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Chapter 14: Problem 18

(II) A tuning fork oscillates at a frequency of 441 \(\mathrm{Hz}\) and the tip of each prong moves 1.5 \(\mathrm{mm}\) to either side of center. Calculate \((a)\) the maximum speed and \((b)\) the maximum acceleration of the tip of a prong.

Short Answer

Expert verified
Maximum speed is approximately 4.16 m/s and maximum acceleration is approximately 11513.5 m/s虏.

Step by step solution

01

Identify the given values

We know the frequency of oscillation is \(f = 441\, \text{Hz}\) and the amplitude of oscillation is \(A = 1.5\, \text{mm} = 0.0015\, \text{m}\). We need to calculate maximum speed and maximum acceleration.
02

Calculate the angular frequency

The angular frequency \(\omega\) can be calculated using the formula:\[ \omega = 2\pi f \]Substitute \(f = 441\, \text{Hz}\):\[ \omega = 2\pi \times 441 \approx 2771.6\, \text{rad/s} \]
03

Calculate the maximum speed

The maximum speed \(v_{\text{max}}\) is given by:\[ v_{\text{max}} = A\omega \]Substitute \(A = 0.0015\, \text{m}\) and \(\omega = 2771.6\, \text{rad/s}\):\[ v_{\text{max}} = 0.0015 \times 2771.6 \approx 4.16\, \text{m/s} \]
04

Calculate the maximum acceleration

The maximum acceleration \(a_{\text{max}}\) is given by:\[ a_{\text{max}} = A\omega^2 \]Substitute \(A = 0.0015\, \text{m}\) and \(\omega = 2771.6\, \text{rad/s}\):\[ a_{\text{max}} = 0.0015 \times (2771.6)^2 \approx 11513.5\, \text{m/s}^2 \]

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Key Concepts

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

Angular Frequency
Angular frequency is a crucial concept in the study of harmonic motion. It describes how fast an object oscillates back and forth, measured in radians per second (rad/s).

Mathematically, you can find angular frequency, often represented by the Greek letter \( \omega \), using the formula:
  • \( \omega = 2\pi f \)
where \( f \) is the regular frequency in Hertz (Hz).

In the case of our tuning fork oscillating at 441 Hz, this formula transforms the frequency from cycles per second to radians per second:
  • \( \omega = 2\pi \times 441 \approx 2771.6 \text{ rad/s} \)
Understanding \( \omega \) is vital because it is directly used to calculate other important properties of oscillating objects, like the maximum speed and acceleration.
Maximum Speed
The maximum speed in harmonic motion refers to the highest velocity reached by the moving object. This speed occurs at the position where the object crosses the equilibrium point.

The formula to determine the maximum speed \( v_{\text{max}} \) is:
  • \( v_{\text{max}} = A\omega \)
where \( A \) is the amplitude and \( \omega \) is the angular frequency.
  • Amplitude \( A \) represent the maximum distance from the center position. In our problem, it鈥檚 0.0015 m (converted from 1.5 mm).
  • Angular frequency \( \omega \) is calculated as 2771.6 rad/s.
By substituting these values into the formula, we have:
  • \( v_{\text{max}} = 0.0015 \times 2771.6 \approx 4.16 \text{ m/s} \)
Thus, the tuning fork's prong reaches a maximum speed of approximately 4.16 m/s.
Maximum Acceleration
Maximum acceleration in harmonic motion is the highest rate of change of velocity that the moving object experiences. It happens when the object is at the maximum displacement from the equilibrium point.

The formula for calculating maximum acceleration \( a_{\text{max}} \) is:
  • \( a_{\text{max}} = A\omega^2 \)
where \( A \) again is the amplitude and \( \omega \) is the angular frequency.
  • In our scenario, the amplitude \( A = 0.0015 \text{ m} \), and angular frequency \( \omega = 2771.6 \text{ rad/s} \).
Inserting these numbers into our equation, we arrive at:
  • \( a_{\text{max}} = 0.0015 \times (2771.6)^2 \approx 11513.5 \text{ m/s}^2 \)
So, the prong of the tuning fork experiences a maximum acceleration of about 11513.5 \( \text{ m/s}^2 \). Understanding this helps in visualizing how quickly the speed can change in oscillating systems.

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Most popular questions from this chapter

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(1I) The amplitude of a driven harmonic oscillator reaches a value of 23.7\(F_{0} / k\) at a resonant frequency of 382 Hz. What is the \(Q\) value of this system?

In some diatomic molecules, the force each atom exerts on the other can be approximated by \(F=-C / r^{2}+D / r^{3}\) where \(r\) is the atomic separation and \(C\) and \(D\) are positive constants. ( \(a\) ) Graph \(F\) vs. \(r\) from \(r=0.8 D / C\) to \(r=4 D / C\). (b) Show that equilibrium occurs at \(r=r_{0}=D / C .\) (c) Let \(\Delta r=r-r_{0}\) be a small displacement from equilibrium, where \(\Delta r \ll r_{0} .\) Show that for such small displacements, the motion is approximately simple harmonic, and \((d)\) determine the force constant. ( \(e\) ) What is the period of such motion? [Hint: Assume one atom is kept at rest.]

(II) A small fly of mass 0.25 \(\mathrm{g}\) is caught in a spider's web. The web oscillates predominately with a frequency of 4.0 \(\mathrm{Hz}\) . (a) What is the value of the effective spring stiffness constant \(k\) for the web? (b) At what frequency would you expect the web to oscillate if an insect of mass 0.50 \(\mathrm{g}\) were trapped?

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