Calculating beta for two moving light sources

[billy92]

Question

Two monochromatic light sources approach each other head-on with equal
speeds, v=\beta c relative to the laboratory. When they pass each other, the
frequency of the light each receives from the other is halved. Show that \beta =3-\sqrt{8}

Attempt at solution

I have tried to solve this a number of different ways using

f={f}'\sqrt{\frac{1-\beta }{1+\beta }} and f=\gamma (1-\beta ){f}'

Any help on how to begin to solve this problem would be appreciated

Thanks

 

[ghwellsjr]

If I understand the problem correctly, the answer should be beta equals 1/3.

Then the Doppler factor while they are approaching is 1.414 and after they pass is 0.707.

Assuming that I am correct, can you now solve the problem?

Where did you get the problem?

EDIT: OK, I see where I went wrong, I was calculating the relative speed between the light sources. You then have to take the velocity addition formula to solve for the laboratory speed.

 

[billy92]

The question is from a worksheet we were set but i wasn't sure how to start the problem.

So i need to calculate the speed of each light source first and then calculate the speed relative to the lab using them?

 

[rude man]

1. You should have available somewhere the formula for Doppler frequency modification for a given relative speed between an observer and a light source. For example, Halliday & Resnick, section 40.5. Doesn't look like you have the right formula (what is γ?) for that.

2. Use that formula to compute the sources' speed relative to each other.

3. Then compute how fast (including sign) each light source "sees" the lab whizzing by. In other words, consider inertal reference frames S1 and S2 as belonging to the two sources, respectively, in your formula (1) above.

 

[rude man]

To clarify: before the sources meet, they see each other's frequency as f. After they pass each other, they each see the other's frequency as f/2. At least, I hope that's what was intended...

PS I came up with the same answer as the one given so all looks correct.

 

[andrien]

from law of addition of velocities,you will have
u=2βc/(1+β^2) because in case of light only relative motion matters unlike sound.
when it is approaching then ,the formula you have written is right. but when it is going away then numerator and denominator interchanged.
so,
√[(1-b)/(1+b)]=(1/2)√[(1+b)/(1-b)] where b=2β/1+β^2.
this gives,
β=(√2-1)/(√2+1)=3-√8

 

[billy92]

from law of addition of velocities,you will have
u=2βc/(1+β^2) because in case of light only relative motion matters unlike sound.
when it is approaching then ,the formula you have written is right. but when it is going away then numerator and denominator interchanged.
so,
√[(1-b)/(1+b)]=(1/2)√[(1+b)/(1-b)] where b=2β/1+β^2.
this gives,
β=(√2-1)/(√2+1)=3-√8

I had worked out both these parts previously which shows i was along the right tracks. However, why do you let the value of u equal to be so that i can be substituted in?