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Chapter 14: Problem 35

Inside a NASA test vehicle, a \(3.50-\) kg ball is pulled along by a horizontal ideal spring fixed to a friction-free table. The force constant of the spring is 225 \(\mathrm{N} / \mathrm{m}\) . The vehicle has a steady acceleration of \(5.00 \mathrm{m} / \mathrm{s}^{2},\) and the ball is not oscillating. Suddenly, when the vehicle's speed has reached \(45.0 \mathrm{m} / \mathrm{s},\) its engines turn off, thus eliminating its acceleration but not its velocity. Find (a) the amplitude and (b) the frequency of the resulting oscillations of the ball. (c) What will be the ball's maximum speed relative to the vehicle?

Short Answer

Expert verified
Amplitude: 0.078 m, Frequency: 1.28 Hz, Maximum speed: 0.625 m/s.

Step by step solution

01

Determine Initial Displacement Due to Acceleration

Before the engines turn off, the ball is in equilibrium under the influence of both the spring force and the inertial force from the acceleration of the vehicle. The spring force is given by Hooke's law, i.e., \( F_s = kx_0 \), where \( k = 225 \ \mathrm{N/m} \) is the spring constant and \( x_0 \) is the initial displacement. Meanwhile, the inertial force is \( F_i = ma \), where \( m = 3.50 \ \mathrm{kg} \) and \( a = 5.00 \ \mathrm{m/s^2} \). Therefore, setting \( F_s = F_i \), we have:\[ kx_0 = ma \]\[ x_0 = \frac{ma}{k} = \frac{3.50 \times 5.00}{225} = 0.0778 \ \mathrm{m} \]
02

Determine Amplitude of Oscillation

Once the engines turn off, the acceleration vanishes, and the ball starts oscillating. The initial displacement due to the inertial force becomes the amplitude of the oscillation. Thus, the amplitude \( A \) is directly equal to the initial displacement calculated in the first step:\[ A = x_0 = 0.0778 \ \mathrm{m} \]
03

Calculate the Natural Frequency of the System

The natural frequency \( \omega \) of the oscillating spring system is given by:\[ \omega = \sqrt{\frac{k}{m}} \]Substituting the given values, we have:\[ \omega = \sqrt{\frac{225}{3.50}} \approx 8.03 \ \mathrm{rad/s} \]The frequency \( f \) is obtained from:\[ f = \frac{\omega}{2\pi} = \frac{8.03}{2\pi} \approx 1.28 \ \mathrm{Hz} \]
04

Find Maximum Speed of Ball Relative to Vehicle

The maximum speed \( v_{\text{max}} \) of the ball relative to the vehicle occurs when the ball passes through the equilibrium position of the oscillation. It is given by:\[ v_{\text{max}} = A\omega \]\[ v_{\text{max}} = 0.0778 \ times \ 8.03 \approx 0.625 \ \mathrm{m/s} \]

