The quartz crystal in a watch executes simple harmonic motion at \(32,768 \mathrm{Hz}\) (This is \(2^{15} \mathrm{Hz}\), chosen so that 15 divisions by 2 give a signal at \(1.00000 \mathrm{Hz}\) ) If each face of the crystal undergoes a maximum displacement of \(100 \mathrm{nm}\), find the maximum velocity and acceleration of the crystal faces.

Short Answer

Expert verified
The maximum velocity and acceleration of the crystal faces are calculated from the given frequency and amplitude of the oscillation, using the expressions for the maximum speed and acceleration in the simple harmonic motion.

Step by step solution

01

Identify given variables

The simple harmonic oscillator is characterized by its frequency of \(32,768 \, \mathrm{Hz}\) and displacement amplitude of \(100 \, \mathrm{nm}\)
02

Calculate angular frequency

The angular frequency, \(\omega\), is the product of \(2\pi\) and the frequency, \(f\). In this case, calculate \(\omega = 2\pi \times 32,768 \, \mathrm{Hz}\)
03

Calculate maximum velocity

The maximum velocity in the simple harmonic motion occurs when the displacement is zero. This is calculated as \(v_{max} = \omega \times \, \text{Amplitude}\). Substitute the calculated angular frequency and given amplitude into this formula to find the maximum velocity.
04

Calculate maximum acceleration

The maximum acceleration in the simple harmonic motion occurs at the maximum displacement amplitude. This is calculated as \(a_{max} = \omega^{2} \times \, \text{Amplitude}\). Substitute the calculated angular frequency and given amplitude into this formula to find the maximum acceleration.

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Key Concepts

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

Understanding Angular Frequency
When studying simple harmonic motion (SHM), knowing the concept of 'angular frequency' is vital. It is closely linked to the frequency of the motion, but involves the inherent geometry of circular movements, hence the term 'angular'.

Angular frequency, denoted as \(\textbf{\(\omega\)}\), is a measure describing how fast something goes through its cycle. It is given by the equation \(\omega = 2\pi f\), where \(f\) is the frequency, and \(2\pi\) is a full revolution in radians. If we think of SHM as a projection of circular motion, this measure gives us an idea of how quickly we are 'sweeping' through the circle.

In our case, the quartz crystal in a watch vibrates at a frequency of \(32,768 \mathrm{Hz}\), an incredibly rapid back-and-forth movement. Calculating its angular frequency involves simply multiplying by \(2\pi\), giving us a much larger number that speaks volumes about the speed of the oscillation. This is especially important when determining other aspects of SHM, such as velocity and acceleration.
Calculating Maximum Velocity in SHM
Maximum velocity is a key concept in simple harmonic motion. It refers to the highest speed achieved by the oscillating object as it moves back and forth. In SHM, this maximum velocity occurs exactly when the object passes through the equilibrium point, where no restoring force is acting upon it.

The formula to calculate maximum velocity in SHM is \(v_{\text{max}} = \omega \times A\), with \(\omega\) as the angular frequency and \(A\) as the amplitude, or the maximum displacement from the equilibrium position.

For a quartz crystal in a watch, we'd use the angular frequency we calculated earlier and multiply it by the maximum displacement, which is given as \(100 \text{nm}\). Despite the tiny size of the displacement, the resulting maximum velocity is quite fast, demonstrating the efficiency and precision needed in timekeeping. This concept ensures the watch's timekeeping elements move quickly enough to keep accurate time, but not so fast that they wear out the mechanism.
Determining Maximum Acceleration in SHM
Maximum acceleration is yet another crucial aspect to grasp when dealing with simple harmonic motion. The object undergoing SHM experiences the greatest acceleration at the points of maximum displacement, where the restoring force is at its peak.

The mathematical representation for this is \(a_{\text{max}} = \omega^2 \times A\), with \(\omega\) being the angular frequency and \(A\) representing the amplitude. For our quartz crystal, this equation allows us to find how quickly the crystal's facing accelerates back toward the equilibrium position once it has reached its furthest point in the oscillation cycle.

This maximum acceleration is pivotal for understanding the forces involved in SHM and ensuring the structural integrity of the oscillating system. It's an intrinsic measure of how 'forceful' the motion is and relates directly to the stresses the material of the crystal faces during its operation.

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Most popular questions from this chapter

An object undergoes simple harmonic motion in two mutually perpendicular directions, its position given by \(\vec{r}=A \sin \omega t \hat{\imath}+\) A cos wit. (a) Show that the object remains a fixed distance from the origin (i.e.. that its path is circular), and find that distance. (b) Find an expression for the object's velocity. (c) Show that the speed remains constant, and find its value. (d) Find the angular speed of the object in its circular path.

Derive the period of a simple pendulum by considering the horizontal displacement \(x\) and the force acting on the bob, rather than the angular displacement and torque.

A massless spring with \(k=74 \mathrm{N} / \mathrm{m}\) hangs from the ceiling. A \(490-\mathrm{g}\) mass is hooked onto the unstretched spring and allowed to drop. Find (a) the amplitude and (b) the period of the resulting

A particle of mass \(m\) has potential energy given by \(U=a x^{2}\) where \(a\) is a constant and \(x\) is the particle's position. Find an expression for the frequency of simple harmonic oscillations this particle undergoes.

You're riding in a friend's \(1400-\mathrm{kg}\) car with bad shock absorbers, bouncing down the highway at \(20 \mathrm{m} / \mathrm{s}\) and executing vertical SHM with amplitude \(18 \mathrm{cm}\) and frequency \(0.67 \mathrm{Hz}\) Concerned about fuel efficiency, your friend wonders what percentage of the car's kinetic energy is tied up in this oscillation. Make an estimate, neglecting the wheels' rotational energy and the fact that not all of the car's mass participates in the oscillation.

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