrm{~m}\) below its equilibrium position to a point \(0.050 \mathrm{~m}\) above it? (d) The motion of the partridge is stopped, and then it is removed from the spring. How much does the spring shorten?
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