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Key Concepts

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

Spring Constant
A spring constant, often represented as \(k\), is a measure of how stiff a spring is. It depicts the force required to compress or extend the spring by a unit length. This characteristic is fundamental when studying harmonic oscillators, as it influences the system's behavior. In this exercise, the spring has a constant of 225 \(\mathrm{N/m}\).
This means that for every meter the spring is compressed or stretched from its equilibrium position, it exerts a force of 225 Newtons in the opposite direction. This property is crucial when calculating the spring force, which is the force exerted by the spring to return to its rest position.
The spring constant directly affects the system's natural frequency. Higher values of \(k\) imply that the spring is stiffer, leading to higher natural frequencies of oscillation. It's essential in calculating how quickly the system will oscillate.
Inertial Force
Inertial force is the apparent force that acts on all masses in an accelerating non-inertial frame of reference, such as the test vehicle in this case. It is not an actual force, but a perceived one due to the acceleration of the frame. Here, it is calculated using \( F_i = ma \) where \( m = 3.50 \ \mathrm{kg} \) (mass of the ball) and \( a = 5.00 \ \mathrm{m/s^2} \) (acceleration of the vehicle).
The inertial force in the exercise amounts to \( 3.50 \times 5.00 = 17.5 \ \mathrm{N} \). This force acts to pull the ball along with the vehicle's acceleration, creating an initial displacement in the spring.
Upon the cessation of vehicle acceleration, the inertial force disappears, but its effects remain, as the ball begins to oscillate. Understanding the inertial force helps grasp how an oscillatory motion initiates in this context.
Natural Frequency
The natural frequency of a system is the rate at which it oscillates when not subjected to any external force initially, other than the restoring force due to its displacement. It's an intrinsic property dependent on the mass-spring arrangement and expressed as \( \omega = \sqrt{\frac{k}{m}} \).
In this case, substituting the spring constant \(k = 225 \ \mathrm{N/m}\) and mass \(m = 3.50 \ \mathrm{kg}\) into the formula gives \( \omega \approx 8.03 \ \mathrm{rad/s} \).
Converting this angular frequency into a regular frequency using \( f = \frac{\omega}{2\pi} \) results in \( f \approx 1.28 \ \mathrm{Hz} \). This frequency tells us how many complete oscillations occur per second, depicting the rhythm of the system's oscillation.
Amplitude of Oscillation
The amplitude of oscillation is the greatest distance a point on the spring (here, the ball) moves from its equilibrium position during its motion. It represents the energy stored in the oscillating system resulting from the initial displacement.
Once the vehicle's acceleration stops, the initial displacement caused by the inertial force becomes the amplitude of oscillation. Here, it is directly obtained as \( A = 0.0778 \ \mathrm{m} \).
This value shows the maximum extent of motion on either side of the equilibrium position. Understanding amplitude is crucial as it influences the energy exchange between potential and kinetic forms during oscillation, although it does not alter the system's frequency.

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Most popular questions from this chapter

When a 0.750 -kg mass oscillates on an ideal spring, the frequency is 1.33 Hz. What will the frequency be if 0.220 kg are (a) added to the original mass and (b) subtracted from the original mass? Try to solve this problem without finding the force constant of the spring.

CP CALC An approximation for the potential energy of a KCl molecule is \(U=A\left[\left(R_{0}^{7} / 8 r^{8}\right)-1 / r\right],\) where \(R_{0}=2.67 \times\) \(10^{-10} \mathrm{m}, A=2.31 \times 10^{-28} \mathrm{J} \cdot \mathrm{m},\) and \(r\) is the distance between the two atoms. Using this approximation: (a) Show that the radial component of the force on each atom is \(F_{r}=A\left[\left(R_{0}^{7} / r^{9}\right)-1 / r^{2}\right].\) (b) Show that \(R_{0}\) is the equilibrium separation. (c) Find the minimum potential energy. (d) Use \(r=R_{0}+x\) and the first two terms of the binomial theorem \(\quad\) (Eq. 14.28\()\) to show that \(F_{r} \approx-\left(7 A / R_{0}^{3}\right) x,\) so that the molecule's force constant is \(k=7 A / R_{0}^{3} .\) (e) With both the \(K\) and \(C 1\) atoms vibrating in opposite directions on opposite sides of the molecule's center of mass, \(m_{1} m_{2} /\left(m_{1}+m_{2}\right)=3.06 \times 10^{-26} \mathrm{kg}\) is the mass to use in cal- culating the frequency. Calculate the frequency of small-amplitude vibrations.

You are watching an object that is moving in SHM. When the object is displaced 0.600 m to the right of its equilibrium position, it has a velocity of 2.20 \(\mathrm{m} / \mathrm{s}\) to the right and an acceleration of 8.40 \(\mathrm{m} / \mathrm{s}^{2}\) to the left. How much farther from this point will the object move before it stops momentarily and then starts to move back to the left?

The Silently Ringing Bell Problem. A large bell is hung from a wooden beam so it can swing back and forth with negligible friction. The center of mass of the bell is 0.60 m below the pivot, the bell has mass 34.0 \(\mathrm{kg}\) , and the moment of inertia of the bell 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 that has length \(L\) and negligible mass. The other end of the rod is attached to the inside of the bell so it 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 for the clapper?

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