Dopple Effect (Frequency vs Time)

  • Thread starter SpringPhysics
  • Start date
  • Tags
In summary, the conversation discussed using the observed frequency vs time graph to calculate the altitude of a jet airplane flying above an observer. The method involves finding two values of time on the graph and using the inverse of their difference to determine the frequency of the jet airplane. It was also mentioned that the jet's velocity can be resolved into two components, with the vertical component affecting the frequency and the horizontal component having no effect. The formula for determining the observed frequency as a function of the initial horizontal distance was also provided.
  • #1
Doppler Effect (Frequency vs Time)

Homework Statement

A jet airplane is flying above an observer parallel to the ground with speed vj. Given the observed frequency vs time graph, would you be able to calculate the altitude of the jet airplane? Explain, also, how to determine observed frequency as a function of x, where x is the initial horizontal distance of the jet airplane to the vertical directly above an observer.

Homework Equations

v = [tex]\lambda[/tex]f
x = vt

The Attempt at a Solution

I am really clueless on how to start this - I've been staring at this question for over an hour.
Is it true that if I find two values of time so that the inverse of the difference is a value of frequency on the graph, then that frequency is the frequency of the jet airplane? (The only problem is that if the jet airplane is initially moving away, this value will not be on the graph.)
Can someone give me a lead please?
Last edited:
Physics news on
  • #2
Resolve vj into two components. vj*sinθ towards the observer.And another component vj*cosθ perpendicular to the line joining jet and the observer. θ is the angle between vertical and line joining the jet and the observer.
vj*sinθ changes the frequency where as vj*cosθ does not affect the frequency.
So f' = [v/(v-vj*sinθ)]*f for approaching jet and f' = [v/(v + vj*sinθ)]*f for receding jet.
Sinθ = x/sqrt(x^2 + h^2) and x = vj*t.
  • #3
Ohh I get it now. Thanks a lot!

1. What is the Doppler Effect?

The Doppler Effect is the change in frequency of a wave, such as sound or light, due to the relative motion between the source of the wave and the observer. This results in a perceived change in the pitch or color of the wave.

2. How does the Doppler Effect work?

The Doppler Effect occurs because as the source of the wave moves closer to the observer, the waves are compressed and have a higher frequency. Conversely, as the source moves away from the observer, the waves are stretched out and have a lower frequency.

3. What is the difference between frequency and time in relation to the Doppler Effect?

Frequency and time are both important factors in understanding the Doppler Effect. Frequency refers to the number of waves that pass by a point in a given amount of time, while time refers to the duration of the wave itself. The Doppler Effect involves a change in frequency over time due to the motion of the source or observer.

4. How is the Doppler Effect used in real life?

The Doppler Effect has many practical applications in various fields. In medicine, it is used in ultrasound technology to measure the speed and direction of blood flow. In astronomy, it helps scientists determine the motion of stars and galaxies. It is also used in radar technology for detecting the speed and position of objects.

5. 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 even water waves. However, the effect may be more pronounced in certain types of waves, such as sound waves, due to their slower speed compared to other types of waves.

Suggested for: Dopple Effect (Frequency vs Time)