Doppler Effect: Wavelengths of Sound Waves

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SUMMARY

The discussion centers on the Doppler Effect as it pertains to sound waves, specifically addressing the relationship between emitted and reflected wavelengths. The user correctly identifies that the wavelength of reflected sound waves (λ2) is shorter than that of emitted waves (λ1) when an object approaches a stationary source. The mathematical representation confirms that λ2 is derived from the equation λ2 = (u-v)λ1/(u+v), demonstrating that λ1 is greater than λ2 when the object is moving towards the source. This clarification resolves the misconception regarding the wavelength of reflected waves.

PREREQUISITES
  • Understanding of the Doppler Effect in sound waves
  • Familiarity with wave frequency and wavelength relationships
  • Basic knowledge of algebraic manipulation
  • Concept of relative velocity in physics
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  • Study the mathematical derivation of the Doppler Effect for sound waves
  • Explore the differences between the Doppler Effect for sound and light waves
  • Learn about applications of the Doppler Effect in radar technology
  • Investigate the impact of medium on sound wave propagation
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Students studying physics, educators teaching wave mechanics, and anyone interested in the practical applications of the Doppler Effect in sound and radar technology.

Karol
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Homework Statement


I read that when a body approaches a stationary one which emits radar waves, the wavelength λ2 of the returned waves is longer than λ1, the wavelength of the emitted ones, but I get the opposite. I suppose this is true for sound waves also.
See picture.

The Attempt at a Solution


Velocity of sound in the air: u
Velocity of approaching object: v
The frequency at which the waves from the stationary source hit the moving object:
f_1=\frac{u+v}{\lambda_1}
The wavelength of the reflected wave:
\lamda_2=\frac{u-v}{f_1}=\frac{(u-v)\lambda_1}{u+v}
Now:
\frac{u-v}{u+v}<1\rightarrow\lambda_1>\lambda_2
 

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Your calculation looks OK. But why would you think that the reflected waves would have a longer wavelength? Where did you read that?
 

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