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Chapter 10: Problem 37

Statement-1: When a source of sound is approaching an observer, the frequency perceived by the observer is greater than the frequency of the sound source. Statement-2: As the sound-source approaches the listener, the wavelength of the sound decreases while its speed relative to the medium remains unchanged.

Short Answer

Expert verified
Both statements are correct; Statement 2 explains Statement 1.

Step by step solution

01

Understanding the Problem

We need to assess the validity of two statements regarding changes in frequency perception when a sound source approaches an observer.
02

Analyzing Statement 1

Statement 1 claims that when a source of sound approaches an observer, the frequency heard by the observer increases compared to the source's frequency. This is an application of the Doppler Effect, where the observed frequency is higher if the source is moving toward the observer. Thus, Statement 1 is correct.
03

Evaluating Statement 2

Statement 2 asserts that as the sound source approaches, the wavelength decreases but the speed relative to the medium stays the same. According to wave theory, the speed of sound in a medium is constant and independent of the source's motion. As the source approaches, the wavelength decreases while the frequency increases, keeping the speed constant. Therefore, Statement 2 is also correct.
04

Conclusion: Relationship Between Statements

Both statements accurately describe aspects of the Doppler Effect related to a sound source moving towards an observer. Statement 2 provides a correct explanation for why the perceived frequency in Statement 1 is greater.

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Key Concepts

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

Frequency Perception
Frequency perception refers to how we perceive the frequency of sound, which is the number of wave cycles per second, usually measured in Hertz (Hz). When a sound source is stationary, the observer hears the frequency emitted by the source, known as the natural frequency of the sound. But, what happens when the sound source starts moving?

This is where the Doppler Effect comes into play. If the sound source approaches the observer, the frequency we perceive increases. Put simply, we hear a higher pitch. Conversely, if the sound source moves away, the frequency we perceive decreases, leading to a lower pitch. This is due to the compression of sound waves as the source moves toward us, resulting in more waves reaching our ears per second.

It's important to realize that the change in frequency perception doesn't mean the sound's actual frequency or speed changes; it's just how we perceive it due to relative movement.
Wavelength Change
Wavelength change is closely linked to the frequency change observed in the Doppler Effect. The wavelength of a sound is the physical distance between two consecutive peaks or troughs in a wave. When a sound source moves towards an observer, the sound waves get compressed.

This compression reduces the wavelength. As the sound waves get crammed closer together, the frequency increases because more compressed waves hit the observer's ear every second. Think of a slinky toy: when you push one end towards the other, the coils get closer. That's similar to how sound wavelengths change when a source approaches the observer.

This change in wavelength does not affect the speed of sound in the medium; it remains constant. It merely adjusts the distance between wave peaks, influencing the frequency that's perceived by the observer.
Wave Theory
Wave theory is fundamental in understanding phenomena like the Doppler Effect. It involves understanding how waves behave and interact with their surroundings. Sound travels through the air (or any medium) in waves, which can be thought of like ripples in a pond after a stone is thrown in.

According to wave theory, the speed of sound in a given medium remains constant, provided the conditions鈥攕uch as temperature and pressure鈥攄o not change. This means that any changes observed in frequency or wavelength can be attributed to the movement of the source relative to the observer, not to changes in the wave鈥檚 speed.

In essence, wave theory helps explain why the pitch of a moving sound source changes for an observer but the speed of sound stays consistent. It's a key principle that provides a foundation for understanding more complex wave phenomena.
Sound Speed in Medium
The speed of sound in a medium, such as air, is determined by the properties of that medium. This includes its temperature, pressure, and density. For air at room temperature, sound travels at approximately 343 meters per second (1235 km/h or 767 mph).

Interestingly, a source moving closer or further away from the observer does not change this speed within the medium. The Doppler Effect may alter our perception of frequency and wavelength, but the speed at which the sound travels remains constant. This constancy allows scientists to use equations involving speed, frequency, and wavelength to predict and explain various acoustic phenomena.

It's worth noting that sound may travel at different speeds in different media鈥攆aster in water than in air, for example. But regardless of the medium, the critical point is: the speed of sound is unaffected by the source鈥檚 motion toward or away from the observer.

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Most popular questions from this chapter

A cylindrical tube, open at one end and closed at the other, is in acoustic unison with an external source of frequency held at the open end of the tube, in its fundamental node. Then (a) the wave from the source gets rellected with a phase change o\lceil \(\pi\) at the closed end (b) the wave from the source gel rellected without a phase change at the closod end (c) the wave rellected from the closed end again gets rellected at the open end (d) the wave reflected from the closed end does not sulfer relloction at the open end

Statement-1 : The fundamental frequency of an open organ pipe increases as the temperature is increased. Statement-2 : As the temperature increases, the velocity of sound increases more rapidly than length of the pipe.

