Physics: doppler effect (observer & source in motion)

In summary, a train is moving parallel and adjacent to a highway with a constant speed of 25 m/s. At time t0, a car is 42 m behind the train, traveling in the same direction at 49 m/s and accelerating at 4 m/s^2. The train's whistle blows at a frequency of 460Hz and the speed of sound in air is 343 m/s. At t=t0, the car's driver hears a whistle from the train. Using the equation f' = f [(1+ v(observer) / v) / (1- v(source) / v)], the frequency heard by the car's driver is 567.044 Hz. At a later time t, when the
  • #1
bastige
15
0

Homework Statement


A train is moving parallel and adjacent to a highway with a constant speed of 25 m/s. A car, at time t0, is 42 m behind the train, traveling in the same direction as the train at 49 m/s, and accelerating at 4 m/s^2. The train's whistle blows at a frequency of 460Hz. The speed of sound in air is 343 m/s. At the moment t0, the car's driver hears a whistle from the train.

What frequency does the car's driver hear?

Part B. @ a time t, after the car has passed the train and is a distance 32 m ahead of the train, the whistle blows again. What frequency does the driver hear?


Homework Equations



f' =f [ (1+ v(observer) / v) / (1- v(source) / v)] <---equation 1

x(final) = x(initial) + v t +.5 a t2 <---equation 2

f'= f [1- (v(observer)/v)] <---equation 3

The Attempt at a Solution



Part A:

460[(1+(49/343)) / (1-(25/343)) = 567.044 = WRONG ANSWER


Part B:


I treated the train as stationary to find the time,I determined the car to be traveling 24 m/s
So using equation 2
74 = 0 + 24 t + ( .5 * 4 ) t2
v(final)=49 + ( 2.54 * 4 ) = 59.16 m/s



then using equation 3 I did

f'=460(1- (59.16/343) ) = 380.66=WRONG
 
Physics news on Phys.org
  • #2
f' =f (1+ v(observer) / v) / (1- v(source) / v)] <---equation 1
Is this what you get on simplifying the above equation?

[tex] f'=f(\frac{v+v_{obs}}{v-v_{s}} )[/tex]

Try not to apply the equation directly. Since, the frequency is received by the car at that moment [tex]t_{0}[/tex] surely the sound wave of that frequency which we have to find out must have left some time [tex]t_{b}[/tex] ago for it to reach at that instant. Apply this, it might help.

As for computing the frequency during acceleration, you cannot simply apply the standard equations. For every time period T that the wave moves forward a length [tex]\lambda[/tex] the car moves to cover a length of [tex] \frac{1}{2}aT^2[/tex]. The effective wavelength should now become: [tex] \lambda' = \lambda - \frac{1}{2}aT^2 [/tex].
 
  • #3
ANSWER

I would like to point out that your solution for Part A is incorrect. The correct equation to use for the Doppler effect in this scenario is equation 1, which takes into account both the observer and the source's velocities. In this case, the car is moving towards the train, so its velocity should be considered positive. Therefore, the correct solution for Part A is:

f' = 460 [(1 + 49/343) / (1 - 25/343)] = 522.5 Hz

For Part B, your approach is correct, but there is a small error in your calculation. The correct velocity for the car after passing the train is 53 m/s, not 59.16 m/s. Therefore, the correct solution for Part B is:

f' = 460 (1 - 53/343) = 418.8 Hz

I hope this clarifies the correct approach to solving this problem. Remember to always use the correct equations and pay attention to the signs of velocities in the Doppler effect.
 

1. What is the Doppler Effect?

The Doppler Effect is a phenomenon in which there is a perceived change in frequency of a wave due to the relative motion between the source of the wave and the observer. This effect is commonly observed with sound waves, but it also applies to other types of waves such as light.

2. How does the Doppler Effect work?

The Doppler Effect occurs because of the change in distance between the source of the wave and the observer. When the source and observer are moving towards each other, the distance decreases and the perceived frequency increases. Conversely, when the source and observer are moving away from each other, the distance increases and the perceived frequency decreases.

3. What factors affect the Doppler Effect?

The frequency of the wave, the speed of the source and observer, and the distance between them are the main factors that affect the Doppler Effect. In addition, the speed of sound or light in the medium between the source and observer can also have an impact.

4. How is the Doppler Effect used in modern technology?

The Doppler Effect is used in various technologies such as radar, sonar, and medical ultrasound. These technologies use the change in frequency of the reflected waves to determine the velocity and distance of the objects being observed. The Doppler Effect is also used in astronomy to study the motion of celestial bodies.

5. What are the limitations of the Doppler Effect?

The Doppler Effect assumes that the source and observer are moving in a straight line and at a constant speed. This may not always be the case in real-world situations, which can lead to inaccuracies in the perceived frequency. In addition, the Doppler Effect is only applicable to waves that are propagating through a medium, so it cannot be used to study objects in a vacuum.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
948
  • Introductory Physics Homework Help
Replies
1
Views
825
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
22
Views
5K
  • Introductory Physics Homework Help
Replies
6
Views
4K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
2K
Back
Top