Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The balance wheel of a watch vibrates with an angular amplitude \(\Theta\), angular frequency \(\omega\), and phase angle \(\phi =\) 0. (a) Find expressions for the angular velocity \(d\theta/dt\) and angular acceleration \(d^2\theta/dt^2\) as functions of time. (b) Find the balance wheel's angular velocity and angular acceleration when its angular displacement is \(\Theta\), and when its angular displacement is \(\Theta\)/2 and \(\theta\) is decreasing. (\(Hint\): Sketch a graph of \(\theta\) versus \(t\).)

Short Answer

Expert verified
Angular velocity is 0 and acceleration is \(-\Theta \omega^2\) at \(\Theta\). At \(\Theta/2\), velocity is \(-\Theta \omega \frac{\sqrt{3}}{2}\) and acceleration is \(-\Theta \omega^2 \frac{1}{2}\).

Step by step solution

01

Understand the Given Parameters

The balance wheel of a watch oscillates with an angular amplitude \( \Theta \), angular frequency \( \omega \), and phase angle \( \phi = 0 \). We need to find expressions for angular velocity \( \frac{d\theta}{dt} \) and angular acceleration \( \frac{d^2\theta}{dt^2} \) as functions of time. These quantities depend on the oscillatory motion described by \( \theta(t) = \Theta \cos(\omega t + \phi) \). Since \( \phi = 0 \), \( \theta(t) = \Theta \cos(\omega t) \).
02

Calculate Angular Velocity

To find angular velocity \( \frac{d\theta}{dt} \), differentiate \( \theta(t) = \Theta \cos(\omega t) \) with respect to time \( t \). The derivative of \( \cos(\omega t) \) is \( -\omega \sin(\omega t) \). Thus,\[\frac{d\theta}{dt} = -\Theta \omega \sin(\omega t).\]
03

Calculate Angular Acceleration

Find angular acceleration \( \frac{d^2\theta}{dt^2} \) by differentiating the expression for angular velocity, \( \frac{d\theta}{dt} = -\Theta \omega \sin(\omega t) \). The derivative of \( \sin(\omega t) \) is \( \omega \cos(\omega t) \). Thus,\[\frac{d^2\theta}{dt^2} = -\Theta \omega^2 \cos(\omega t).\]
04

Evaluate Angular Velocity at Maximum Displacement \( \Theta \)

At angular displacement \( \theta = \Theta \), the term inside the cosine is zero, \( \omega t = 0 \). From \( \frac{d\theta}{dt} = -\Theta \omega \sin(\omega t) \), since \( \sin(0) = 0 \), the angular velocity is 0. So, the angular velocity is zero when the displacement is at a maximum.
05

Evaluate Angular Acceleration at Maximum Displacement \( \Theta \)

At angular displacement \( \theta = \Theta \), substitute in \( \omega t = 0 \) for \( \frac{d^2\theta}{dt^2} = -\Theta \omega^2 \cos(\omega t) \). Since \( \cos(0) = 1 \), the angular acceleration is:\[\frac{d^2\theta}{dt^2} = -\Theta \omega^2.\]
06

Evaluate Kinematic Parameters at Half Displacement \( \Theta/2 \)

For angular displacement \( \theta = \Theta/2 \), we solve \( \Theta \cos(\omega t) = \Theta/2 \), which gives \( \cos(\omega t) = 1/2 \). This occurs at \( \omega t = \pi/3 \). Evaluating angular velocity \( \frac{d\theta}{dt} = -\Theta \omega \sin(\omega t) \), with \( \sin(\pi/3) = \sqrt{3}/2 \), gives\[\frac{d\theta}{dt} = -\Theta \omega \frac{\sqrt{3}}{2}.\]For angular acceleration \( \frac{d^2\theta}{dt^2} = -\Theta \omega^2 \cos(\omega t) \), plugging \( \cos(\pi/3) = 1/2 \) gives\[\frac{d^2\theta}{dt^2} = -\Theta \omega^2 \frac{1}{2}.\]

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 Vaia!

Key Concepts

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

Oscillatory Motion
Oscillatory motion refers to a repetitive back-and-forth movement over the same path, much like a swing on a playground or a pendulum in a clock. It's characterized by its regularity and periodic nature, meaning the motion occurs at regular intervals. In physics, this can be seen in systems like pendulums, springs, or any system that returns to an equilibrium state after being disturbed.

For example, in the case of a balance wheel in a watch, the wheel experiences oscillatory motion as it vibrates back and forth. This motion can be described mathematically using a sine or cosine function, which accounts for its periodic behavior. Here, the angular displacement function is given as \[\theta(t) = \Theta \cos(\omega t)\] where \(\Theta\) is the angular amplitude, reflecting the maximum displacement from the equilibrium position. This function gives us insight into how the wheel moves over time and is key to understanding how oscillatory systems function in general.
Angular Velocity
Angular velocity is a measure of how quickly an object rotates or revolves relative to another point, often the center of a circle. It tells us the rate of change of angular displacement (\(\theta\)). In mathematical terms, it's expressed as the derivative of angular displacement with respect to time.

