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Chapter 16: Problem 62

A traffic policeman standing on a road sounds a whistle emitting the main frequency of \(2 \cdot 00 \mathrm{kHz}\). What could be the appparent frequency heard by a scooter-driver approaching the policeman at a speed of \(36 \cdot 0 \mathrm{~km} / \mathrm{h}\) ? Speed of sound in air \(=340 \mathrm{~m} / \mathrm{s}\)

Short Answer

Expert verified
The apparent frequency is approximately 2058.8 Hz.

Step by step solution

01

Convert Speed to m/s

First, we need to convert the speed of the scooter-driver from km/h to m/s. We know that 1 km/h is equal to \(\frac{1}{3.6}\) m/s. Therefore, \(36\, \text{km/h} = 36 \times \frac{1}{3.6} = 10\, \text{m/s}\).
02

Write the Doppler Effect Formula for Approaching Source

The Doppler effect formula for the apparent frequency \(f'\) when the source is stationary and the observer is moving towards it is given by \(f' = f \left( \frac{v + v_o}{v} \right)\), where \(f\) is the emitted frequency, \(v\) is the speed of sound, and \(v_o\) is the speed of the observer.
03

Substitute Values into the Formula

We have \(f = 2000\, \text{Hz}\), \(v = 340\, \text{m/s}\), and \(v_o = 10\, \text{m/s}\). Substitute these values into the Doppler effect formula: \[f' = 2000 \left( \frac{340 + 10}{340} \right)\]
04

Calculate the Apparent Frequency

Simplify and calculate the expression. \[f' = 2000 \left( \frac{350}{340} \right) = 2000 \times \left(1.0294\right) \approx 2058.8\, \text{Hz}\]The apparent frequency heard by the scooter-driver is approximately \(2058.8\, \text{Hz}\).

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Key Concepts

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

Apparent Frequency
The apparent frequency is the frequency that is measured by an observer when they are moving relative to the source of the wave. In simpler terms, it鈥檚 what you actually hear or perceive, not necessarily the noise that's being directly produced by the source. This concept is fundamental when considering how we perceive sounds when either we or the sound source is in motion.

For instance:
  • An ambulance siren seems higher pitched as it approaches, and lower as it moves away.
  • The actual sound produced doesn't change, but the frequency that reaches your ears does.
Understanding apparent frequency is crucial for grasping how the Doppler effect impacts real-world situations like traffic, astronomy, and even medical imaging.
Speed of Sound
The speed of sound is how quickly sound waves travel through a medium. In air at room temperature, it typically travels at about 340 meters per second (m/s). This speed can vary depending on factors like:
  • Temperature: Sound travels faster in warmer air because the molecules are more active and bump into each other more easily.
  • Medium: Sound travels differently through water, air, or solids. For example, it moves faster in water than in air.
The calculations of apparent frequency and Doppler effect heavily rely on knowing the speed of sound, so it鈥檚 vital to use the correct value based on your specific conditions.
Frequency Conversion
Frequency conversion in the context of speed isn鈥檛 about changing the frequency itself, but more about translating units so they work in equations. When we deal with problems involving motion and sound, we often convert speeds to a common unit to simplify calculations.
  • For example, converting kilometers per hour (km/h) to meters per second (m/s) is frequently necessary. The conversion factor is: 1 km/h = \(\frac{1}{3.6}\) m/s.
  • This ensures that all variables in the Doppler equation align, making it easier to work out the apparent frequency accurately.
Correct unit conversion is fundamental to achieving the precise results expected in scientific calculations.
Doppler Effect Formula
The Doppler effect formula gives us the mathematical relationship to compute how the frequency of a wave changes relative to a moving observer or source. When the observer is moving towards a stationary source, the formula is:\[ f' = f \left( \frac{v + v_o}{v} \right) \]Here:
  • \( f' \) is the apparent frequency detected by the observer.
  • \( f \) is the emitted frequency of the wave.
  • \( v \) is the speed of sound in the medium.
  • \( v_o \) is the speed of the observer moving towards the source.
The Doppler effect formula is essential for calculating what an observer hears in scenarios involving moving objects and sound. It's widely applicable, from radar technology to understanding the universe's expansion.

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Most popular questions from this chapter

A tuning fork of frequency \(256 \mathrm{~Hz}\) produces 4 beats per second with a wire of length \(25 \mathrm{~cm}\) vibrating in its fundamental mode. The beat frequency decreases when the length is slightly shortened. What could be the minimum length by which the wire be shortened so that it produces no beats with the tuning fork?

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 \(C_{p}=3 \cdot 5 R\)

A tuning fork of unknown frequency makes 5 beats per second with another tuning fork which can cause a closed organ pipe of length \(40 \mathrm{~cm}\) to vibrate in its fundamental mode. The beat frequency decreases when the first tuning fork is slightly loaded with wax. Find its original frequency. The speed of sound in air is \(320 \mathrm{~m} / \mathrm{s}\)

The intensity of sound from a point source is \(1 \cdot 0 \times 10^{-8} \mathrm{~W} \mathrm{~m}^{-2}\) at a distance of \(5 \cdot 0 \mathrm{~m}\) from the source. What will be the intensity at a distance of \(25 \mathrm{~m}\) from the source?

Two coherent narrow slits emitting sound of wavelength \(\lambda\) in the same phase are placed parallel to each other at a small separation of \(2 \lambda\). The sound is detected by moving a detector on the screen \(\Sigma\) at a distance \(D(\gg>)\) from the slit \(S_{1}\) as shown in figure (16-E6). Find the distance \(x\) such that the intensity at \(P\) is equal to the intensity at \(O\).
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