/*! 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 69 \(A\) quartz watch contains a cr... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\(A\) quartz watch contains a crystal oscillator in the form of a block of quartz that vibrates by contracting and expanding. Two opposite faces of the block, \(7.05 \mathrm{~mm}\) apart, are antinodes, moving alternately toward and away from each other. The plane halfway between these two faces is a node of the vibration. The speed of sound in quartz. is \(3.70 \mathrm{~km} / \mathrm{s}\). Find the frequency of the vibration. An oscillating electric voltage accompanies the mechanical

Short Answer

Expert verified
The frequency of the vibration of the crystal oscillator in the quartz watch is approximately \(2.62 \times 10^5 \mathrm{Hz}\) or 262 kHz.

Step by step solution

01

Understand and Identify the Given Values

The exercise presents us with a quartz watch that contains a crystal oscillator. The oscillator is a block of quartz vibrating by contracting and expanding. Two opposite faces of the block, which are 7.05 mm apart, are antinodes and the plane midway between these faces is the node. The given speed of sound in quartz is 3.70 km/s.
02

Convert Distance into Meters

As SI units are used, we need to convert the given distance of 7.05 mm into meters. As \(1 \mathrm{mm} = 1 \times 10^{-3} \mathrm{m}\), the distance between the antinodes is \(7.05 \times 10^{-3}\) meters or \(7.05 \mathrm{mm} = 0.00705 \mathrm{m}\).
03

Identify What Represents One Complete Wave

In a wave, the distance between two consecutive nodes or antinodes represents one complete wave (wavelength). But according to the explanation, the distance of 7.05 mm is the distance between the two antinodes, passing through the nodal plane which is the midpoint. Therefore, the given distance represents only half of the wavelength (i.e. \( \lambda/2\)).
04

Calculate the Full Wavelength

As the distance 0.00705 m represents only a half wavelength, we need to calculate the full wavelength by doubling the distance \( \lambda = 2 \times 0.00705 \mathrm{m} = 0.0141 \mathrm{m}\).
05

Calculate the Frequency

We know the formula linking speed, frequency and wavelength is \(v = f \lambda\), where \(v\) is the speed, \(f\) is the frequency and \(\lambda\) is the wavelength. From this, we can isolate \(f = v / \lambda\) to find the frequency. Substituting the given speed of sound in quartz (which is 3.70 km/s or 3700 m/s) and the calculated wavelength (0.0141 m) gives us \(f = 3700 / 0.0141\).

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.

Frequency Calculation
Frequency is a fundamental concept when it comes to understanding how sound waves operate. It tells us how often the peaks of the wave pass a certain point per second. For the quartz oscillator in our exercise, we are interested in calculating how many times it vibrates per second. This value is known as the frequency and is measured in Hertz (Hz).
To find the frequency, we use the formula that links speed of sound, wavelength, and frequency:
  • Speed of sound (\(v\)) = Frequency (\(f\)) × Wavelength (\(\lambda\))
By rearranging this formula, we find the frequency with:
  • Frequency, \(f = \frac{v}{\lambda}\)
In our exercise, the speed of sound in quartz is given as 3700 m/s. By calculating the wavelength correctly, which is 0.0141 m, we divide speed by wavelength using the formula to get the frequency:\[ f = \frac{3700}{0.0141} \approx 262411.35\text{ Hz}\]This tells us the quartz crystal oscillates about 262,411 times per second!
Wavelength
Wavelength is the distance between two consecutive similar points of a wave, like peak to peak or trough to trough. In the context of our quartz oscillator, the wavelength corresponds to the length between two consecutive antinodes. However, the given distance between the antinodes in the exercise is only half of the full wavelength because the wave is also passing through a node in between.
For the quartz oscillator example, the provided distance is 7.05 mm, but because this measures only from antinode to antinode, it represents half of the wave. This midpoint, known as a node, divides the wave into two equal halves. Therefore, to get the full wavelength, we double that distance.
  • Half wavelength (\(\lambda/2\)) = 0.00705 m
  • Full wavelength (\(\lambda\)) = 2 × 0.00705 m = 0.0141 m
Having determined the correct full wavelength, it becomes simple to further calculate the frequency of vibration or to understand other wave properties.
Quartz Oscillator
A quartz oscillator is a fascinating device, integral to how many electronic clocks and watches keep time. Quartz crystals have this unique ability to oscillate or vibrate at a constant frequency when subjected to mechanical stress – a property known as piezoelectricity.
In the exercise, we use a quartz block oscillator oscillating by contraction and expansion. The specific conditions caused by sound waves within the quartz help this oscillation maintain accuracy and consistency. The crystal features two crucial areas known as antinodes, where the displacement of air particles is maximum, and nodes where there is minimal to no movement. By ensuring these vibrations remain steady, quartz watches can keep incredibly accurate time, significantly better than older mechanical watch technologies.
Vibrations
Vibrations are a key aspect of wave behavior, especially when discussing how quartz oscillators work. When you hear sound, you've experienced vibrations traveling through a medium (like quartz or air). In our quartz block, vibrations imply a back-and-forth movement of the block particles between the antinodes. This movement might be too brief and rapid for human perception.
Quartz's ability to vibrate comes from its piezoelectric properties. When mechanical stress such as pressure is applied to quartz, its internal electrical charges deform and generate an electric field. This field can lead to mechanical wave vibrations within the crystal.
  • At antinodes, particles of quartz oscillate with maximum amplitude.
  • At the node, there is minimal movement.
These characteristics allow the useful conversion of mechanical energy into electrical energy and vice versa, enabling modern watches to keep time and electronic circuits to function properly.

