Doppler question - find the observed frequency

In summary, the conversation discusses calculating the final velocity by multiplying time and acceleration, with a specific example resulting in an incorrect answer. The speaker also mentions other mistakes such as not giving units and rounding incorrectly. They then ask about calculating final velocity without knowing the distance of the drop. To solve this, two equations can be constructed and solved for the time it takes to fall.
  • #1
JoeyBob
256
29
Homework Statement
see attached
Relevant Equations
fobs= ((v-vo)/(v-vs))fs
So first I calculate the final velocity by multiplying the time by the acceleration, 9.8, to get 88.2 m/s.

Now I use the equation. (343/(343-(-88.2))*108.3 = 86.1477.

But the answer should be 88.47. What am I doing wrong here?
 

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  • #2
9.0s is not the time it takes to fall. It is the time it takes to fall plus the time the sound takes to travel to the surface. So the time it takes to fall is less than 9.0s.

Other mistakes are:
- stating the acceleration is 9.8 without giving the unit;
- giving values for the final frequency and the 'official answer' frequency without giving the unit;
- rounding your answer to an incorrect number of significant figures.
 
  • #3
Steve4Physics said:
9.0s is not the time it takes to fall. It is the time it takes to fall plus the time the sound takes to travel to the surface. So the time it takes to fall is less than 9.0s.

Other mistakes are:
- stating the acceleration is 9.8 without giving the unit;
- giving values for the final frequency and the 'official answer' frequency without giving the unit;
- rounding your answer to an incorrect number of significant figures.
How would one calculate the final velocity then if distance of the drop isn't given?
 
  • #4
JoeyBob said:
How would one calculate the final velocity then if distance of the drop isn't given?
If the time it takes to fall is T, then the time for the sound to travel back up is (9.0-T).

You can construct 2 equations:
- one equation for the distance covered by the noisemaker in time T;
- one equation for the distance covered by the sound in time (9.0 - T).
Since these 2 distances are the same you get a quadratic equation for T which you solve..
 

1. How does the Doppler effect affect observed frequency?

The Doppler effect is a phenomenon where the observed frequency of a wave is altered by the relative motion between the source of the wave and the observer. If the source and observer are moving closer together, the observed frequency will increase, and if they are moving farther apart, the observed frequency will decrease.

2. How is the observed frequency calculated using the Doppler effect?

The observed frequency can be calculated using the formula fobs = fs (v±vo) / (v±vs), where fs is the frequency of the source, v is the speed of the wave, vo is the speed of the observer, and vs is the speed of the source. The ± sign depends on whether the source and observer are moving closer together or farther apart.

3. How does the speed of the source affect the observed frequency?

The speed of the source affects the observed frequency by altering the wavelength of the wave. If the source is moving towards the observer, the wavelength will be shortened and the observed frequency will increase. If the source is moving away from the observer, the wavelength will be lengthened and the observed frequency will decrease.

4. What is the difference between the observed frequency and the actual frequency?

The observed frequency is the frequency of a wave that is perceived by an observer, taking into account the relative motion between the source and the observer. The actual frequency is the frequency of the wave emitted by the source, without any alteration due to the Doppler effect.

5. How is the Doppler effect used in real-life applications?

The Doppler effect has numerous real-life applications, such as in radar and sonar systems, where it is used to measure the speed and direction of moving objects. It is also used in astronomy to determine the speed and direction of celestial bodies. Additionally, the Doppler effect is utilized in medical imaging techniques, such as ultrasound, to create images of internal structures in the body.

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