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A friend of yours is loudly singing a single note at \(400 \mathrm{Hz}\) while racing toward you at \(25.0 \mathrm{m} / \mathrm{s}\) on a day when the speed of sound is \(340 \mathrm{m} / \mathrm{s}\) a. What frequency do you hear? b. What frequency does your friend hear if you suddenly start singing at \(400 \mathrm{Hz} ?\)

Short Answer

Expert verified
For part a, the frequency that the stationary observer hears is approximately \(431.7 \, \text{Hz}\). For part b, the frequency that the moving observer hears is approximately \(429.4 \, \text{Hz}\).

Step by step solution

01

Calculate the frequency heard by stationary observer (Part a)

To calculate the frequency that the stationary observer hears, use the formula for the Doppler effect: \( f' = f \, \frac{v + v_o}{v - v_s}\) where \(f\) is the original frequency, \(f'\) is the perceived frequency, \(v\) is the speed of sound, \(v_o\) is the speed of the observer, and \(v_s\) is the speed of the source. Note the observer is stationary so \(v_o = 0\), thus the formula simplifies to \(f' = f \, \frac{v}{v - v_s}\). Now substitute the given values into the formula: \(f' = 400 \, \frac{340}{340 - 25}\).
02

Solve for the new frequency (Part a)

You calculate this part simply by solving the equation that results in \(f' = \frac{400 \cdot 340}{315}\), which further simplifies to \(f' \approx 431.7 \, \text{Hz}\). This is the frequency that the stationary observer hears.
03

Calculate the frequency heard by moving observer (Part b)

The observer in this scenario is the one moving, so use the formula for the Doppler effect when the observer is moving towards the source: \(f' = f \, \frac{v + v_o}{v}\). Now substitute the given values into the formula: \(f' = 400 \, \frac{340 + 25}{340}\).
04

Solve for the new frequency (Part b)

You calculate this part simply by solving the equation that results in \(f' = \frac{400 \cdot 365}{340}\), which simplifies to \(f' \approx 429.4 \, \text{Hz}\). This is the frequency that the moving observer hears.

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Key Concepts

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

Frequency Calculation
In the problem of calculating frequency changes due to motion, the Doppler Effect plays a critical role. It's a fascinating phenomenon where the frequency of a wave changes based on the relative motion between a wave source and an observer. This is what happens when a car honking its horn drives toward you, and its pitch sounds higher than when it moves away.

When determining the frequency a stationary observer hears, we use the formula: \[ f' = f \frac{v}{v - v_s} \]Here,
  • \( f \) is the original frequency
  • \( f' \) is the perceived frequency
  • \( v \) represents the speed of sound
  • \( v_s \) is the speed of the sound source
The observer's speed \( v_o \) is zero, simplifying the equation. By plugging in the values, you can determine how the frequency is altered from 400 Hz to about 431.7 Hz. The formula accounts for the compression of sound waves as the source approaches, resulting in a higher pitch perceived by the observer.

Understanding calculations concerning changing frequencies helps us connect mathematical formulas with real-world experiences, such as interpreting how fast sound travels and alters under different conditions.
Speed of Sound
The speed of sound is a crucial element of the Doppler Effect, enabling us to comprehend how sound waves move through different mediums. On any given day, the speed of sound can be affected by factors such as air temperature and pressure. However, in many problems, it is estimated at around 340 m/s in air at room temperature.

The speed of sound defines the rate at which sound waves propagate through the medium. In the context of the problem, knowing that sound travels at 340 m/s allows us to compute how the motion of the source affects the sound wave's frequency. Remember, sound speed is essential for calculating how motion alters our perception of frequencies via the Doppler Effect.
  • Speeds greater than the speed of sound result in phenomena like sonic booms.
  • Understanding the speed of sound helps develop technologies in audio, acoustics, and even sonar systems.

Recognizing how sound speed interacts with moving sources enhances your intuition on wave behaviors in real-world scenarios, from car engines to jet planes.
Wave Motion
Wave motion describes how vibrations travel through a medium, transferring energy from one point to another without transporting matter. Sound is a perfect example of wave motion where air molecules vibrate, creating pressure waves our ears interpret as sound. The speed and frequency at which these waves move are central to understanding principles like the Doppler Effect.
Every wave comprises crests (high points) and troughs (low points), and the distance between them is the wavelength.
- The frequency of a wave, such as in sound, indicates how many waves pass a point per second.- The speed of the wave is a product of its frequency and wavelength, expressed as \( v = f \lambda \).In simpler terms, high frequencies imply more waves in a given timeframe and are often perceived as higher-pitched sounds, while low frequencies correspond to fewer waves and lower-pitched sounds.

Understanding wave motion is central when analyzing how sounds compress and stretch with movement, as experienced in the Doppler Effect. Every sound, be it music or a siren, traces back to these fundamental principles of wave propagation.

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Most popular questions from this chapter

A concert loudspeaker suspended high above the ground emits 35 W of sound power. A small microphone with a \(1.0 \mathrm{cm}^{2}\) area is \(50 \mathrm{m}\) from the speaker. a. What is the sound intensity at the position of the microphone? b. How much sound energy impinges on the microphone each second?

The displacement of a wave traveling in the positive \(x\) direction is \(D(x, t)=(3.5 \mathrm{cm}) \sin (2.7 x-124 t),\) where \(x\) is in m and \(t\) is in \(s\). What are the (a) frequency, (b) wavelength, and (c) speed of this wave?

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One cue your hearing system uses to localize a sound ( \(1 . e,\) to tell where a sound is coming from) is the slight difference in the arrival times of the sound at your ears. Your ears are spaced approximately \(20 \mathrm{cm}\) apart. Consider a sound source \(5.0 \mathrm{m}\) from the center of your head along a line \(45^{\circ}\) to your right. What is the difference in arrival times? Give your answer in microseconds. Hint: You are looking for the difference between two numbers that are nearly the same. What does this near equality imply about the necessary precision during intermediate stages of the calculation?

A spherical wave with a wavelength of \(2.0 \mathrm{m}\) is emitted from the origin. At one instant of time, the phase at \(r=4.0 \mathrm{m}\) is \(\pi\) rad. At that instant, what is the phase at \(r=3.5 \mathrm{m}\) and at \(r=4.5 \mathrm{m} ?\)

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