Specifically, for oscillatory motion described by \(\theta(t) = \Theta \cos(\omega t)\), its angular velocity is calculated by differentiating this function with respect to time:\[\frac{d\theta}{dt} = -\Theta \omega \sin(\omega t)\]

The result shows that angular velocity is dependent on the sine of the angular frequency multiplied by time. This expression indicates how the speed of rotation varies over time, reaching its maximum magnitude when the displacement is zero and slowing down to zero when displacement reaches a maximum.
Angular Acceleration
Angular acceleration is the rate of change of angular velocity over time. It provides insights into how quickly an object's rotational speed is changing at any given moment. Critical in oscillatory systems, angular acceleration helps us understand the dynamics of motion.

From the expression of angular velocity, \(\frac{d\theta}{dt} = -\Theta \omega \sin(\omega t)\), angular acceleration can be derived through further differentiation:\[\frac{d^2\theta}{dt^2} = -\Theta \omega^2 \cos(\omega t)\]

This result indicates that angular acceleration depends on a cosine function of time and is proportional to the square of the angular frequency. It shows that when angular displacement reaches its maximum or minimum value, the acceleration is at peak magnitude, exerting a restoring force that pulls the system back toward equilibrium.
Phase Angle
The phase angle, often denoted as \(\phi\), indicates the initial angle of the system at time \(t = 0\), influencing the starting point of the oscillation. The phase angle can shift the entire wave left or right on a graph.

In the case of simple harmonic motion like that of a balance wheel with \(\phi = 0\), this simplifies the angular displacement function:\[\theta(t) = \Theta \cos(\omega t)\]

A phase angle of zero means that the balance wheel starts its motion at its maximum displacement. The absence of a phase shift makes it straightforward to analyze the system's motion since its sinusoidal behavior begins at the peak of its cycle, directly related to the amplitude.
Trigonometric Functions
Trigonometric functions, such as sine and cosine, are pivotal in describing oscillatory motion. They capture the periodic nature of motion through their wave patterns.

For the balance wheel example, the functions \(\cos(\omega t)\) and \(\sin(\omega t)\) help us describe the changing displacement, velocity, and acceleration over time:
  • \(\cos(\omega t)\) describes how displacement changes, varying from maximum to minimum in a smooth, predictable manner.
  • \(\sin(\omega t)\) is used to represent angular velocity, highlighting how velocities change throughout the cycle.
These functions showcase the inherent symmetry and predictability in oscillatory systems. They help visualize how objects move back and forth as part of a broader cycle, crucial for accurately predicting future motion dynamics.

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

An object is undergoing \(\textbf{SHM}\) with period 0.900 s and amplitude 0.320 m. At \(t =\) 0 the object is at \(x =\) 0.320 m and is instantaneously at rest. Calculate the time it takes the object to go (a) from \(x =\) 0.320 m to \(x =\) 0.160 m and (b) from \(x =\) 0.160 m to \(x =\) 0.

While on a visit to Minnesota ("Land of 10,000 Lakes"), you sign up to take an excursion around one of the larger lakes. When you go to the dock where the 1500-kg boat is tied, you find that the boat is bobbing up and down in the waves, executing simple harmonic motion with amplitude 20 cm. The boat takes 3.5 s to make one complete up-and-down cycle. When the boat is at its highest point, its deck is at the same height as the stationary dock. As you watch the boat bob up and down, you (mass 60 kg) begin to feel a bit woozy, due in part to the previous night's dinner of lutefisk. As a result, you refuse to board the boat unless the level of the boat's deck is within 10 cm of the dock level. How much time do you have to board the boat comfortably during each cycle of up-and-down motion?

A building in San Francisco has light fixtures consisting of small \(2.35-\mathrm{kg}\) bulbs with shades hanging from the ceiling at the end of light, thin cords \(1.50 \mathrm{~m}\) long. If a minor earthquake occurs, how many swings per second will these fixtures make?

A 0.0200-kg bolt moves with SHM that has an amplitude of 0.240 m and a period of 1.500 s. The displacement of the bolt is \(+\)0.240 m when \(t =\) 0. Compute (a) the displacement of the bolt when \(t =\) 0.500 s; (b) the magnitude and direction of the force acting on the bolt when \(t =\) 0.500 s; (c) the minimum time required for the bolt to move from its initial position to the point where \(x = -\)0.180 m; (d) the speed of the bolt when \(x = -\)0.180 m.

A 40.0-N force stretches a vertical spring 0.250 m. (a) What mass must be suspended from the spring so that the system will oscillate with a period of 1.00 s? (b) If the amplitude of the motion is 0.050 m and the period is that specified in part (a), where is the object and in what direction is it moving 0.35 s after it has passed the equilibrium position, moving downward? (c) What force (magnitude and direction) does the spring exert on the object when it is 0.030 m below the equilibrium position, moving upward?

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.

Sign-up for free