Paying 50fg per parts.

Equations



#1
Two coherent sources of radio waves, A and B, are 5.00 meters apart. Each source emits waves with wavelength 6.00 meters. Consider points along the line connecting the two sources.

Part A
At what distance from source A is there constructive interference between points A and B?
Express your answer in meters.

Part B
At what distances from source A is there destructive interference between points A and B?
Note that there will be two separate interference fringes between point A and point B. Enter your answers in ascending order separated by a comma.


#2
A train is traveling at 30.0 m/s relative to the ground in still air. The frequency of the note emitted by the train whistle is 262 Hz.
The speed of sound in air should be taken as 344 m/s.

Part A
What frequency fapproach is heard by a passenger on a train moving at a speed of 18.0 m/s relative to the ground in a direction opposite to the first train and approaching it?
Express your answer in hertz.

Part B
What frequency frecede is heard by a passenger on a train moving at a speed of 18.0 m/s relative to the ground in a direction opposite to the first train and receding from it?
Express your answer in hertz.

This post was edited by Stolem on Nov 21 2013 08:20pm

ill do for all ur fg

Come on..

#1
Assuming they produce similar waves, we can assume that they both produce waves that are positive from 0 to 3 meters from their points of origin because the wavelength is 6 meters.

A: Because they are 5 meters apart, there is a one meter section in the middle (from 2 meters to 3 meters away from either station) in which there is constructive interference.
So the answer here would be between 2 and 3 meters.
B: For the same reason, all other points (from 0 to 2 meters from either station) has destructive interference.
The answer here would be from 0 to 2 meters and from 3 to 5 meters.

#2
Use doppler effect equations
A
f' = (1+uobs/uwave)f
f = 262 Hz
uwave = 374 m/s <- train + speed of sound
uobs = +18
f' = (1+18/374)*262Hz
f' = 274.51 Hz

B
For the second case, everything is the same except uobs is negative because the observer is moving away from the train.
f' = (1-18/374)*262Hz
f' = 249.39 Hz

This post was edited by Dontrunaway on Nov 21 2013 08:58pm

#5

This is what i did too but i got the wrong answer.

Approach F' = (1+ 48 m/s / 344 m/s)
F' = 276

Part B
F' = (1-48 m/s / 344 m/s ) f = 225 Hz


#3
Part A
Correct answer is 2.5 m

Part B
i got 4m but its wrong..

This post was edited by Stolem on Nov 21 2013 09:11pm

The equations you posted are not adequate for the problem, which is why we were both wrong.

Following equations from: http://formulas.tutorvista.com/physics/doppler-shift-formula.html

When both moving toward each other:

f' = (v+vo)/(v-vs)*f
f' = (344+18)/(344-30)*262
f' = 302.05
When both moving away from each other:

f' = (v-vo)/(v+vs)*f
f' = (344-18)/(344+30)*262
f' = 228.37

Those should be correct.

This post was edited by Dontrunaway on Nov 21 2013 09:13pm

#1 Answer was..

2.5

and

1,4

Thanks.. Done.