What time interval between wave fronts in S and S'?

AI Thread Summary
The discussion focuses on calculating the time interval between wave fronts arriving at observer C' from a moving train, which is 1.0 km long and traveling at 150 km/h. The solution involves using the formula Δt = t(back) - t(front), incorporating the train's speed and the speed of light. Participants clarify that while the train's speed is a small fraction of light speed, special relativity (SR) is not necessary for this scenario. They emphasize understanding the derivation of the formula and the impact of length contraction on the distances involved. The conversation concludes with a reminder that the formula accounts for the distance the train moves while the waves propagate.
Breadsticks
Messages
16
Reaction score
1

Homework Statement



1-9: Assume the train is 1.0km long as measured by the observer at C' and is moving at 150km/h. What time interval between the arrival of the wave fronts at C' is measured by the observer at C in S?
media%2F46a%2F46acd9ee-28b9-49f5-81fe-33901a4069ac%2Fphppv6pAB.png


[/B]

Homework Equations

The Attempt at a Solution


The solution from the manual is Δt=t(back)-t(front)= 500/(c-(150/3.6m/s))-500/(c+(150/3.6m/s))=4.6*10^-13

How can it use the distance it travels in S'? It doesn't travel that distance in S.
 
Physics news on Phys.org
Breadsticks said:
How can it use the distance it travels in S'? It doesn't travel that distance in S.
The speed of the train is a tiny fraction of the speed of light. No need for SR here.
 
Doc Al said:
The speed of the train is a tiny fraction of the speed of light. No need for SR here.

Thanks for the reply. If the train was moving at a non-negligible fraction of the speed of light, I would then simply find where the waves meet to figure distance for each then use the above procedure with the modified distances?
 
Breadsticks said:
If the train was moving at a non-negligible fraction of the speed of light, I would then simply find where the waves meet to figure distance for each then use the above procedure with the modified distances?
I would use the exact procedure with the only difference being the modified distances (the length of the train in frame S).

Be sure you understand how that formula is derived.
 
Doc Al said:
I would use the exact procedure with the only difference being the modified distances (the length of the train in frame S).

Be sure you understand how that formula is derived.
Right, there would be length contraction but what about the distance the train has moved while the wave is moving?
 
Breadsticks said:
Right, there would be length contraction but what about the distance the train has moved while the wave is moving?
That is covered by the formula you quoted. (Understand how it is derived!)
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top