Doppler Effect: Wavelengths of Sound Waves

AI Thread Summary
When a moving object approaches a stationary source emitting waves, the wavelength of the reflected waves is shorter than that of the emitted waves, contrary to the initial assumption that it would be longer. The calculations confirm that the wavelength of the reflected wave, λ2, is derived from the relationship between the velocities of sound and the moving object. The formula shows that λ2 is indeed less than λ1, indicating a compression of the waves. The confusion may stem from a misunderstanding of the Doppler Effect, which states that approaching sources increase frequency and decrease wavelength. Understanding these principles clarifies the behavior of sound waves in motion.
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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