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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 small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the amplitude of the motion is 0.090 m, it takes the block 2.70 s to travel from \(x =\) 0.090 m to \(x = -\)0.090 m. If the amplitude is doubled, to 0.180 m, how long does it take the block to travel (a) from \(x =\) 0.180 m to \(x = -\)0.180 m and (b) from \(x =\) 0.090 m to \(x = -\)0.090 m?

You are watching an object that is moving in SHM. When the object is displaced 0.600 m to the right of its equilibrium position, it has a velocity of 2.20 m/s to the right and an acceleration of 8.40 m/s\(^2\) to the left. How much farther from this point will the object move before it stops momentarily and then starts to move back to the left?

This procedure has been used to "weigh" astronauts in space: A 42.5-kg chair is attached to a spring and allowed to oscillate. When it is empty, the chair takes 1.30 s to make one complete vibration. But with an astronaut sitting in it, with her feet off the floor, the chair takes 2.54 s for one cycle. What is the mass of the astronaut?

An object with mass 0.200 kg is acted on by an elastic restoring force with force constant 10.0 N/m. (a) Graph elastic potential energy \(U\) as a function of displacement \(x\) over a range of \(x\) from \(-\)0.300 m to \(+\)0.300 m. On your graph, let 1 cm \(=\) 0.05 J vertically and 1 cm \(=\) 0.05 m horizontally. The object is set into oscillation with an initial potential energy of 0.140 J and an initial kinetic energy of 0.060 J. Answer the following questions by referring to the graph. (b) What is the amplitude of oscillation? (c) What is the potential energy when the displacement is one-half the amplitude? (d) At what displacement are the kinetic and potential energies equal? (e) What is the value of the phase angle \(\phi\) if the initial velocity is positive and the initial displacement is negative?

A mass is oscillating with amplitude \(A\) at the end of a spring. How far (in terms of \(A\)) is this mass from the equilibrium position of the spring when the elastic potential energy equals the kinetic energy?

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