/*! 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 74 A loudspeaker diaphragm is produ... [FREE SOLUTION] | 91Ó°ÊÓ

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A loudspeaker diaphragm is producing a sound for 2.5 s by moving back and forth in simple harmonic motion. The angular frequency of the motion is \(7.54 \times 10^{4}\) rad/s. How many times does the diaphragm move back and forth?

Short Answer

Expert verified
The diaphragm moves back and forth 30,000 times.

Step by step solution

01

Understand the Problem

We need to determine how many full oscillations, or cycles, the diaphragm completes within a given time, given its angular frequency.
02

Recall the Relationship Between Angular Frequency and Cycles

The angular frequency \( \omega \) is related to the frequency \( f \) by the equation \( \omega = 2 \pi f \). Therefore, \( f = \frac{\omega}{2 \pi} \), where \( f \) is the frequency in cycles per second.
03

Calculate the Frequency

Using the equation from Step 2, calculate the frequency: \( f = \frac{7.54 \times 10^{4} \text{ rad/s}}{2 \pi} \approx 1.2 \times 10^{4} \text{ Hz} \). This means the diaphragm moves back and forth \(1.2 \times 10^{4}\) times each second.
04

Determine the Total Number of Oscillations

Multiply the frequency by the time duration to find the total number of cycles: \( \text{Total Oscillations} = f \times \text{time} = (1.2 \times 10^{4} \text{ Hz}) \times (2.5 \text{ s}) = 3.0 \times 10^{4} \).

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Key Concepts

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

Understanding Angular Frequency
Angular frequency is a key concept in simple harmonic motion. It represents the rate at which an object moves through its cycle of motion. In simpler terms, angular frequency tells us how fast the oscillations or cycles are happening. It's measured in radians per second (rad/s). This is because angular frequency considers the whole circular path, measured in radians, rather than just the number of cycles per second.

In the context of a diaphragm, such as in a loudspeaker, the angular frequency tells us how quickly the diaphragm is moving back and forth. For example, an angular frequency of \(7.54 \times 10^{4}\) rad/s means that during each second, the diaphragm covers a path that's equivalent to \(7.54 \times 10^{4}\) radians. This value helps us understand the speed and pattern of the diaphragm's motion.
Exploring Oscillations
Oscillations refer to any motion that repeats itself at regular intervals. When we talk about a diaphragm vibrating to produce sound, we're talking about oscillations. Each complete back-and-forth movement from its original starting position and back constitutes one oscillation or cycle.
  • Oscillations are characterized by regularity. They occur at consistent time intervals.
  • The frequency of oscillations tells us how many complete cycles happen in one second. It's measured in hertz (Hz).
  • High frequency means a high number of oscillations per second, creating higher pitch sounds in terms of audio devices.
In our problem, the diaphragm completes a large number of oscillations in just a few seconds, thanks to its high angular frequency. Understanding how oscillations work is crucial for interpreting the function and performance of devices like loudspeakers.
The Role of the Diaphragm in Sound Production
The diaphragm in a loudspeaker plays a pivotal role in sound production. Simple harmonic motion allows the diaphragm to move back and forth rapidly, creating sound waves that we can hear. Here's how it works:
  • The diaphragm acts as a flexible membrane, moving in and out as electrical signals pass through the loudspeaker.
  • This movement vibrates the air particles around it, making pressure waves—that's essentially sound.
  • The frequency of these vibrations determines the pitch of the sound. Higher frequencies result in higher pitch sounds, while lower frequencies produce deeper sounds.
In summary, the diaphragm's oscillations transform electrical energy into mechanical energy and then into sound energy. This simple harmonic motion is essential for turning electric impulses into the music or voices you hear from speakers.

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Most popular questions from this chapter

\(\mathrm{A} 1.00 \times 10^{-2}-\mathrm{kg}\) block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring constant is 124 \(\mathrm{N} / \mathrm{m}\) . The block is shoved parallel to the spring axis and is given an initial speed of 8.00 \(\mathrm{m} / \mathrm{s}\) , while the spring is initially unstrained. What is the amplitude of the resulting simple harmonic motion?

Astronauts on a distant planet set up a simple pendulum of length 1.2 \(\mathrm{m}\) . The pendulum executes simple harmonic motion and makes 100 complete vibrations in 280 s. What is the magnitude of the acceleration due to gravity on this planet?

An archer, about to shoot an arrow, is applying a force of \(+240 \mathrm{N}\) to a drawn bowstring. The bow behaves like an ideal spring whose spring constant is 480 \(\mathrm{N} / \mathrm{m}\) . What is the displacement of the bowstring?

A spring is resting vertically on a table. A small box is dropped onto the top of the spring and compresses it. Suppose the spring has a spring constant of 450 \(\mathrm{N} / \mathrm{m}\) and the box has a mass of 1.5 \(\mathrm{kg}\) . The speed of the box just before it makes contact with the spring is 0.49 \(\mathrm{m} / \mathrm{s}\) . (a) Determine the magnitude of the spring's displacement at an instant when the acceleration of the box is zero. (b) What is the magnitude of the spring's displacement when the spring is fully compressed?

Depending on how you fall, you can break a bone easily. The severity of the break depends on how much energy the bone absorbs in the accident, and to evaluate this let us treat the bone as an ideal spring. The maximum applied force of compression that one man's thighbone can endure without breaking is \(7.0 \times 10^{4} \mathrm{N}\) . The minimum effective cross-sectional area of the bone is \(4.0 \times 10^{-4} \mathrm{m}^{2}\) , its length is \(0.55 \mathrm{m},\) and Young's modulus is \(Y=9.4 \times 10^{9} \mathrm{N} / \mathrm{m}^{2}\) . The mass of the man is 65 \(\mathrm{kg}\) . He falls straight down without rotating, strikes the ground stiff-legged on one foot, and comes to a halt without rotating. To see that it is easy to break a thighbone when falling in this fashion, find the maximum distance through which his center of gravity can fall without his breaking a bone.

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