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Chapter 12: Problem 81

The maximum speed of a \(3.1-\mathrm{kg}\) mass attached to a spring is \(0.68 \mathrm{m} / \mathrm{s},\) and the maximum force exerted on the mass is \(11 \mathrm{N}\). (a) What is the amplitude of motion for this mass? (b) What is the force constant of the spring? (c) What is the frequency of this system?

Short Answer

Expert verified
(a) Amplitude \( A = 0.34 \) m; (b) Spring constant \( k = 32.35 \) N/m; (c) Frequency \( f = 0.92 \) Hz.

Step by step solution

01

Introduction to Amplitude

To find the amplitude of motion, we can use the formula for the maximum speed in simple harmonic motion: \( v_{max} = \omega A \), where \( \omega \) is the angular frequency and \( A \) is the amplitude.
02

Using Angular Frequency Relationship

First, we need to express angular frequency in terms of given parameters. Angular frequency \( \omega \) is given by \( \omega = \sqrt{\frac{k}{m}} \), where \( k \) is the spring constant and \( m \) is the mass. We will substitute \( \omega \) in the amplitude equation later.
03

Finding the Force Constant

Given the maximum force \( F = k \cdot A = 11 \text{ N} \) and knowing \( F = kA \), we solve for \( k \) by substituting \( A = \frac{v_{max}}{\omega} \).
04

Solve for Amplitude

From previous calculations, \( v_{max} = 0.68 \mathrm{m/s} \) and \( F_{max} = 11 \mathrm{N} \). Using \( k = m\cdot\omega^2 \), substitute \( k \) back into \( F = kA \) to solve for \( A \). So, \( A = \frac{F_{max}}{k} = \frac{11}{k} \).
05

Calculating Angular Frequency

Once you know the amplitude, \( k = \frac{11 \text{ N}}{A} \). Substitute in \( \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{11}{3.1 A}} \).
06

Calculate Frequency

The frequency \( f \) is related to angular frequency by \( f = \frac{\omega}{2\pi} \). Use the value of \( \omega \) from previous steps to find this. Plug in the values to calculate \( f \).

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Key Concepts

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

Amplitude of Motion
In simple harmonic motion, the amplitude (A) refers to the maximum displacement of a mass from its equilibrium position. It's like the highest point a swing reaches when you push it. For a spring-mass system, we determine the amplitude using the relation between maximum speed (v_{max}) and angular frequency (\omega):
  • Maximum speed is given by the equation \( v_{max} = \omega A \).
  • Angular frequency is \( \omega = \sqrt{\frac{k}{m}} \), where \( k \) is the spring constant and \( m \) is mass.
  • Rearrange these formulas to find \( A = \frac{v_{max}}{\omega} \).
To calculate the amplitude, you need to first find the angular frequency using given information about the spring constant and mass. After determining \omega, use it to solve for A. Remember that the amplitude tells us how far the object moves from the midpoint of its path, a key feature of oscillatory systems.
Force Constant
The force constant (k), or spring constant, measures a spring's stiffness. It specifies how much force the spring exerts per unit of displacement. When you push down on a spring and feel resistance, that resistance is related to the spring constant. To find \( k \) with simple harmonic motion:
  • The formula is \( F_{max} = kA \), where \( F_{max} \)is the maximum force.
  • Rearrange to find \( k = \frac{F_{max}}{A} \).
  • Hence, given \( F_{max} = 11 \, \mathrm{N} \), you can find \( k \)if you know \( A \).
The larger the \( k \), the stiffer the spring. It is crucial to solve for \( k \)and \( A \)iteratively, given they depend on each other, to explore the spring's behavior in a mechanical system. This interdependent relationship is key in mechanical and physical system analyses.
Frequency Calculation
Once the angular frequency (\omega) is determined, calculating the frequency (f) of a simple harmonic oscillator becomes straightforward. Frequency is how often the motion repeats per unit of time, typically measured in cycles per second or Hertz (Hz). Follow these steps:
  • Use the relationship \( f = \frac{\omega}{2\pi} \)to convert angular frequency to frequency in Hertz.
  • With \( \omega = \sqrt{\frac{k}{m}} \), plug in the force constant and mass values.
Remember:
  • Higher frequency implies faster oscillations.
  • The frequency is independent of the amplitude.
Calculating the frequency helps predict how quickly systems respond to external forces, fundamental in designing systems that must resonate at specific frequencies.

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Most popular questions from this chapter

A 0.321 -kg mass is attached to a spring with a force constant of \(13.3 \mathrm{N} / \mathrm{m}\). If the mass is displaced \(0.256 \mathrm{m}\) from equilibrium and released, what is its speed when it is \(0.128 \mathrm{m}\) from equilibrium?

A mass moves back and forth in simple harmonic motion with amplitude \(A\) and period \(T .\) (a) In terms of \(T\), how long does it take for the mass to move through a total distance of \(2 A ?\) (b) How long does it take for the mass to move through a total distance of \(3 A ?\)

Suspended from the ceiling of an elevator is a simple pendulum of length \(L .\) What is the period of this pendulum if the elevator \((a)\) accelerates upward with an acceleration \(a,\) or \((b)\) accelerates downward with an acceleration whose magnitude is greater than zero but less than \(g\) ? Give your answer in terms of \(L, g\), and \(a\).

When a mass \(m\) is attached to a vertical spring with a force constant \(k,\) it stretches the spring by the amount \(L\). Calculate (a) the period of this mass and (b) the period of a simple pendulum of length \(L\)

A grandfather clock keeps correct time at sea level. If the clock is taken to the top of a nearby mountain, (a) would you expect it to keep correct time, run slow, or run fast? (b) Choose the best explanation from among the following: I. Gravity is weaker at the top of the mountain, leading to a greater period of oscillation. II. The length of the pendulum is unchanged, and therefore its period remains the same. III. The extra gravity from the mountain causes the period to decrease.
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