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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 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?

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 m from its equilibrium position and released with zero initial speed, then after 0.800 s its displacement is found to be 0.120 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.

A 0.500-kg glider, attached to the end of an ideal spring with force constant \(k =\) 450 N/m, undergoes SHM with an amplitude of 0.040 m. Compute (a) the maximum speed of the glider; (b) the speed of the glider when it is at \(x = -\)0.015 m; (c) the magnitude of the maximum acceleration of the glider; (d) the acceleration of the glider at \(x = -\)0.015 m; (e) the total mechanical energy of the glider at any point in its motion.

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 50.0 cm. She finds that the pendulum makes 100 complete swings in 136 s. What is the value of \(g\) on this planet?

A block with mass \(m =\) 0.300 kg is attached to one end of an ideal spring and moves on a horizontal frictionless surface. The other end of the spring is attached to a wall. When the block is at \(x = +\)0.240 m, its acceleration is \(a_x = -\)12.0 m/s\(^{2}\) and its velocity is \(v_x = +\)4.00 m/s. What are (a) the spring's force constant \(k\); (b) the amplitude of the motion; (c) the maximum speed of the block during its motion; and (d) the maximum magnitude of the block's acceleration during its motion?

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