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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 \(7.54 \times 10^{4} \mathrm{rad} / \mathrm{s} .\) How many times does the diaphragm move back and forth?

Short Answer

Expert verified
30,000 cycles.

Step by step solution

01

Understanding Simple Harmonic Motion

In simple harmonic motion, the diaphragm moves between two extreme positions, completing one full cycle when it returns to the original position. The frequency of the motion determines how often these cycles occur.
02

Calculating the Frequency

Frequency (\(f\)) is the number of cycles per second and is given by the formula \(f = \frac{\omega}{2\pi}\), where \(\omega\) is the angular frequency. Here, \(\omega = 7.54 \times 10^{4} \mathrm{rad}/\mathrm{s}.\) Substitute into the equation:\[f = \frac{7.54 \times 10^{4}}{2\pi}\approx 1.20 \times 10^{4} \, \text{Hz}.\]
03

Determining Total Cycles in Given Time

The number of cycles the diaphragm completes in a given time can be found by multiplying the frequency by the time duration. Number of cycles = Frequency \(\times\) Time \[= 1.20 \times 10^4 \, \text{Hz} \times 2.5 \, \text{s} = 3.00 \times 10^4 \, \text{cycles}.\]
04

Conclusion

The diaphragm moves back and forth 30,000 times in the time duration of 2.5 seconds.

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Key Concepts

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

Angular Frequency
Angular frequency is a key concept in understanding simple harmonic motion. It is the rate of change of the angle in radians per unit time.
The angular frequency, often represented by the symbol \( \omega \), is directly related to the speed of an oscillating object.
In simple terms, it tells us how fast something is oscillating around a circular path. In mathematical terms, angular frequency \( \omega \) is measured in radians per second \( \mathrm{rad/s} \). It is important to note that angular frequency is different from frequency (measured in Hertz), as it uses radians instead of cycles.
This is essential in physics and engineering, especially when analyzing waveforms and sound waves like those produced by a loudspeaker diaphragm. Knowing the angular frequency allows us to calculate the frequency of motion, which is crucial for many practical applications.
Frequency of Motion
Frequency of motion is used to describe how some components of a system repeat over time in simple harmonic motion.
It is represented by \( f \) and is measured in Hertz (Hz), telling us the number of complete cycles per second. Frequency is critical to understand because it indicates how often an event repeats itself in a given time period.The relationship between angular frequency \( \omega \) and frequency can be expressed by the formula:
\[ f = \frac{\omega}{2\pi} \]
In this equation, \( \omega \) is the angular frequency in radians per second. This formula helps us convert the angular frequency, which is in terms of radians, to the frequency of motion in cycles per second.
For example, in the case of a loudspeaker diaphragm, determining the frequency of motion helps us know how the sound waves are generated and perceived.
Cycles Per Second
Cycles per second is another term for frequency represented by the unit Hertz (Hz).
This concept is part of the bigger picture when dealing with waves and oscillations, like sound waves produced by a speaker. In this context, understanding cycles per second allows us to determine how many times the loudspeaker diaphragm moves back and forth or completes a cycle in one second.
This measurement is crucial not only in acoustics but also in various fields such as electronics and mechanics.
Knowing the cycles per second gives us insights into the properties of the sound wave, such as its pitch and timbre, which are vital in sound engineering and acoustics. The number of cycles is directly proportional to frequency, which means if you know the frequency, you can easily calculate how many cycles occur in a specific timeline.
This is particularly useful in calculating the total number of cycles in a given duration, as shown in the original problem.
Loudspeaker Diaphragm Oscillation
The loudspeaker diaphragm is a crucial component of a speaker that converts electrical signals into sound waves.
When a current passes through the voice coil of a speaker, the diaphragm moves back and forth, creating changes in air pressure.
This movement is known as oscillation and is typically in the form of simple harmonic motion. In simple harmonic motion, the diaphragm moves between two extremes, creating a sound wave through compression and rarefaction of air molecules.
The characteristics of this oscillation, such as its frequency and amplitude, determine the sound's volume and pitch.
Understanding the diaphragm's oscillation helps in designing speakers with desired qualities and improving sound reproduction. When dealing with the oscillation of a diaphragm, the terms angular frequency and frequency come into play.
  • Angular frequency gives us an indication of how fast the diaphragm swings through its range.
  • Frequency tells us how many complete oscillations occur per second, influencing how we perceive the sound.
  • These concepts are used to analyze how efficiently a loudspeaker can reproduce sound.

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    Most popular questions from this chapter

    A person who weighs \(670 \mathrm{~N}\) steps onto a spring scale in the bathroom, and the spring compresses by \(0.79 \mathrm{~cm}\). (a) What is the spring constant? (b) What is the weight of another person who compresses the spring by \(0.34 \mathrm{~cm}\) ?

    Multiple-Concept Example 6 presents a model for solving this problem. As far as vertical oscillations are concerned, a certain automobile can be considered to be mounted on four identical springs, each having a spring constant of \(1.30 \times 10^{5} \mathrm{~N} / \mathrm{m}\). Four identical passengers sit down inside the car, and it is set into a vertical oscillation that has a period of \(0.370 \mathrm{~s}\). If the mass of the empty car is \(1560 \mathrm{~kg}\), determine the mass of each passenger. Assume that the mass of the car and its passengers is distributed evenly over the springs.

    A \(30.0\) -kg block is resting on a flat horizontal table. On top of this block is resting a \(15.0\) kg block, to which a horizontal spring is attached, as the drawing illustrates. The spring constant of the spring is \(325 \mathrm{~N} / \mathrm{m}\). The coefficient of kinetic friction between the lower block and the table is \(0.600\), and the coefficient of static friction between the two blocks is \(0.900\). A horizontal force \(\overrightarrow{\mathbf{F}}\) is applied to the lower block as shown. This force is increasing in such a way as to keep the blocks moving at a constant speed. At the point where the upper block begins to slip on the lower block, determine (a) the amount by which the spring is compressed and (b) the magnitude of the force \(\overrightarrow{\mathbf{F}}\).

    Objects of equal mass are oscillating up and down in simple harmonic motion on two different vertical springs. The spring constant of spring 1 is \(174 \mathrm{~N} / \mathrm{m}\). The motion of the object on spring 1 has twice the amplitude as the motion of the object on spring \(2 .\) The magnitude of the maximum velocity is the same in each case. Find the spring constant of spring 2

    The length of a simple pendulum is \(0.79 \mathrm{~m}\) and the mass of the particle (the "bob") at the end of the cable is \(0.24 \mathrm{~kg}\). The pendulum is pulled away from its equilibrium position by an angle of \(8.50^{\circ}\) and released from rest. Assume that friction can be neglected and that the resulting oscillatory motion is simple harmonic motion. (a) What is the angular frequency of the motion? (b) Using the position of the bob at its lowest point as the reference level, determine the total mechanical energy of the pendulum as it swings back and forth. (c) What is the bob's speed as it passes through the lowest point of the swing?

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