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

Short Answer

Expert verified
The period is approximately 0.00114 s, and the angular frequency is about 5520.34 rad/s.

Step by step solution

01

Understand the Given Information

We know that the tuning fork goes through 440 vibrations in 0.500 seconds. Each vibration corresponds to one full cycle of the wave motion.
02

Calculate the Frequency

Frequency (\( f \)) is the number of complete vibrations per second. Calculate it as follows:\[f = \frac{\text{Number of vibrations}}{\text{Time (s)}} = \frac{440}{0.5} = 880 \, \text{Hz}\]
03

Determine the Period of the Motion

The period (\( T \)) is the reciprocal of the frequency. It indicates the time for one complete vibration:\[T = \frac{1}{f} = \frac{1}{880} \approx 0.00114 \, \text{s}\]
04

Find the Angular Frequency

The angular frequency (\( \omega \)) is related to the frequency by:\[\omega = 2\pi f = 2\pi \times 880 \approx 5520.34 \, \text{rad/s}\]

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Key Concepts

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

Tuning Fork Vibrations
A tuning fork is a simple instrument that helps us understand sound vibrations. When struck, it creates a wave by vibrating at a specific frequency. Each time the tuning fork vibrates, it completes a full cycle of motion. In the exercise, it was noted that the tuning fork completes 440 such cycles in half a second. This repetitive motion is what produces the sound we perceive.

Vibrations in a tuning fork occur very rapidly. Each vibration represents one complete oscillation or cycle. During this cycle, the fork’s prongs move from their initial position, reach maximum displacement, and return to the starting position. This motion, repeated many times per second, generates the sound waves we hear. The regularity of these cycles makes tuning forks ideal for experiments and demonstrations related to sound and vibration.
Frequency Calculation
Frequency is a fundamental concept in understanding oscillations and waves. It measures how many cycles are completed in one second and is expressed in hertz (Hz). In our example, we calculate it using the number of vibrations divided by the time:
  • Number of vibrations: 440
  • Time: 0.500 seconds
  • Frequency, \( f = \frac{440}{0.5} = 880 \, \text{Hz} \)
This calculation shows that the tuning fork produces 880 cycles every second. A higher frequency means that more cycles occur per second, which typically produces a higher-pitched sound.

Understanding frequency is crucial in various fields, from music to electronics. In music, different instrument notes have distinct frequencies, contributing to their unique tones.
Period of Motion
The period of motion is another key concept when studying waves and vibrations. The period, denoted by \( T \), is the time taken to complete one full vibration or cycle. It is the reciprocal of frequency, meaning the two quantities are inversely related. In mathematical terms, \( T = \frac{1}{f} \).

For our tuning fork, given a frequency of 880 Hz, the period calculation is simple:
  • \( T = \frac{1}{880} \approx 0.00114 \, \text{s} \)
This tells us that each cycle of the tuning fork's motion takes approximately 0.00114 seconds. Understanding the period of motion helps in visualizing how quickly the oscillations are happening. Whether it's determining the pulse of a wave or analyzing the motion of a pendulum, the period provides insight into the timing characteristics of repetitive motion.

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Most popular questions from this chapter

A small sphere with mass \(m\) is attached to a massless rod of length \(L\) that is pivoted at the top, forming a simple pendulum. The pendulum is pulled to one side so that the rod is at an angle \(\Theta\) from the vertical, and released from rest. (a) In a diagram, show the pendulum just after it is released. Draw vectors representing the forces acting on the small sphere and the acceleration of the sphere. Accuracy counts! At this point, what is the linear acceleration of the sphere? (b) Repeat part (a) for the instant when the pendulum rod is at an angle \(\Theta / 2\) from the vertical. (c) Repeat part (a) for the instant when the pendulum rod is vertical. At this point, what is the linear speed of the sphere?

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

A Pendulum on Mars. A certain simple pendulum has a period on the earth of 1.60 s. What is its period on the surface of Mars, where \(g=3.71 \mathrm{m} / \mathrm{s}^{2} ?\)

A harmonic oscillator has angular frequency \(\omega\) and amplitude \(A\) . (a) What are the magnitudes of the displacement and velocity when the elastic potential energy is equal to the kinetic energy? (Assume that \(U=0\) at equilibrium, (b) How often does this occur in each cycle? What is the time between occurrences? (c) At an instant when the displacement is equal to \(A / 2,\) what fraction of the total energy of the system is kinetic and what fraction is potential?

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.

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