Moving Source, Observer at Rest, derivation for Doppler effect

In summary, the question asks whether the equation assumes that ##\frac{v}{f} ≥ \frac{v_S}{f}## in order for the wavelength to not be negative. The response clarifies that the object would be moving faster than the sound speed, indicating a supersonic speed which is a different physical regime.
  • #1
member 731016
Homework Statement
Please see below
Relevant Equations
Please see below
For this,
1685661674611.png

Does someone please know whether they assume for the equation highlighted that ##\frac{v}{f} ≥ \frac{v_S}{f}## since otherwise the wavelength would be negative (which I assume is impossible)?

Many thanks!
 
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  • #2
ChiralSuperfields said:
Homework Statement: Please see below
Relevant Equations: Please see below

For this,
View attachment 327342
Does someone please know whether they assume for the equation highlighted that ##\frac{v}{f} ≥ \frac{v_S}{f}## since otherwise the wavelength would be negative (which I assume is impossible)?

Many thanks!
The object would be moving faster than the sound speed; i.e., supersonic which is a different physical regime.
 
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  • #3
Frabjous said:
The object would be moving faster than the sound speed; i.e., supersonic which is a different physical regime.
Thank you for your reply @Frabjous!
 

Related to Moving Source, Observer at Rest, derivation for Doppler effect

What is the Doppler effect for a moving source and a stationary observer?

The Doppler effect for a moving source and a stationary observer describes the change in frequency or wavelength of a wave (such as sound or light) as perceived by an observer due to the motion of the source relative to the observer. If the source is moving towards the observer, the frequency appears higher (blue shift), and if the source is moving away, the frequency appears lower (red shift).

How is the Doppler effect formula derived for a moving source and a stationary observer?

The Doppler effect formula for a moving source and a stationary observer can be derived using the wave equation and the relative velocities of the source and the observer. The formula is given by \( f' = f \left(\frac{v + v_o}{v - v_s}\right) \), where \( f' \) is the observed frequency, \( f \) is the emitted frequency, \( v \) is the speed of the wave in the medium, \( v_o \) is the velocity of the observer (which is zero for a stationary observer), and \( v_s \) is the velocity of the source. Simplifying for a stationary observer, the formula becomes \( f' = f \left(\frac{v}{v - v_s}\right) \).

What assumptions are made in the derivation of the Doppler effect for a moving source?

The key assumptions made in the derivation include: 1) The medium through which the wave travels is uniform and stationary. 2) The velocities of the source and observer are much less than the speed of the wave in the medium. 3) The motion of the source is along the line connecting the source and the observer (i.e., linear motion).

How does the speed of the source affect the observed frequency in the Doppler effect?

If the source is moving towards the observer, the observed frequency increases because the waves are compressed, leading to a higher frequency (blue shift). Conversely, if the source is moving away from the observer, the observed frequency decreases because the waves are stretched, leading to a lower frequency (red shift). The change in frequency is directly proportional to the speed of the source relative to the speed of the wave in the medium.

Can the Doppler effect be observed with all types of waves?

Yes, the Doppler effect can be observed with all types of waves, including sound waves, light waves, and electromagnetic waves. However, the specific characteristics of the Doppler effect, such as the magnitude of the frequency shift, depend on the type of wave and the medium through which it travels. For example, the Doppler effect for sound waves is influenced by the speed of sound in air, while for light

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