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A microphone is attached to a spring that is suspended from the ceiling, as the drawing indicates. Directly below on the floor is a stationary 440-Hz source of sound. The microphone vibrates up and down in simple harmonic motion with a period of 2.0 s. The difference between the maximum and minimum sound frequencies detected by the microphone is 2.1 Hz. Ignoring any reflections of sound in the room and using 343 m/s for the speed of sound, determine the amplitude of the simple harmonic motion.

Short Answer

Expert verified
The amplitude of simple harmonic motion is approximately 0.258 meters.

Step by step solution

01

Understand the Situation

A microphone moves up and down in simple harmonic motion (SHM). A stationary source of sound with a frequency of 440 Hz is below it. Due to the Doppler effect, the frequency detected by the microphone changes as it moves. The problem gives the difference in detected frequencies as 2.1 Hz.
02

Use the Doppler Effect Formula

The frequency detected by the microphone as it approaches the source is given by \(f' = \frac{v}{v - v_0}f\) where \v_0\ is the velocity of the microphone toward the source. When moving away, \(f'' = \frac{v}{v + v_0}f\). The maximum frequency difference detected is \(f' - f'' = 2.1\, \text{Hz}\).
03

Apply Simple Harmonic Motion Velocity

For SHM, the velocity is \(v_0 = A\omega\sin(\omega t)\), with angular frequency \(\omega = \frac{2\pi}{T}\), and period \(T = 2\,\text{s}\). Calculate \(\omega = \frac{2\pi}{2} = \pi\, \text{rad/s}\). Thus, maximum velocity \(v_0 = A\pi\).
04

Set Up and Solve for Amplitude

Use Doppler effect formulas: \(f' - f'' = 2.1 = \frac{2fv_0}{v^2 - v_0^2}\). Assume \(v_0 << v\), so approximate \(\frac{2fv_0}{v}\) gives \(2.1 = \frac{2 \cdot 440 \cdot A\pi}{343}\). Solve for \(A\): \(A = \frac{2.1 \cdot 343}{2 \cdot 440 \cdot \pi}\).
05

Calculate Amplitude

Substitute values into the equation: \(A = \frac{2.1 \times 343}{2 \times 440 \times \pi} \approx 0.258 \, \text{m}\). Thus, the amplitude is about 0.258 meters.

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Key Concepts

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

Doppler Effect
The Doppler Effect is a fascinating phenomenon involving a change in the frequency of a wave in relation to an observer moving relative to the source of the wave. In simpler terms, it explains why a siren sounds different as it moves past you. In our exercise, the microphone experiences frequency changes due to its motion in simple harmonic motion (SHM). As the microphone swings towards the sound source, the frequency increases, resulting in what is known as the Doppler shift.
  • The closer the microphone moves to the source, the higher the frequency it detects.
  • When the microphone moves away, the frequency decreases.
  • The frequency change detected in the exercise was 2.1 Hz.
This change helps us understand the velocity of the microphone at any point in its motion. Think of it like how quickly things seem to shift and swirl as you move past them while driving in a car. That's the Doppler Effect in action!
Sound Frequency
Sound frequency refers to how fast sound waves vibrate per unit time. It is measured in Hertz (Hz) and directly influences the pitch of the sound we hear. Imagine a musical note; the frequency of that note will determine whether it sounds high or low to our ears. In this exercise, a sound frequency of 440 Hz is used, which corresponds to the musical note "A" above Middle C in a music standard pitch.
  • A higher frequency means a higher pitch.
  • Lower frequency results in a lower pitch.
  • The microphone detects frequency shifts due to its movement.
This means as the microphone moves in SHM, it experiences these alterations in sound frequency due to the Doppler Effect, allowing us to calculate sound wave behavior in dynamic settings.
Amplitude Calculation
Amplitude is a measure of how far something moves from its equilibrium position. In simple harmonic motion, it tells us how high or low the motion reaches. In this exercise, the amplitude is the maximum displacement of the microphone from its resting point as it vibrates up and down. To find the amplitude, we used the relationship between maximum velocity in SHM and the Doppler Effect formulas.
  • The maximum velocity for SHM is given by the equation: \( v_0 = A\omega \)
  • Doppler form: \( f' - f'' = \frac{2f \cdot v_0}{v^2 - v_0^2} \approx \frac{2f \cdot v_0}{v} \)
  • Solving for amplitude: \( A = \frac{2.1 \times 343}{2 \times 440 \times \pi} \approx 0.258 \text{ m} \)
So, in this scenario, the microphone has an amplitude of 0.258 meters, meaning that's as far as it moves from its position while vibrating.
Angular Frequency
Angular frequency, denoted \( \omega \), is a concept used to describe the rate of rotation or vibration in circular motion or oscillations in SHM. It provides a way to express how rapidly cycles or oscillations happen. In the context of SHM, it is crucial for calculating both velocity and displacement. The angular frequency \( \omega \) is calculated using the period \( T = 2 \text{ s} \) from the exercise, where:
  • \( \omega = \frac{2\pi}{T} \)
  • Plugging the period into it, we get \( \omega = \frac{2\pi}{2} = \pi \text{ rad/s} \)
  • It tells us that the microphone completes each cycle of motion it makes in this SHM setup in a precise time interval.
This value of \( \omega \) plays a vital role in determining how rapidly the Doppler shifts occur as the microphone moves towards and away from the sound source. By understanding angular frequency, one gets a better handle on the intricacies of oscillatory motion.

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Most popular questions from this chapter

An ultrasonic ruler, such as the one discussed in Example 4 in Section 16.6, displays the distance between the ruler and an object, such as a wall. The ruler sends out a pulse of ultrasonic sound and measures the time it takes for the pulse to reflect from the object and return. The ruler uses this time, along with a preset value for the speed of sound in air, to determine the distance. Suppose that you use this ruler under water, rather than in air. The actual distance from the ultrasonic ruler to an object is 25.0 m. The adiabatic bulk modulus and density of seawater are \(B_{\mathrm{ad}}=2.37 \times 10^{9} \mathrm{Pa}\) and \(\rho=1025 \mathrm{kg} / \mathrm{m}^{3}\) , respectively. Assume that the ruler uses a preset value of 343 \(\mathrm{m} / \mathrm{s}\) for the speed of sound in air. Determine the distance reading that the ruler displays.

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