Why Does T0 Appear in the Doppler Effect Equation?

AI Thread Summary
The discussion centers on the appearance of T0 in the Doppler effect equation, specifically questioning why it is included in the time calculation for wave propagation from a moving source. The user explains the relationship between frequency, period, and the additional time required for waves emitted from a moving source to reach an observer. They illustrate this with an analogy of a car beeping its horn while moving away, emphasizing that the time between beeps increases due to the car's velocity. The conversation highlights the importance of understanding how the period changes with motion, suggesting that focusing on period rather than frequency may clarify the concept. Overall, the discussion aims to clarify the mathematical derivation and implications of the Doppler effect in wave mechanics.
logearav
Messages
329
Reaction score
0

Homework Statement



Revered members,
I have attached the image of Doppler effect explanation.

Homework Equations



I have the following doubts
1)t1 = L/v
2) t2 = T0 +( L + vsT0)/v
But why T0 comes here, instead of ( L + vsT0)/v

The Attempt at a Solution


 

Attachments

  • doppler effect.jpg
    doppler effect.jpg
    43.5 KB · Views: 539
Physics news on Phys.org
A stationary source emits waves of frequency F and period 1/F. The period is the time it takes successive wave fronts to reach you. How does the period change when the source moves away from you? First you need to know how far the source moves in the time 1/F, that is simply 1/F times the source velocity v_s. Now how much is the period lengthened? By the extra time it takes wave to go the extra distance v_s/F which is delta_t =v_s/Fv where v is the wave velocity. So the new period between wave fronts of a moving source is

1/F' = 1/F + v_s/Fv = 1/F(1+ v_s/v)

Makes more sense to me thinking of the period then the frequency?

If you are troubled by what is written derive the formula yourself.
 
A stationary source emits waves of frequency F and period 1/F. The period is the time it takes successive wave fronts to reach you. How does the period change when the source moves away from you? First you need to know how far the source moves in the time 1/F, that is simply 1/F times the source velocity v_s.
Spinnor said:
Now how much is the period lengthened? By the extra time it takes wave to go the extra distance v_s/F which is delta_t =v_s/Fv where v is the wave velocity.

Thanks for the reply Spinnor. I don't understand the line given in quote.
 
Lets say you are in a car and beep your horn once a second. Everyone who is at rest relative to the car will also hear the horn beep once a second. Now suppose you drive away from me at some large velocity, say 1/6 the speed of sound all the while beeping your horn once a second. With this information you should be able to compute the amount of time I hear between beeps. Yes? It will be longer then one second because the beep has longer to travel. You only need three numbers, the period of the beeps, the velocity of the car, and the velocity of sound in air.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top