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A tuning fork labeled 392 Hz has the tip of each of its two prongs vibrating with an amplitude of 0.600 \(\mathrm{mm}\) . (a) What is the maximum speed of the tip of a prong? (b) A housefly (Musca domestica) with mass 0.0270 \(\mathrm{g}\) is holding on to the tip of one of the prongs. As the prong vibrates, what is the fly's maximum kinetic energy? Assume that the fly's mass has a negligible effect on the frequency of oscillation.

Short Answer

Expert verified
(a) Calculate \( v_{max} \) using \( \omega \times A \). (b) Use \( KE = \frac{1}{2}mv^2 \).

Step by step solution

01

Understand the Relation

The speed of the vibrating prong can be described by simple harmonic motion, where the maximum speed is given by the formula \( v_{max} = \, \omega \times A \). Here, \( \omega \) is the angular frequency and \( A \) is the amplitude of vibration.
02

Calculate Angular Frequency

The angular frequency \( \omega \) can be calculated using the formula \( \omega = 2\pi f \), where \( f = 392 \) Hz. Thus, \( \omega = 2 \times 3.1416 \times 392 \; \mathrm{rad/s} \).
03

Calculate Maximum Speed

Substitute \( \omega \) and \( A = 0.600 \; \mathrm{mm} = 0.0006 \; \mathrm{m} \) into the equation \( v_{max} = \omega \times A \). Calculate \( v_{max} \).
04

Find the Fly's Maximum Kinetic Energy

Use the formula for kinetic energy, \( KE_{max} = \frac{1}{2}mv_{max}^2 \). First, convert the mass of the fly from grams to kilograms: \( 0.0270 \; \mathrm{g} = 0.000027 \; \mathrm{kg} \). Then, substitute \( m \) and \( v_{max} \) to find the maximum kinetic energy.

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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 key concept when dealing with simple harmonic motion. It describes how quickly an object undergoes oscillation and is often denoted by the Greek letter \( \omega \). In the context of a tuning fork's vibration, determining the angular frequency is crucial for understanding how fast each prong vibrates.

To calculate angular frequency, the formula \( \omega = 2\pi f \) is used, where \( f \) is the frequency of the vibration in Hertz (Hz). For instance, a tuning fork labeled as 392 Hz has a frequency \( f = 392 \; \mathrm{Hz} \). Using the formula, we find:
  • \( \omega = 2 \times 3.1416 \times 392 \; \mathrm{rad/s} \)
  • This simplifies to \( \omega \approx 2461 \; \mathrm{rad/s} \)
Understanding angular frequency allows us to compute other important quantities in simple harmonic motion, such as the maximum speed and kinetic energy of an oscillating object.
Kinetic Energy
Kinetic energy plays a pivotal role in understanding the dynamic behavior of objects in motion. For a vibrating tuning fork, the kinetic energy at its maximum is when the prong is moving at its fastest.

The formula used to calculate kinetic energy is \( KE = \frac{1}{2} mv^2 \), where \( m \) is the mass, and \( v \) is the velocity. In the case where a small object, like a housefly with mass \( 0.0270 \; \mathrm{g} \), holds onto a prong, it too will experience motion.

First, one must convert the mass of the fly to kilograms \( (0.000027 \; \mathrm{kg}) \). Then, using the maximum speed already calculated from the angular frequency and amplitude, substitute the values into the kinetic energy formula:
  • Suppose \( v_{max} \) is calculated as some value in \( \mathrm{m/s} \)
  • The resulting kinetic energy \( KE \) can be found by plugging these numbers in, providing insight into the energy experienced by the fly.
This kinetic energy is impressive for its small mass, showcasing the power of motion imparted by the vibrating prong.
Amplitude of Vibration
Amplitude of vibration is another fundamental aspect of simple harmonic motion. It represents the maximum extent of displacement from the rest position.

For the tuning fork, the amplitude is given as 0.600 mm, which can be conveniently transformed into meters (0.0006 m) for calculation purposes. Amplitude is crucial in the equation that determines the maximum speed \( (v_{max}) \) of the vibrating object, which is expressed as \( v_{max} = \omega \times A \).

In this context, amplitude refers to how far the prong moves in its vibration, and is a critical factor in determining both the speed and energy of the prong's motion.
  • A larger amplitude means more potential energy stored in the system due to its greater displacement potential.
  • This affects the overall speed calculations, since energy and speed are interconnected in simple harmonic motions.
Comprehending the amplitude helps us understand the rhythmic nature of the oscillating system and its potential influences on attached objects, like the housefly in the exercise scenario.

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Most popular questions from this chapter

If an object on a horizontal, frictionless surface is attached to a spring, displaced, and then released, it will oscillate. If it is displaced 0.120 \(\mathrm{m}\) from its equilibrium position and released with zero initial speed, then after 0.800 \(\mathrm{s}\) its displacement is found to be 0.120 \(\mathrm{m}\) on the opposite side, and it has passed the equilibrium position once during this interval. Find (a) the amplitude; (b) the period; (c) the frequency.

Two identical atoms in a diatomic molecule vibrate as harmonic oscillators. However, their center of mass, midway between them, remains at rest. (a) Show that at any instant, the momenta of the atoms relative to the center of mass are \(\vec{p}\) and \(-\vec{p} .\) (b) Show that the total kinetic energy \(K\) of the two atoms at any instant is the same as that of a single object with mass \(m / 2\) with a momentum of magnitude \(p .\) (Use \(K=p^{2} / 2 m . )\) This result shows why \(m / 2\) should be used in the expression for \(f\) in have masses \(m_{1}\) and \(m_{2},\) show that the result of part (a) still holds and the single object's mass in part (b) is \(m_{1} m_{2} /\left(m_{1}+m_{2}\right) .\) The quantity \(m_{1} m_{2} J\left(m_{1}+m_{2}\right)\) is called the reducedmass of the system.

The balance wheel of a watch vibrates with an angular amplitude \(\Theta,\) angular frequency \(\omega\) , and phase angle \(\phi=0\) . (a) Find expressions for the angular velocity \(d \theta / d t\) and angular acceleration \(d^{2} \theta / d t^{2}\) as functions of time. (b) Find the balance wheel's angular velocity and angular acceleration when its angular displacement is \(\Theta,\) and when its angular displacement is \(\Theta / 2\) and \(\theta\) is decreasing. (Hint: Sketch a graph of \(\theta\) versus \(t\) )

You want to find the moment of inertia of a complicated machine part about an axis through its center of mass. You suspend it from a wire along this axis. The wire has a torsion constant of 0.450 \(\mathrm{N} \cdot \mathrm{m} / \mathrm{rad}\) You twist the part a small amount about this axis and let it go, timing 125 oscillations in 265 \(\mathrm{s}\) . What is the moment of inertia you want to find?

A \(0.0200-\mathrm{kg}\) bolt moves with SHM that has an amplitude of 0.240 \(\mathrm{m}\) and a period of 1.500 \(\mathrm{s}\) . The displacement of the bolt is \(+0.240 \mathrm{m}\) when \(t=0 .\) Compute (a) the displacement of the bolt when \(t=0.500 \mathrm{s} ;(\mathrm{b})\) the magnitude and direction of the force acting on the bolt when \(t=0.500 \mathrm{s} ;(\mathrm{c})\) the minimum time required for the bolt to move from its initial position to the point where \(x=-0.180 \mathrm{m} ;(\mathrm{d})\) the speed of the bolt when \(x=-0.180 \mathrm{m} .\)

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