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A bullet passes past a person at a speed of \(220 \mathrm{~m} / \mathrm{s}\). Find the fractional change in the frequency of the whistling sound heard by the person as the bullet crosses the person. Speed of sound in air \(=330 \mathrm{~m} / \mathrm{s}\).

Short Answer

Expert verified
The fractional change in frequency is \( \frac{12}{5} \).

Step by step solution

01

Understanding the Doppler Effect

To solve this problem, we use the Doppler Effect, which describes how the frequency of a wave changes for an observer moving relative to the source of the wave. The formula is given by \( f' = f \frac{v + v_o}{v + v_s} \), where \( v \) is the speed of sound in air, \( v_o \) is the observer's speed, \( v_s \) is the source's speed, \( f' \) is the observed frequency, and \( f \) is the emitted frequency.
02

Relating the Problem to the Formula

In the problem, the bullet acts as the source, and the person is the observer. As the person is stationary, \( v_o = 0 \). Given: Speed of bullet \( v_s = 220 \, \text{m/s} \) and speed of sound \( v = 330 \, \text{m/s} \).
03

Calculating Frequency When Approaching

As the bullet approaches the person, the frequency \( f' \) heard is calculated using \( f' = f \frac{v}{v - v_s} = f \frac{330}{330 - 220} = f \frac{330}{110} = 3f \).
04

Calculating Frequency When Receding

When the bullet is moving away, the frequency \( f'' \) is calculated using \( f'' = f \frac{v}{v + v_s} = f \frac{330}{330 + 220} = f \frac{330}{550} = \frac{3}{5}f \).
05

Finding the Fractional Change in Frequency

The fractional change in frequency as the bullet crosses the person is computed by taking the ratio of the difference in frequencies to the initial frequency. This is given by \( \frac{f' - f''}{f} = \frac{3f - \frac{3}{5}f}{f} = \frac{15f - 3f}{5f} = \frac{12f}{5f} = \frac{12}{5} \).

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Key Concepts

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

Fractional Change in Frequency
The fractional change in frequency is an important concept when understanding the Doppler Effect. It measures how much the frequency of a sound wave changes compared to its original frequency. In the exercise, we can determine this by using the change in frequency as perceived by a stationary observer when a sound-emitting object, like a bullet, passes by.

IIn this scenario, as the bullet approaches, the frequency seems to increase, and as it moves away, the frequency seems to decrease. The formula to find the fractional change is given by:
  • Frequency approaching: \( f' = f \frac{v}{v - v_s} \)
  • Frequency receding: \( f'' = f \frac{v}{v + v_s} \)
Using these, the fractional change is given by \( \frac{f' - f''}{f} \). This formula helps isolate how much the sound frequency is altered as the object moves relative to the observer, emphasizing the observable impact of velocity on sound.

In the provided solution, this leads to a fractional change of \( \frac{12}{5} \), indicating a considerable change in perceived sound as the bullet moves past the listener.
Speed of Sound
The speed of sound is a crucial factor in calculating the perceived frequency of a sound wave due to the Doppler Effect. Sound travels at a constant speed in air, which is given in the exercise as \( 330 \, \text{m/s} \).

This constant speed allows us to predict how sound waves will behave as they travel from a source to an observer, whether the source or the observer is in motion.When calculating the frequency change due to the Doppler Effect, knowing the speed of sound is essential because it provides the baseline for measuring how the source's velocity alters the wave frequency relative to a stationary observer. In simple terms, the speed of sound determines how fast and how far the sound waves can travel in a given medium, making it a fixed reference point in our calculations.

By using the speed of sound in the exercises, we can determine how drastic the frequency change will be when the source, like our bullet, moves at speeds close to or vastly different from this constant speed.
Wave Frequency
Wave frequency is at the heart of understanding the Doppler Effect. It refers to the number of wave cycles that pass a given point per unit of time. In this problem, the bullet emits a frequency, perceived differently by a person standing still when the bullet moves towards and away from them due to changes in the relative motion.

By changing the speed of the source (like our bullet), the wave frequency changes, explaining why an approaching bullet sounds different compared to when it is receding. The emitted frequency from the source remains constant, but the relative motion causes an apparent increase or decrease in this frequency. This occurrence is integral to grasping the mechanics of both the Doppler Effect and sound wave behavior.

Essentially, as the source approaches, the frequency appears higher than normal; as it recedes, the frequency appears lower. Understanding these frequency changes can help us interpret phenomena in everyday life relating to all sorts of waves, from sound to light.

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Most popular questions from this chapter

Two sources of sound, \(S_{1}\) and \(S_{2}\), emitting waves of equal wavelength \(20^{\circ} 0 \mathrm{~cm}\), are placed with a separation of \(20 \cdot 0 \mathrm{~cm}\) between them. A detector can be moved on a line parallel to \(S_{1} S_{2}\) and at a distance of \(20 \cdot 0 \mathrm{~cm}\) from it. Initially, the detector is equidistant from the two sources. Assuming that the waves emitted by the sources are in phase, find the minimum distance through which the detector should be shifted to detect a minimum of sound.

Two point sources of sound are kept at a separation of \(10 \mathrm{~cm} .\) They vibrate in phase to produce waves of wavelength \(5 \cdot 0 \mathrm{~cm} .\) What would be the phase difference between the two waves arriving at a point \(20 \mathrm{~cm}\) from one source (a) on the line joining the sources and (b) on the perpendicular bisector of the line joining the sources?

Calculate the bulk modulus of air from the following data about a sound wave of wavelength \(35 \mathrm{~cm}\) travelling in air. The pressure at a point varies between \(\left(1 \cdot 0 \times 10^{5} \pm 14\right) \mathrm{Pa}\) and the particles of the air vibrate in simple harmonic motion of amplitude \(5 \cdot 5 \times 10^{-6} \mathrm{~m}\).

A bat emitting an ultrasonic wave of frequency \(4 \cdot 5 \times 10^{4} \mathrm{~Hz}\) flies at a speed of \(6 \mathrm{~m} / \mathrm{s}\) between two parallel walls. Find the two frequencies heard by the bat and the beat frequency between the two. The speed of sound is \(330 \mathrm{~m} / \mathrm{s}\).

A violin player riding on a slow train plays a \(440 \mathrm{~Hz}\) note. Another violin player standing near the track plays the same note. When the two are close by and the train approaches the person on the ground, he hears \(4 \cdot 0\) beats per second. The speed of sound in air= \(340 \mathrm{~m} / \mathrm{s}\). (a) Calculate the speed of the train. (b) What beat frequency is heard by the player in the train?

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