\(\delta=\frac{m}{L}=\frac{32}{4} \times 10^{2}=\frac{4}{5} \times 10^{2} \mathrm{~kg} / \mathrm{m}\) Now, According to question, \(v=v_{o}=220 \mathrm{II} z\) \(\therefore \frac{v}{2 L}=220 \mathrm{~Hz} \quad[\quad L=40 \mathrm{~cm}\) \(m=3.2 \mathrm{gm}\) \(\sqrt{\frac{T}{\delta}} \times \frac{1}{2 L}=220 \mathrm{II} z\) 1. \(\sqrt{\frac{T}{\rho}}=220 \times 2 L\) \(\therefore \quad \sqrt{T}=\sqrt{\rho} \times 220 \times 2 L$$$ \begin{aligned} &\therefore \quad T=\rho \times(220 \times 2 L)^{2}=\frac{4}{5} \times 10^{-2}(22 \times 2 \times 4)^{2} \\ &=\frac{4}{5} \times 10^{-2} \times 30976=2478.8 \times 10^{-2} \\ &\text { Now suress }=\frac{F}{\Lambda}=\frac{T}{\Lambda}=\frac{2478.8 \times 10^{-2}}{\left(10^{-3}\right)^{2}} \\ &=2478.8 \times 10^{4}=\mathrm{N} / \mathrm{m}^{2} \end{aligned} $$ and strain $$ =\frac{0.05}{40} \times \frac{10^{2}}{10^{2}}=\frac{5}{40} \times 10^{-2}=\frac{1}{8} \times 10^{-2} $$ \)\therefore$ (Youngs modulus) $$ \begin{aligned} Y &=\frac{\text { stress }}{\text { strain }}=\frac{2478.8 \times 10^{4}}{10^{2} / 8} \\ &=1.98 \times 101^{\prime} \mathrm{N} / \mathrm{m}^{2} \end{aligned} $$

Statement-1: If the amplitude of the oscillation is doubled the maximum kinetic energy becomes four times. Statement-2: If the amplitude of the oscillation is doubled the minimum potential energy also becomes four times.

Lel us consider source is of point or sphelical nature then, \(I \propto \frac{1}{x^{2}}\) \(L_{1}=120 d B \rightarrow I_{1}, x_{1}=3 \mathrm{~m}\) \(L_{2}=100 \mathrm{~dB} \rightarrow I_{2}, x_{2}=?\) \(L_{3}=10 d B \rightarrow I_{3}, x_{3}=?\) (a) \(\frac{I_{1}}{I_{2}}=\frac{x_{2}^{2}}{x_{1}^{2}}\) also \(L=10 \log \frac{I}{l_{v}}\) \(\therefore \quad L_{1}-L_{2}=10 \log \frac{I_{1}}{I_{n}}-10 \log \frac{I_{2}}{l_{0}}\) \(\therefore \quad 120-100=10 \log \left(\frac{l_{1}}{I_{2}}\right)\) \(\therefore \frac{I_{1}}{l_{2}}=10^{2}\) (2) \(\therefore \quad\) From \((1)\) and \((2)\), \(\frac{x_{2}^{2}}{x_{1}^{2}}=10^{2} \quad \therefore \quad x_{2}^{2}=x_{1}^{2} \times 10^{2}=(3 \times 10)^{2}\) \(\therefore \quad x_{2}=30 \mathrm{~m}\)(b) \(\frac{I_{1}}{l_{3}}=\frac{x_{3}^{2}}{x_{2}^{2}}\) \(\Rightarrow L_{2}-L_{3}=10 \log \left(\frac{I_{2}}{I_{3}}\right)\) \(\Rightarrow 90=10 \log \left(\frac{I_{2}}{I_{3}}\right) \Rightarrow \frac{I_{2}}{I_{3}}=10^{\circ}\) \(\therefore \quad\) From (1) and (2), \(\frac{x_{3}^{2}}{x_{2}^{2}}=10^{9}\) \(\Rightarrow x_{3}^{2}=x_{2}^{2} \times 10^{9}=\left(30 \times 10^{4} \sqrt{10}\right)^{2}\) \(\therefore \quad x_{3}=30 \times 3.163 \times 10^{4}=9.49 \times 10^{5}\)
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