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Chapter 16: Problem 64

A person riding a car moving at \(72 \mathrm{~km} / \mathrm{h}\) sounds a whistle emitting a wave of frequency \(1250 \mathrm{~Hz}\). What frequency will be heard by another person standing on the road (a) in front of the car (b) behind the car ? Speed of sound in air \(=340 \mathrm{~m} / \mathrm{s}\).

Short Answer

Expert verified
The frequency heard in front of the car is about 1328 Hz, and behind the car is about 1181 Hz.

Step by step solution

01

Convert Car Speed to m/s

First, we need to convert the speed of the car from kilometers per hour to meters per second. The conversion is done as follows: \[ 1 ext{ km/h} = \frac{1000}{3600} ext{ m/s} = \frac{5}{18} ext{ m/s} \]Thus, the speed of the car is:\[ 72 ext{ km/h} \times \frac{5}{18} = 20 ext{ m/s} \]
02

Understand the Doppler Effect Formula

The Doppler effect formula for a source moving towards a stationary observer is given by:\[ f' = \frac{f \cdot (v + v_o)}{v - v_s} \]where:- \( f' \) is the observed frequency,- \( f \) is the emitted frequency (1250 Hz),- \( v \) is the speed of sound in air (340 m/s),- \( v_o \) is the speed of the observer (0 m/s since the observer is stationary),- \( v_s \) is the speed of the source (20 m/s).
03

Calculate Frequency in Front of the Car

Use the Doppler effect formula when the car is moving towards the observer:\[ f' = \frac{1250 \cdot (340 + 0)}{340 - 20} \]Simplify and calculate:\[ f' = \frac{1250 \cdot 340}{320} = 1328.125 \text{ Hz} \]Therefore, in front of the car, the frequency heard is approximately 1328 Hz.
04

Calculate Frequency Behind the Car

Use the Doppler effect formula when the car is moving away from the observer:\[ f' = \frac{1250 \cdot (340 + 0)}{340 + 20} \]Simplify and calculate:\[ f' = \frac{1250 \cdot 340}{360} = 1180.56 \text{ Hz} \]Thus, behind the car, the frequency heard is approximately 1181 Hz.

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Key Concepts

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

Wave Frequency
Let's take a closer look at wave frequency. Frequency refers to the number of wave cycles that pass a point per unit of time. It is measured in hertz (Hz), where one hertz equals one cycle per second. In our example, a car emits a sound wave with a frequency of 1250 Hz. This means that 1250 wave cycles are emitted every second by the whistle on the car.

When the source of the sound moves, there can be changes in how this frequency is perceived by an observer. The reason behind these changes is the Doppler Effect, a physical phenomenon that describes the alteration of a wave's frequency due to the motion of its source relative to an observer.

Here's something exciting about wave frequency: it stays the same as the source emits it, but when the source moves towards or away from an observer, the observer hears a different frequency. This effect is crucial in understanding the scenarios when the car approaches and then passes the stationary person on the road.
Speed of Sound
The speed of sound plays a vital role in how we hear waves from moving sources. In the air, sound travels at a speed of approximately 340 meters per second (m/s). This speed can vary slightly depending on factors like temperature and humidity.

Understanding the speed of sound helps in determining how fast sound waves travel through the air, serving as a constant in our calculations. It is the foundation upon which the Doppler Effect formula rests.

When we talk about the Doppler Effect, the speed of sound ( v ) is crucial. It appears in the denominator of the formula we use to calculate perceived frequency. Knowing that sound travels consistently at 340 m/s allows us to make accurate predictions about how the frequency shifts as the car moves relative to a stationary observer.
Frequency Calculation
Calculating frequency changes due to the Doppler Effect involves a straightforward but essential formula: \[ f' = \frac{f \cdot (v + v_o)}{v - v_s} \]where:
  • \( f' \) is the frequency observed by the listener.
  • \( f \) is the original frequency emitted (1250 Hz here).
  • \( v \) is the speed of sound in air (340 m/s).
  • \( v_o \) is the observer's speed (0 m/s as the observer is stationary).
  • \( v_s \) is the source's speed (20 m/s, the speed of the car).
For an observer in front of the moving car, the speed of the source reduces the relative speed of sound, increasing the perceived frequency to about 1328 Hz. Conversely, when the car moves away, the relative speed increases, lowering the perceived frequency to roughly 1181 Hz.

The effective use of this formula not only allows us to quantify the changes in frequency but also aids in grasping why such changes occur, offering an insightful perspective into wave behavior and the practical application of the Doppler Effect.

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Most popular questions from this chapter

Calculate the speed of sound in oxygen from the following data. The mass of \(22 \cdot 4\) litre of oxygen at STP \(\left(T=273 \mathrm{~K}\right.\) and \(p=1 \cdot 0 \times 10^{5} \mathrm{Nm}^{-2}\) ) is \(32 \mathrm{~g}\), the molar heat capacity of oxygen at constant volume is \(C_{V}=2 \cdot 5\) \(R\) and that at constant pressure is \(C_{p}=3 \cdot 5 R\)

Find the greatest length of an organ pipe open at both ends that will have its fundamental frequency in the normal hearing range \((20-20,000 \mathrm{~Hz}) .\) Speed of sound in air \(=340 \mathrm{~m} \mathrm{~s}^{-1}\).

Two audio speakers are kept some distance apart and are driven by the same amplifier system. A person is sitting at a place \(6 \cdot 0 \mathrm{~m}\) from one of the speakers and \(6 \cdot 4 \mathrm{~m}\) from the other. If the sound signal is continuously varied from \(500 \mathrm{~Hz}\) to \(5000 \mathrm{~Hz}\), what are the frequencies for which there is a destructive interference at the place of the listener? Speed of sound in air \(=320 \mathrm{~m} \mathrm{~s}^{-1}\).

In Quincke's experiment, the sound intensity has a minimum value \(I\) at a particular position. As the sliding tube is pulled out by a distance of \(16: 5 \mathrm{~mm}\), the intensity increases to a maximum of \(9 I .\) Take the speed of sound in air to be \(330 \mathrm{~m} \mathrm{~s}^{-1}\). (a) Find the frequency of the sound source. (b) Find the ratio of the amplitudes of the two waves arriving at the detector assuming that it does not change much between the positions of minimum intensity and maximum intensity.

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 \(=330 \mathrm{~m} / \mathrm{s}\).
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