/*! 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 65 A loudspeaker diaphragm is oscil... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A loudspeaker diaphragm is oscillating in simple harmonic motion with a frequency of \(440 \mathrm{~Hz}\) and a maximum displacement of \(0.75 \mathrm{~mm}\). What are the (a) angular frequency, (b) maximum speed, and (c) magnitude of the maximum acceleration?

Short Answer

Expert verified
(a) 2764.6 rad/s; (b) 2.073 m/s; (c) 5729.9 m/s².

Step by step solution

01

Understand the Problem

A loudspeaker diaphragm is moving in simple harmonic motion with a frequency of 440 Hz and a maximum displacement of 0.75 mm (which is 0.00075 meters). We'll find the angular frequency, maximum speed, and magnitude of the maximum acceleration.
02

Calculate Angular Frequency

The angular frequency \( \omega \) is related to the frequency \( f \) by the formula \( \omega = 2\pi f \). Substituting the given \( f = 440 \text{ Hz} \), we have \( \omega = 2\pi \times 440 \approx 2764.6 \text{ radians/second} \).
03

Calculate Maximum Speed

The maximum speed \( v_{max} \) is given by \( v_{max} = \omega A \), where \( A \) is the maximum displacement. Substitute \( \omega = 2764.6 \text{ radians/second} \) and \( A = 0.00075 \text{ meters} \): \( v_{max} = 2764.6 \times 0.00075 \approx 2.073 \text{ meters/second} \).
04

Calculate Maximum Acceleration

The magnitude of the maximum acceleration \( a_{max} \) is given by \( a_{max} = \omega^2 A \). Substitute \( \omega = 2764.6 \text{ radians/second} \) and \( A = 0.00075 \text{ meters} \). Calculate: \( a_{max} = 2764.6^2 \times 0.00075 \approx 5729.9 \text{ meters/second}^2 \).

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 91Ó°ÊÓ!

Key Concepts

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

Angular Frequency
Understanding angular frequency is crucial when analyzing simple harmonic motion. Angular frequency, often denoted by \( \omega \), is a measure of how fast an object rotates or oscillates in a circular path. It is closely related to the standard frequency \( f \), which is the number of cycles per second. To calculate angular frequency, we use the formula: \[ \omega = 2 \pi f \] This formula comes from the fact that one complete cycle corresponds to an angle of \( 2\pi \) radians. Let's consider our example of a loudspeaker diaphragm oscillating at \( 440 \text{ Hz} \). By applying the formula, we find that: \[ \omega = 2 \pi \times 440 \approx 2764.6 \text{ radians/second} \] In this context:
  • \( \omega \) translates the linear metric of frequency into a rotational aspect, needed for understanding oscillatory systems.
  • It's a fundamental property of your oscillating system, providing insights into its behavior.
  • This measure is vital for predictive models of oscillations, like calculating derived quantities such as speed and acceleration.
Maximum Speed
In simple harmonic motion, the maximum speed of an oscillating object tells us how fast the object moves at its fastest point. This maximum speed, denoted by \( v_{\text{max}} \), occurs when the object passes through its equilibrium position. At this point, all potential energy is converted into kinetic energy, resulting in the highest velocity. To find \( v_{\text{max}} \), we use: \[ v_{\text{max}} = \omega A \] Here, \( A \) is the maximum displacement, or amplitude, of the motion. Substituting our known values of \( \omega = 2764.6 \text{ radians/second} \) and \( A = 0.00075 \text{ meters} \): \[ v_{\text{max}} = 2764.6 \times 0.00075 \approx 2.073 \text{ meters/second} \] Key points to remember:
  • The maximum speed offers a direct insight into how rapidly energy is exchanged in the system.
  • At maximum speed, the motion is purely kinetic, with no potential energy stored.
  • It helps predict how fast an object will move at its central position when oscillating.
Maximum Acceleration
The maximum acceleration of an object in simple harmonic motion is a measure of how quickly it can change its velocity. The acceleration reaches its peak when the object is at its maximum displacement, turning point. At this moment, the force restoring the object to its equilibrium position is greatest, hence accelerating the object most. This maximum acceleration, \( a_{\text{max}} \), is calculated using: \[ a_{\text{max}} = \omega^2 A \] Substituting the known values of \( \omega = 2764.6 \text{ radians/second} \) and \( A = 0.00075 \text{ meters} \): \[ a_{\text{max}} = (2764.6)^2 \times 0.00075 \approx 5729.9 \text{ meters/second}^2 \] A few takeaways:
  • Maximum acceleration is essential for understanding the limits of force exertion in oscillating systems.
  • It highlights the extreme forces the system undergoes at its boundaries of motion.
  • This concept aids in designing systems that must withstand high accelerations, like sound wave generators or suspension systems.

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

Although California is known for earthquakes, it has large regions dotted with precariously balanced rocks that would be easily toppled by even a mild earthquake. The rocks have stood this way for thousands of years, suggesting that major earthquakes have not occurred in those regions during that time. If an earthquake were to put such a rock into sinusoidal oscillation (parallel to the ground) with a frequency of \(2.2 \mathrm{~Hz}\), an oscillation amplitude of \(1.0 \mathrm{~cm}\) would cause the rock to topple. What would be the magnitude of the maximum acceleration of the oscillation, in terms of \(g\) ?

A \(1000 \mathrm{~kg}\) car carrying four \(82 \mathrm{~kg}\) people travels over a "washboard" dirt road with corrugations \(4.0 \mathrm{~m}\) apart. The car bounces with maximum amplitude when its speed is \(16 \mathrm{~km} / \mathrm{h}\). When the car stops, and the people get out, by how much does the car body rise on its suspension?

A thin uniform rod (mass \(=0.50 \mathrm{~kg}\) ) swings about an axis that passes through one end of the rod and is perpendicular to the plane of the swing. The rod swings with a period of \(1.5\) s and an angular amplitude of \(10^{\circ} .\) (a) What is the length of the rod? (b) What is the maximum kinetic energy of the rod as it swings?

A \(2.00 \mathrm{~kg}\) block hangs from a spring. A \(300 \mathrm{~g}\) body hung below the block stretches the spring \(2.00 \mathrm{~cm}\) farther. (a) What is the spring constant? (b) If the \(300 \mathrm{~g}\) body is removed and the block is set into oscillation, find the period of the motion.

The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of \(0.76 \mathrm{~m}\). If the piston moves with simple harmonic motion with an angular frequency of \(180 \mathrm{rev} / \mathrm{min}\), what is its maximum speed?

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.