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The tip of a tuning fork goes through 440 complete vibrations in 0.500 s. Find the angular frequency and the period of the motion.

Short Answer

Expert verified
The angular frequency is \( 1760\pi \) rad/s and the period is approximately 0.001136 s.

Step by step solution

01

Understand the Problem

We are given that a tuning fork vibrates 440 times in 0.500 seconds. We need to find the angular frequency and the period of this motion. Angular frequency is related to the frequency of vibration, while the period is the time it takes to complete one full cycle.
02

Calculate Frequency

Frequency is the number of vibrations per second. It is calculated using the formula \( f = \frac{N}{t} \), where \( N \) is the number of vibrations and \( t \) is the time in seconds. Here, \( N = 440 \) and \( t = 0.500 \) seconds.\[ f = \frac{440}{0.500} = 880 \text{ Hz} \]
03

Calculate Angular Frequency

Angular frequency \( \omega \) is related to the frequency \( f \) by the formula \( \omega = 2\pi f \). Thus, we substitute \( f = 880 \text{ Hz} \) into this formula to get:\[ \omega = 2 \pi \times 880 = 1760 \pi \text{ rad/s} \]
04

Calculate the Period

The period \( T \) is the reciprocal of the frequency, calculated as \( T = \frac{1}{f} \). Substituting \( f = 880 \text{ Hz} \):\[ T = \frac{1}{880} \approx 0.001136 \text{ s} \]

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Key Concepts

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

Angular Frequency
Angular frequency, often denoted by the symbol \( \omega \), represents how quickly an object rotates in a circular path per unit time. It is a key concept in harmonic motion, where it describes the rate of oscillation. Angular frequency is measured in radians per second (rad/s), which considers the angle that an object covers over a period of time.When dealing with harmonic motion like the vibrations of a tuning fork, we often relate angular frequency to the standard frequency, \( f \), which is measured in hertz (Hz). The relationship is given by the formula:
  • \( \omega = 2\pi f \)
This formula allows us to convert between the more tangible concept of frequency (vibrations per second) to the angular terms, offering a complete picture of the vibration's dynamics.
Vibration Period
The vibration period, noted as \( T \), is the time it takes for one complete cycle of vibration to occur. A cycle is any repeating unit in an oscillation, like the back-and-forth movement of a pendulum.It's crucial to understand that the period is inversely related to frequency. This means the higher the frequency, the shorter the period. The formula for calculating the period \( T \) from frequency \( f \) is:
  • \( T = \frac{1}{f} \)
Knowing the period of a vibration helps us predict the timing of subsequent cycles. For instance, with a high-frequency tuning fork, the period will be very short, indicating rapid oscillations. This immediate understanding is essential for applications like sound synthesis or analyzing the vibrations of different materials.
Frequency Calculation
Frequency calculation is one of the fundamental steps in analyzing harmonic motion. Frequency \( f \) is defined as the number of complete vibrations, or cycles, that occur in a unit of time, typically expressed in hertz (Hz).To calculate frequency in simple scenarios like our tuning fork example, you use the formula:
  • \( f = \frac{N}{t} \)
where \( N \) is the total number of vibrations, and \( t \) is the time in seconds over which these vibrations occur.Understanding how to calculate frequency provides a basis for determining other properties like the angular frequency and the period of motion. Both the precision of measurement and a grasp of these connections are important, as they allow you to transition fluidly between different metrics of vibration and harmonic motion.

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Most popular questions from this chapter

A uniform beam is suspended horizontally by two identical vertical springs that are attached between the ceiling and each end of the beam. The beam has mass 225 kg, and a 175-kg sack of gravel sits on the middle of it. The beam is oscillating in SHM with an amplitude of 40.0 cm and a frequency of 0.600 cycle/s. (a) The sack falls off the beam when the beam has its maximum upward displacement. What are the frequency and amplitude of the subsequent SHM of the beam? (b) If the sack instead falls off when the beam has its maximum speed, repeat part (a).

A building in San Francisco has light fixtures consisting of small \(2.35-\mathrm{kg}\) bulbs with shades hanging from the ceiling at the end of light, thin cords \(1.50 \mathrm{~m}\) long. If a minor earthquake occurs, how many swings per second will these fixtures make?

A uniform, solid metal disk of mass 6.50 kg and diameter 24.0 cm hangs in a horizontal plane, supported at its center by a vertical metal wire. You find that it requires a horizontal force of 4.23 N tangent to the rim of the disk to turn it by 3.34\(^\circ\), thus twisting the wire. You now remove this force and release the disk from rest. (a) What is the torsion constant for the metal wire? (b) What are the frequency and period of the torsional oscillations of the disk? (c) Write the equation of motion for \(\theta(t)\) for the disk.

A 0.500-kg mass on a spring has velocity as a function of time given by \({v_x}(t) = -\)(3.60 cm/s) sin[ (4.71 rad/s)\(t - \pi\)/2) ]. What are (a) the period; (b) the amplitude; (c) the maximum acceleration of the mass; (d) the force constant of the spring?

An object is undergoing \(\textbf{SHM}\) with period 0.900 s and amplitude 0.320 m. At \(t =\) 0 the object is at \(x =\) 0.320 m and is instantaneously at rest. Calculate the time it takes the object to go (a) from \(x =\) 0.320 m to \(x =\) 0.160 m and (b) from \(x =\) 0.160 m to \(x =\) 0.

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