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Chapter 15: Problem 69

In the engine of a locomotive, a cylindrical piece known as a piston oscillates in \(\mathrm{SHM}\) in a cylinder head (cylindrical chamber) with an angular frequency of 180 rev/min. Its stroke (twice the amplitude) is \(0.76 \mathrm{~m}\). What is its maximum speed?

Short Answer

Expert verified
The maximum speed of the piston is approximately 7.16 m/s.

Step by step solution

01

Convert Angular Frequency

The angular frequency \( \omega \) is given in revolutions per minute, which we need to convert to radians per second for use in calculations. Given: \( \omega = 180 \) revolutions per minute.Since \( 1 \) revolution equals \( 2\pi \) radians, and there are \( 60 \) seconds in a minute, we convert as follows:\[ \omega = 180 \times \frac{2\pi}{60} = 6\pi \text{ rad/s} \]
02

Determine Amplitude

The stroke of the piston is the full distance the piston travels back and forth and is equal to twice the amplitude. We are given that the stroke is \(0.76 \mathrm{~m}\).Thus, the amplitude \(A\) is:\[ A = \frac{0.76}{2} = 0.38 \text{ m} \]
03

Calculate Maximum Speed

The maximum speed \( v_{max} \) in simple harmonic motion is given by the formula \( v_{max} = A\omega \).Plug in the values we have:\[ v_{max} = 0.38 \times 6\pi = 2.28\pi \text{ m/s} \]Calculate to find the numerical value:\[ v_{max} = 2.28 \times 3.14159 \approx 7.16 \text{ m/s} \]

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Key Concepts

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

Angular Frequency
Angular frequency provides a way to describe how fast an oscillation occurs within simple harmonic motion (SHM). In the context of the locomotive's piston, it is initially expressed as 180 revolutions per minute (rev/min). However, for calculations in physics, it's preferable to use radians per second (rad/s).
  • Conversion involves recognizing that one complete revolution is equivalent to a full circle, or \( 2\pi \) radians.
  • There are 60 seconds in a minute, which serves as a time conversion factor.
  • Thus, converting 180 rev/min to rad/s involves the equation: \( \omega = 180 \times \frac{2\pi}{60} = 6\pi \text{ rad/s} \).
It's crucial because angular frequency \( \omega \) is used to find other key properties in SHM, such as maximum speed and force dynamics. This conversion simplifies utilizing other SHM equations.
Amplitude
In simple harmonic motion, the amplitude \(A\) represents the maximum displacement from the equilibrium position. Imagine how far you swing a pendulum from its rest position; that's the amplitude.
  • Given the exercise, twice the amplitude is known as the stroke of the piston, which is \(0.76 \text{ m}\).
  • This makes the amplitude \(A\) half of the stroke: \(\frac{0.76}{2} = 0.38\text{ m}\).
Amplitude is vital in determining the extreme points of oscillation where the speed of the oscillating particle changes direction. It reflects the energy in SHM, influencing how forcefully a system might behave under oscillation.
Maximum Speed
Maximum speed in SHM occurs when the oscillating object passes through its equilibrium position. At this point, all the system's energy is kinetic, and the position-related potential energy is minimal.
  • The formula to calculate maximum speed is \( v_{max} = A\omega \), where \(A\) is the amplitude and \(\omega\) is the angular frequency.
  • From the exercise, the amplitude \(A\) is \(0.38 \text{ m}\) and \(\omega\) is \(6\pi \text{ rad/s}\).
  • Substitute these values into the formula: \( v_{max} = 0.38 \times 6\pi = 2.28\pi \text{ m/s} \).
  • After calculating, this results in approximately \( v_{max} \approx 7.16 \text{ m/s} \).
This shows the swiftest speed the piston achieves during its cycle, an essential aspect in designing mechanical components to cope with the forces during high-speed operations.

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Most popular questions from this chapter

A (hypothetical) large slingshot is stretched \(2.30 \mathrm{~m}\) to launch a \(170 \mathrm{~g}\) projectile with speed sufficient to escape from Earth \((11.2 \mathrm{~km} / \mathrm{s})\). Assume the elastic bands of the slingshot obey Hooke's law. (a) What is the spring constant of the device if all the elastic potential energy is converted to kinetic energy? (b) Assume that an average person can exert a force of \(490 \mathrm{~N}\). How many people are required to stretch the elastic bands?

A physical pendulum consists of a meter stick that is pivoted at a small hole drilled through the stick a distance \(d\) from the \(50 \mathrm{~cm}\) mark. The period of oscillation is \(2.5 \mathrm{~s} .\) Find \(d\)

A 10 g particle undergoes SHM with an amplitude of 2.0 mm, a maximum acceleration of magnitude \(8.0 \times 10^{3} \mathrm{~m} / \mathrm{s}^{2},\) and an unknown phase constant \(\phi .\) What are (a) the period of the motion, (b) the maximum speed of the particle, and (c) the total mechanical energy of the oscillator? What is the magnitude of the force on the particle when the particle is at (d) its maximum displacement and (e) half its maximum displacement?

\(\cdot 45\) A A performer seated on a trapeze is swinging back and forth with a period of \(8.85 \mathrm{~s}\). If she stands up, thus raising the center of mass of the trapeze \(+\) performer system by \(35.0 \mathrm{~cm}\), what will be the new period of the system? Treat trapeze + performer as a simple pendulum.

A block of mass \(M=5.4 \mathrm{~kg},\) at rest on a horizontal frictionless table, is attached to a rigid support by a spring of constant \(k=6000 \mathrm{~N} / \mathrm{m} .\) A bullet of mass \(m=9.5 \mathrm{~g}\) and velocity \(\vec{v}\) of magnitude \(630 \mathrm{~m} / \mathrm{s}\) strikes and is embedded in the block (Fig. 15-40). Assuming the compression of the spring is negligible until the bullet is embedded, determine (a) the speed of the block immediately after the collision and (b) the amplitude of the resulting simple harmonic motion.
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