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Chapter 16: Problem 9

9\. Electric Shaver In an electric shaver, the blade moves back and forth over a distance of \(2.0 \mathrm{~mm}\) in simple harmonic motion, with frequency \(120 \mathrm{~Hz}\). Find (a) the amplitude, (b) the maximum blade speed, and (c) the magnitude of the maximum blade acceleration.

Short Answer

Expert verified
Amplitude: 1.0 mm, Max Speed: 0.754 m/s, Max Acceleration: 181 m/s虏.

Step by step solution

01

- Understanding the Problem

Given: the distance over which the blade moves is 2.0 mm and the frequency of motion is 120 Hz. Required: (a) amplitude of motion, (b) maximum blade speed, and (c) maximum blade acceleration.
02

- Calculating the Amplitude

The amplitude is half of the total distance of motion. Given the total distance is 2.0 mm, the amplitude A is: \[ A = \frac{2.0\, \text{mm}}{2} = 1.0\, \text{mm} = 1.0 \times 10^{-3} \text{ m} \]
03

- Calculating the Maximum Blade Speed

The maximum speed (\(v_{max}\)) in simple harmonic motion can be calculated using the formula:\[ v_{max} = A \omega \] where \( \omega \) is the angular frequency. Angular frequency \( \omega \) is given by: \[ \omega = 2 \pi f \] Substituting the values: \[ \omega = 2 \pi \times 120 \mathrm{~Hz} = 240 \pi \mathrm{~rad/s} \] Therefore, \[ v_{max} = 1.0 \times 10^{-3} \text{ m} \times 240 \pi \mathrm{~rad/s} = 0.24 \pi \text{ m/s} \approx 0.754 \text{ m/s} \]
04

- Calculating the Maximum Blade Acceleration

The maximum acceleration (\(a_{max}\)) in simple harmonic motion is given by the formula: \[ a_{max} = A \omega^2 \] Using the previously calculated \( \omega \):\[ a_{max} = 1.0 \times 10^{-3} \text{ m} \times (240 \pi \mathrm{~rad/s})^2 \]Substitute and simplify:\[ a_{max} = 1.0 \times 10^{-3} \times 57,600 \pi^2 \mathrm{~m/s^2} \approx 181 \text{ m/s}^2 \]

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Key Concepts

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

Amplitude Calculation
In simple harmonic motion (SHM), the amplitude represents the maximum displacement from the equilibrium position. For this problem, the blade moves a total distance of 2.0 mm. Since the amplitude is half of this total distance, we can calculate the amplitude as follows:

Given: total distance = 2.0 mm.
Amplitude, A, is: \( A = \frac{2.0 \text{ mm}}{2} = 1.0 \text{ mm} = 1.0 \times 10^{-3} \text{ m}\)

This means the blade's maximum displacement from its equilibrium position is 1.0 mm.
Maximum Speed in SHM
The maximum speed of an object in simple harmonic motion occurs as it passes through the equilibrium position. This speed can be determined using the formula:

\( v_{max} = A \omega \)
where:
  • \( A \) is the amplitude
  • \( \omega \) is the angular frequency

First, we need the angular frequency, which is given by:

\( \omega = 2 \pi f \)
Given: frequency \( f = 120 \text{ Hz} \)
Therefore, \( \omega = 2 \pi \times 120 \text{ Hz} = 240 \pi \text{ rad/s} \)

Now we can find the maximum speed:

\( v_{max} = 1.0 \times 10^{-3} \text{ m} \times 240 \pi \text{ rad/s} = 0.24 \pi \text{ m/s} \approx 0.754 \text{ m/s} \)

This means the blade's maximum speed is approximately 0.754 m/s.
Angular Frequency
Angular frequency is a measure of how quickly an object moves through its cycle in simple harmonic motion. It is given by the formula:

\ \omega = 2 \pi f \
where:
  • \( \omega \) is the angular frequency
  • \( \pi \) is a constant (approximately 3.14159)
  • \( f \) is the frequency of motion

For this problem, the frequency is given as \ f = 120 \text{ Hz} \.

Calculating the angular frequency:

\ \omega = 2 \pi \times 120 \text{ Hz} = 240 \pi \text{ rad/s} \.
Angular frequency tells us how many radians per second the blade covers as it moves in its oscillating pattern. In this case, it's \ 240 \pi \text{ rad/s} \.
Maximum Acceleration in SHM
The maximum acceleration in simple harmonic motion occurs at the endpoints of the motion. It can be calculated using the formula:

\ a_{max} = A \omega^2 \
where:
  • \( a_{max} \) is the maximum acceleration
  • \( A \) is the amplitude
  • \( \omega \) is the angular frequency

With our previously calculated values:

\ a_{max} = 1.0 \times 10^{-3} \text{ m} \times (240 \pi \text{ rad/s})^2 = 1.0 \times 10^{-3} \times 57,600 \pi^2 \text{ m/s}^2 \approx 181 \text{ m/s}^2 \

This means the blade experiences a maximum acceleration of approximately \ 181 \text{ m/s}^2 \. Understanding how to calculate this is crucial for analyzing the dynamics of simple harmonic motion.

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Most popular questions from this chapter

70\. What's Wrong with cos? Observation of the oscillation of a mass on the end of a spring reveals that the detailed structure of the position as a function of time is fit very well by a function of the form $$ x(t)=X \cos \left(\omega t+\phi_{0}\right) $$ Yet subsequent observations give convincing evidence that this cannot be a good representation of the motion for long time periods. Explain what observation leads to this conclusion and resolve the apparent contradiction.

49\. Mechanical Energy Find the mechanical energy of a blockspring system having a spring constant of \(1.3 \mathrm{~N} / \mathrm{cm}\) and an oscillation amplitude of \(2.4 \mathrm{~cm}\).

14\. BMMD Astronauts sometimes use a device called a body-mass measuring device (BMMD). Designed for use on orbiting space vehicles, its purpose is to allow astronauts to measure their mass in the "weightless" conditions in Earth orbit. The BMMD is a springmounted chair; an astronaut measures his or her period of oscillation in the chair; the mass follows from the formula for the period of an oscillating block-spring system. (a) If \(M\) is the mass of the astronaut and \(m\) the effective mass of that part of the BMMD that also oscillates, show that $$ M=\left(k / 4 \pi^{2}\right) T^{2}-m $$ where \(T\) is the period of oscillation and \(k\) is the spring constant. (b) The spring constant was \(k=605.6 \mathrm{~N} / \mathrm{m}\) for the BMMD on Skylab Mission Two; the period of oscillation of the empty chair was \(0.90149 \mathrm{~s}\). Calculate the effective mass of the chair. (c) With an astronaut in the chair, the period of oscillation became \(2.08832 \mathrm{~s}\). Calculate the mass of the astronaut.

46\. Angular Amplitude For a simple pendulum, find the angular amplitude \(\Theta\) at which the restoring torque required for simple harmonic motion deviates from the actual restoring torque by \(1.0 \%\). (See "Trigonometric Expansions" in Appendix E.)

31\. Balance Wheel The balance wheel of a watch oscillates with a rotational amplitude of \(\pi\) rad and a period of \(0.500 \mathrm{~s}\) Find (a) the maximum rotational speed of the wheel, (b) the rotational speed of the wheel when its displacement is \(\pi / 2 \mathrm{rad}\), and (c) the magnitude of the rotational acceleration of the wheel when its displacement is \(\pi / 4 \mathrm{rad}\).
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