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Chapter 16: Problem 13

13\. Piston in Cylinder The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of \(0.76 \mathrm{~m}\). If the piston moves with simple harmonic motion with a frequency of 180 rev/min, what is its maximum speed?

Short Answer

Expert verified
7.166 m/s

Step by step solution

01

Determine the Frequency in SI Units

Frequency should be in hertz (Hz) for our calculations. Given the frequency is 180 revolutions per minute (rev/min), convert this to hertz. First, convert minutes to seconds: 180 rev/min 脳 (1 min / 60 s) = 3 Hz. Thus, the frequency is 3 Hz.
02

Calculate the Amplitude

Amplitude is half of the stroke. Given the stroke is 0.76 m, calculate the amplitude. Amplitude = Stroke / 2 = 0.76 m / 2 = 0.38 m.
03

Use the Formula for Maximum Speed

The formula for the maximum speed of an object in simple harmonic motion is given by: \[ v_{max} = 2 \pi f A \] where f is the frequency in Hz (3 Hz) A is the amplitude (0.38 m).
04

Plug in the Values

Now substitute the values of f and A into the formula: \[ v_{max} = 2 \pi \times 3 \times 0.38 \] \[ v_{max} 鈮� 7.166 m/s \] Thus, the maximum speed of the piston is approximately 7.166 m/s.

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Key Concepts

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

maximum speed
Understanding the maximum speed of a piston in simple harmonic motion is crucial. The maximum speed (usually denoted as \(v_{max}\)) occurs when the piston is at its equilibrium point. The piston鈥檚 speed is highest here because the entire kinetic energy is at its peak. To find the maximum speed, we use the formula:
\[ v_{max} = 2 \pi f A \]
In this formula:
- \( f \) is the frequency in hertz (Hz)
- \( A \) is the amplitude
In our problem:
- \( f = 3 \, \text{Hz} \)
- \( A = 0.38 \, \text{m} \)
Plugging in these values, we get:
\[ v_{max} 鈮� 7.166 \, \text{m/s} \]
This means the piston reaches a top speed of approximately 7.166 meters per second.
frequency conversion
Frequency conversion is often necessary to ensure calculations use consistent units. Here, we convert frequency from revolutions per minute (rev/min) to hertz (Hz).
Frequency given:
  • 180 rev/min

Swap minutes with seconds by dividing by 60 (since there are 60 seconds in a minute):
\[ 180 \, \text{rev/min} \times \left( \frac{1 \, \text{min}}{60 \, \text{s}} \right) = 3 \, \text{Hz} \]
Thus, the frequency in hertz (Hz) is 3. This is because 1 Hz represents one complete cycle per second.
amplitude calculation
Amplitude refers to the maximum extent of vibration or oscillation, measured from the central position (equilibrium). Here, the exercise provides the stroke of the piston, which is twice the amplitude. To find the amplitude:
  • Given: Stroke = 0.76 m

    • Dividing by 2 gives the amplitude:
  • Amplitude = Stroke / 2 = 0.76 m / 2 = 0.38 m

So, the amplitude of the piston's motion is 0.38 meters. This value is used to calculate the maximum speed of the piston in simple harmonic motion.
piston motion
Piston motion in this context follows the principles of simple harmonic motion (SHM). SHM describes the motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement. In the case of a piston:
  • The piston moves back and forth within the cylinder head of the locomotive.
  • Its motion can be described using sine or cosine functions.
  • The piston鈥檚 position varies sinusoidally with time.
  • It reaches its maximum speed at the equilibrium position.
  • The acceleration is greatest at the extreme positions (amplitude).
Since the piston鈥檚 motion is periodic and regular, understanding its speed and position at different times involves applying SHM principles and equations effectively.

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Most popular questions from this chapter

15\. Harbor At a certain harbor, the tides cause the ocean surface to rise and fall a distance \(d\) (from highest level to lowest level) in simple harmonic motion, with a period of \(12.5 \mathrm{~h} .\) How long does it take for the water to fall a distance \(d / 4\) from its highest level?

62\. Washboard Road A \(1000 \mathrm{~kg}\) car carrying four \(82 \mathrm{~kg}\) people travels over a rough "washboard" dirt road with corrugations \(4.0 \mathrm{~m}\) apart, which cause the car to bounce on its spring suspension. The car bounces with maximum amplitude when its speed is \(16 \mathrm{~km} / \mathrm{h}\). The car now stops, and the four people get out. By how much does the car body rise on its suspension due to this decrease in mass?

20\. Simple Harmonic Oscillator A simple harmonic oscillator consists of a block of mass \(2.00 \mathrm{~kg}\) attached to a spring of spring constant \(100 \mathrm{~N} / \mathrm{m} .\) When \(t=1.00 \mathrm{~s}\), the position and velocity of the block are \(x=0.129 \mathrm{~m}\) and \(v=3.415 \mathrm{~m} / \mathrm{s}\). (a) What is the amplitude of the oscillations? What were the (b) position and (c) velocity of the block at \(t=0 \mathrm{~s}\) ?

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.)

16\. Two Blocks In Fig. 16-32 two blocks \((m=1.0 \mathrm{~kg}\) and \(M=\) \(10 \mathrm{~kg}\) ) and a spring \((k=200 \mathrm{~N} / \mathrm{m})\) are arranged on a horizontal, frictionless surface. The coefficient of static friction between the two blocks is \(0.40\). What amplitude of simple harmonic motion of the spring-blocks system puts the smaller block on the verge of slipping over the larger block?
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