/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 The tip of a tuning fork goes th... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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
Frequency: 880 Hz, Period: 0.001136 s, Angular frequency: 5529.2 rad/s.

Step by step solution

01

Understand the Problem

We are given the number of complete vibrations (440) that occur in a time period (0.500 seconds). We need to find the angular frequency and the period of the motion.
02

Calculate the Frequency

Frequency (\( f \)) is the number of vibrations per second. It is calculated as the total number of vibrations divided by the total time in seconds. \[\text{Frequency} \ (f) = \frac{440 \ \text{vibrations}}{0.500 \ \text{s}} = 880 \ \text{Hz}\]
03

Calculate the Period

The period (\( T \)) is the reciprocal of the frequency. It represents the time taken for one complete cycle of vibration.\[T = \frac{1}{f} = \frac{1}{880 \ \text{Hz}} \approx 0.001136 \ \text{s}\]
04

Calculate the Angular Frequency

Angular frequency (\( \omega \)) is related to the frequency by the formula \( \omega = 2\pi f \).\[\omega = 2\pi \times 880 \ \text{Hz} \approx 5529.2 \ \text{rad/s}\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Understanding Angular Frequency
Angular frequency is a crucial concept in harmonic motion, especially when examining oscillatory movements like those of a tuning fork. It represents how quickly something oscillates in a circular motion.
Think of it as measuring how swiftly the oscillation moves through its cycle. Angular frequency is denoted by the symbol \( \omega \) and is measured in radians per second (rad/s). This measurement tells us how many radians the vibrations complete in one second, giving a more complete picture of the oscillation speed beyond linear frequency.
To calculate angular frequency, we use the formula:
  • \[ \omega = 2\pi f \]
Here, \( \pi \) is a constant (approximately 3.14159), and \( f \) is the frequency in hertz (Hz), or cycles per second. This formula shows that angular frequency is directly proportional to linear frequency. Therefore, if you know the frequency, you can quickly find the angular frequency and understand the oscillatory nature of the motion.
Exploring the Period of Motion
The period of motion in harmonic motion is a fundamental concept that tells us how much time it takes for one complete cycle of vibration. Imagine a tuning fork vibrating back and forth; the period measures the time it takes to go from one extreme point, through the equilibrium, to the opposite extreme, and back again.
The period is denoted by \( T \) and is measured in seconds (s). A shorter period means that the vibrations occur more rapidly, while a longer period indicates slower oscillations. This relationship between frequency and period is expressed simply by the formula:
  • \[ T = \frac{1}{f} \]
Since the period is the inverse of frequency, understanding one allows you to find the other. For example, if you know a tuning fork's frequency is 880 Hz (which means it completes 880 cycles each second), calculating its period can help you grasp how quick each cycle is.
A lower period value suggests a swiftly vibrating tuning fork, which is a key characteristic in musical acoustics, as it influences the pitch and tonal quality of the sound produced.
The Role of a Tuning Fork in Harmonic Motion
A tuning fork is an excellent example of harmonic motion in everyday life. When struck, it creates sound waves that vibrate at a specific frequency. This precision makes it invaluable in tuning musical instruments and in acoustical experiments.
The structure of a tuning fork consists of two prongs (or tines) atop a handle. When the tines are struck against a surface, they begin to vibrate back and forth rapidly, creating resonant sound waves. These vibrations are a practical demonstration of uniform harmonic motion, occurring at a specific natural frequency unique to the fork’s design and material.
Typically, tuning forks are designed to vibrate at a set frequency, such as 440 Hz, which is often used as a standard pitch ('A' above middle 'C') in music. This standardization assures a consistent reference point for tuning different instruments, ensuring they produce harmonious sound during a performance.
The interaction of a tuning fork with harmonic motion principles illustrates concepts like frequency, period, and angular frequency, providing a tangible way to understand abstract physical concepts.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

On a horizontal, frictionless table, an open-topped \(5.20-\mathrm{kg}\) box is attached to an ideal horizontal spring having force constant 375 \(\mathrm{N} / \mathrm{m}\) . Inside the box is a 3.44 -kg stone. The system is oscillating with an amplitude of 7.50 \(\mathrm{cm} .\) When the box has reached its maximum speed, the stone is suddenly plucked vertically out of the box without touching the box. Find (a) the period and (b) the amplitude of the resulting motion of the box. (c) Without doing any calculations, is the new period greater or smaller than the original period? How do you know?

BIO Weighing a Virus. In February \(2004,\) scientists at Purdue University used a highly sensitive technique to measure the mass of a vaccinia virus (the kind used in smallpox vaccine). The procedure involved measuring the frequency of oscillation of a tiny sliver of silicon (just 30 \(\mathrm{nm}\) long with a laser, first without the virus and then after the virus had attached itself to the silicon. The difference in mass caused a change in the frequency. We can model such a process as a mass on a spring. (a) Show that the ratio of the frequency with the virus attached \(\left(f_{\mathrm{S}+\mathrm{v}}\right)\) to the frequency without the virus \(\left(f_{\mathrm{S}}\right)\) is given by the formula \(\frac{f_{\mathrm{S}+\mathrm{V}}}{f_{\mathrm{S}}}=\frac{1}{\sqrt{1+\left(m_{\mathrm{V}} / m_{\mathrm{S}}\right)}}\) where \(m_{\mathrm{V}}\) is the mass of the virus and \(m_{\mathrm{S}}\) is the mass of the silicon sliver. Notice that is not necessary to know or measure the force constant of the spring. (b) In some data, the silicon sliver has a mass of \(2.10 \times 10^{-16} \mathrm{g}\) and a frequency of \(2.00 \times 10^{15} \mathrm{Hz}\) without the virus and \(2.87 \times 10^{14} \mathrm{Hz}\) with the virus. What is the mass of the virus, in grams and in femtograms?

CP SHM of a Floating Object. An object with height \(h,\) mass \(M,\) and a uniform cross-sectional area \(A\) floats upright in a liquid with density \(\rho\) (a) Calculate the vertical distance from the surface of the liquid to the bottom of the floating object at equilibrium. (b) A downward force with magnitude \(F\) is applied to the top of the object. At the new equilibrium position, how much farther below the surface of the liquid is the bottom of the object than it was in part (a)? (Assume that some of the object remains above the surface of the liquid.) (c) Your result in part (b) shows that if the force is suddenly removed, the object will oscillate up and down in SHM. Calculate the period of this motion in terms of the density \(\rho\) of the liquid, the mass \(M,\) and the cross-sectional area \(A\) of the object. You can ignore the damping due to fluid friction (see Section 14.7\()\) .

BIO Weighing Astronauts. This procedure has actually 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?

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 \(\mathrm{m} / \mathrm{s}\) to the right and an acceleration of 8.40 \(\mathrm{m} / \mathrm{s}^{2}\) 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?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.