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Statement-1: When a sound source moves towards observer, then frequency of sound increases. Statement-2: Wavclength of sound in medium towards obscrver is decreased.

Short Answer

Expert verified
Both statements are true and related through the Doppler Effect.

Step by step solution

01

Understand Statement-1

Statement-1 suggests that the frequency of the sound increases as the source approaches the observer. This is a known phenomenon called the Doppler Effect, which occurs because the relative motion affects the frequency perceived by the observer.
02

Analyze Statement-2

Statement-2 claims that the wavelength of sound decreases in the medium when the source moves towards the observer. According to the Doppler Effect, as the source approaches, the sound waves are compressed, reducing their wavelength.
03

Correlate the Statements with Doppler Effect

Both statements are linked through the Doppler Effect. As the sound source moves towards the observer, the waves in front of the source are compressed, leading to a shorter wavelength (as per Statement-2) and higher frequency (as per Statement-1).
04

Conclusion

Both statements are true and complement each other in explaining the effect of a moving sound source on frequency and wavelength. Statement-1 is explained by the outcome described in Statement-2 due to the Doppler Effect.

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Key Concepts

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

Frequency Change
Imagine you're standing on the sidewalk, and a car approaches with a blaring horn. Have you noticed how the pitch of the horn seems to increase as the car gets closer? This change in pitch or frequency is a beautiful phenomenon known as the Doppler Effect.

**What Causes Frequency Change?** When a sound source, like the car horn, moves toward you, the sound waves in front of it get "squeezed." This means more wave peaks hit your ear per second, leading to an increase in frequency - you hear a higher pitch.
  • Sound waves compress as they travel towards the observer.
  • This compression causes the waves to reach the ear more frequently.
  • As a result, the frequency perceived by the observer rises.

Just remember - as the source moves towards you, think of frequency as the number of waves per time unit that are squishing closer together, raising the pitch you perceive.
Wavelength Compression
While the notion of frequency might seem abstract, wavelength is more tangible. When we talk about wavelength compression, we're discussing how the physical distance between wave peaks changes when a sound source moves.

**How Does Wavelength Compression Work?** If you held a slinky and pushed one end rapidly, you'd notice how the rings get closer - this is what happens with sound waves when the source moves closer to you. The waves "bunch up" or compress.
  • Wavelength is the distance between consecutive peaks of the wave.
  • As the source moves closer, this distance shortens.
  • This compression causes a higher frequency since more waves are packed in the same space.
Think of wavelength compression as the physical shortening of sound waves in the air in front of the moving source. It's like a vehicle getting into a tighter parking spot, making everything feel snug and closer.
Sound Waves
Sound waves are all around us and fundamentally simple yet fascinating. Sound travels as waves through air by vibrating particles. If you've ever thrown a pebble into a pond, you've seen how waves ripple outward - sound waves behave similarly in the air as they move away from the source.

**Understanding Sound Waves** Think of sound as energy that moves through a medium, such as air, water, or any solid object.
  • Sound waves are longitudinal, meaning they move in the same direction as their energy transfer.
  • They rely on particles to vibrate and carry the sound.
  • Without a medium like air, sound cannot travel; space is silent!
A fun way to imagine sound waves is as invisible ripples spreading through an ocean of air molecules. That's why when a sound source moves toward you, those ripples squish closer together and create the Doppler Effect's impressive frequency change.

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Most popular questions from this chapter

If two orthogonal SHM of same frequency having initial phase difference of \(\pi / 2\) act simultaneously on particle free to move. The particle will move in (a) Straight line (b) Parabola (c) Cirele (d) Ellipse

In case of forced vibrations (a) displacement varies with the frequency of force (b) displacement is not in phase with force (c) amplitude decreases exponentially with time (d) amplitude becomes infinity if the frequency of force coincides with that of the body

The driver of a motor van approaching a vertical wall notices that the frequency of his van's horn changes from \(440 \mathrm{Ilz}\) to \(480 \mathrm{lIz}\) when it gets reflected from the wall. Find the speed of the van. (The specd of sound \(=330 \mathrm{~m} / \mathrm{s}\) )

As a wave propagates (a) the wave intensity remains constant for a plane wave. (b) the wave intensity decreases as the inverse of the distance from the source for a spherical wave. (c) the wave intensity decreases as the inverse square of the distance from the source for a spherical wave. (d) total intensity of the spherical wave over the spherical surface centered at the sources remains constant at all the times.

\(\delta=\frac{m}{L}=\frac{32}{4} \times 10^{2}=\frac{4}{5} \times 10^{2} \mathrm{~kg} / \mathrm{m}\) 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) $$ \begin{aligned} Y &=\frac{\text { stress }}{\text { strain }}=\frac{2478.8 \times 10^{4}}{10^{2} / 8} \\ &=1.98 \times 101^{\prime} \mathrm{N} / \mathrm{m}^{2} \end{aligned} $$

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