The Student Room Group

Dopper effect question on IsaacPhysics

The question is :
https://isaacphysics.org/questions/conveying_the_doppler_effect?board=81ad395a-23fb-474a-a133-6891fd9266cb&stage=all

So far, I've tried to use the equation f' = f(V + Vo)/(V - Vs) which I got the difference as 162875Hz. I've also tried to use the equation f' = f(1 - v/c) which didnt get me anywhere either...
If anyone could possible help, it'd be greatly appreciated!!!!
did you remember to take the radial velocity of the belt?
Reply 2
Original post by Joinedup
did you remember to take the radial velocity of the belt?

In the question, am I right in saying that point B is the observer? And the source is the laser, which is travelling towards the conveyor belt. Therefore I think that the radial velocity is 0.11cos(75) + 3x10^8. Is this correct?
Original post by SGDumpling
In the question, am I right in saying that point B is the observer? And the source is the laser, which is travelling towards the conveyor belt. Therefore I think that the radial velocity is 0.11cos(75) + 3x10^8. Is this correct?

No, I was thinking of the rate of change of the displacement of a peanut (for instance) at B and the laser.
it's just 0.11 cos(75).... 2.85 m/s

when the radial speed is very much less than the wave speed the approximation given here is appropriate

https://isaacphysics.org/concepts/cp_doppler_effect?stage=all

Δf/f Δv/c

(because adding very small things to very large things can run you into problems with precision)

you'll also need a factor of 2 because the light is being reflected off the moving belt at B - section 1 of the question is hinting at this.
Reply 4
Original post by Joinedup
No, I was thinking of the rate of change of the displacement of a peanut (for instance) at B and the laser.
it's just 0.11 cos(75).... 2.85 m/s
when the radial speed is very much less than the wave speed the approximation given here is appropriate
https://isaacphysics.org/concepts/cp_doppler_effect?stage=all
Δf/f Δv/c
(because adding very small things to very large things can run you into problems with precision)
you'll also need a factor of 2 because the light is being reflected off the moving belt at B - section 1 of the question is hinting at this.

OH MY LORD, thank you so much!!!! You are a saviour!!! I completely overcomplicated things and didn't realize it was asking for reflected light... I got the right answer now, but I am still confused about some things in the question. In part a, I am a little confused why the frequency change is 2ΔF. Is it due to the fact that the conveyer belt is travelling towards the source, distance d, so the phase difference is 2d... but I'm not sure how that can immediately tell us that the frequency change is 2ΔF... but I can see the wavelength probably decreases, so frequency increases.
In part B, I dont really understand why the radial velocity is only 0.285 (0.11cos(75)) and not that + 3x10^8. My original thinking was that the source is stationary and the conveyor belt is as if moving towards the source, so i thought that that was the radial velocity. But why do we not take 3x10^8 into the radial velocity? as its like adding them together..

I'm really sorry about the questions, but thank you so much anyways!
Well you wouldn't multiply by 2 if the source was the laser and the observer was on the conveyor belt... but in this question we're interested in light that's bounced back to the laser source, so the laser source is also the position of the observer.

you could think about looking at yourself in a mirror, if you stand 2 meters in front of a mirror you see an image of yourself that looks like it's 2 m behind the mirror... i.e. 4 m away from you. if you walk forward by 1 meter so that you're now 1 m in front of the mirror the image now appears 1 m behind the mirror or 2 m away from you. if you'd been getting closer to the mirror at a rate of (say) 1 m/s your image would appear to be getting closer to you at a rate of 2 m/s
The same would occur if you stood still and a friend moved the mirror towards you - the image would appear to get closer to you at twice the speed the mirror is getting closer to you.


---

That's just the definition of radial velocity. Its the component of the velocity of an object along the straight line between the object and the observer. It doesn't depend on the method used to measure it. With the same setup the radial velocity would still be the same is the laser was turned off or if the laser was replaced by some ultrasound doppler detector at the same position in relation to the belt.
Reply 6
Original post by Joinedup
Well you wouldn't multiply by 2 if the source was the laser and the observer was on the conveyor belt... but in this question we're interested in light that's bounced back to the laser source, so the laser source is also the position of the observer.
you could think about looking at yourself in a mirror, if you stand 2 meters in front of a mirror you see an image of yourself that looks like it's 2 m behind the mirror... i.e. 4 m away from you. if you walk forward by 1 meter so that you're now 1 m in front of the mirror the image now appears 1 m behind the mirror or 2 m away from you. if you'd been getting closer to the mirror at a rate of (say) 1 m/s your image would appear to be getting closer to you at a rate of 2 m/s
The same would occur if you stood still and a friend moved the mirror towards you - the image would appear to get closer to you at twice the speed the mirror is getting closer to you.
---
That's just the definition of radial velocity. Its the component of the velocity of an object along the straight line between the object and the observer. It doesn't depend on the method used to measure it. With the same setup the radial velocity would still be the same is the laser was turned off or if the laser was replaced by some ultrasound doppler detector at the same position in relation to the belt.

Ok thank you so much, that has cleared things up for me now!!!! Thank you so much!!!

Quick Reply

Latest