Answer:
1) an observer in B 'sees the two simultaneous events
2)observer B sees that the events are not simultaneous
3) Δt = Δt₀ /√ (1 + v²/c²)
Explanation:
This is an exercise in simultaneity in special relativity. Let us remember that the speed of light is the same in all inertial systems
1) The events are at rest in the reference system S ', so as they advance at the speed of light which is constant, so it takes them the same time to arrive at the observation point B' which is at the point middle of the two events
Consequently an observer in B 'sees the two simultaneous events
2) For an observer B in system S that is fixed on the Earth, see that the event in A and B occur at the same instant, but the event in A must travel a smaller distance and the event in B must travel a greater distance since the system S 'moves with velocity + v. Therefore, since the velocity is constant, the event that travels the shortest distance is seen first.
Consequently observer B sees that the events are not simultaneous
3) let's calculate the times for each event
Δt = Δt₀ /√ (1 + v²/c²)
where t₀ is the time in the system S' which is at rest for the events
Answer:
b) 252 Hz or 260 Hz
c) 0.25 s
Explanation:
b) The frequency of the beats is 4 Hz, and one tuning fork has a frequency of 256 Hz. Therefore, the second tuning fork is either 4 Hz lower or 4 Hz higher.
f = 252 Hz or 260 Hz
c) Period is the inverse of frequency.
T = 1/f
T = 1 / (4 Hz)
T = 0.25 s
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