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

A skyrocket explodes \(100 \mathrm{~m}\) above the ground (Fig. P14.22). Three observers are spaced \(100 \mathrm{~m}\) apart, with the first (A) directly under the explosion. (a) What is the ratio of the sound intensity heard by observer \(\mathrm{A}\) to that heard by observer \(\mathrm{B}\) ? (b) What is the ratio of the intensity heard be observer A to that heard by observer C?

A student stands several meters in front of a smooth reflecting wall. holding a board on which a wire is fixed at each end. The wire, vibrating in its third harmonic. is \(75.0 \mathrm{~cm}\) long, has a mass of \(2.25 \mathrm{~g}, \mathrm{and}\) is under a tension of \(400 \mathrm{~N}\). A second student, moving towards the wall. hears \(8.30\) beats per second. What is the speed of the suudent approaching the wall? Use \(340 \mathrm{~m} / \mathrm{s}\) as the speed of sound in air.

Earthquakes at fault lines in Earth's crust create seismic waves, which are longitudinal (P-waves) or transverse (Swaves). The P-waves have a speed of about \(7 \mathrm{~km} / \mathrm{s}\). Estimate the average bulk modulus of Earth's crust given that the density of rock is about \(2500 \mathrm{~kg} / \mathrm{m}^{3}\).

One of the loudest sounds in recent history was that made by the explosion of Krakatoa on August \(26-27,1889\). According to barometric measurements, the sound had a decibel level of \(180 \mathrm{~d} \mathrm{~B}\) at a distance of \(161 \mathrm{~km}\). Assuming the intensity falls off as the inverse of the distance squared, what was the decibel level on Rodriguez Island, \(4800 \mathrm{~km}\) away?

ecp A standing wave is set up in a string of variable length and tension by a vibrator of variable frequency. Both ends of the string are fixed. When the vibrator has a frequency \(f_{A}\), in a string of length \(L_{4}\) and under tension \(T_{1}, n_{A}\) antinodes are set up in the string. (a) Write an expression for the frequency \(f_{A}\) of a standing wave in terms of the number \(n_{A}\), length \(L_{\cdot+}\), tension \(T_{A}\), and linear density \(\mu_{A}\). (b) If the length of the string is doubled to \(L_{h}=2 L_{A}\), what frequency \(f_{B}\) (written as a multiple of \(f_{A}\) ) will result in the same number of antinodes? Assume the tension and linear density are unchanged. Hint Make a ratio of expressions for \(f_{b}\) and \(f_{A^{\prime}}(\mathrm{c})\) If the frequency and length are held constant, what tension \(T_{g}\) will produce \(n_{A}+1\) antinodes? (d) If the frequency is tripled and the length of the string is halved. by what factor should the tension be changed so that twice as many antinodes are produced:

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.