Final answer:
The phase difference from t=2 sec to t=4 sec for an oscillator with an 8-sec period is π/2 radians because each second corresponds to π/4 radians, and the question covers a span of 2 seconds.
The correct answer is option B). π/2.
Step-by-step explanation:
The phase difference for an oscillator can be calculated knowing that the time period of a complete oscillation corresponds to a phase change of 2π radians.
In this question, the time period T is given as 8 seconds, which means a complete cycle corresponds to this duration. Therefore, every second corresponds to a phase change of π/4 radians since 2π radians is spread over 8 seconds.
For a duration of 2 seconds (from t=2 sec to t=4 sec), the phase change would be twice that of one second:
- Phase change per second = 2π / 8 = π/4 radians
- Phase change in 2 seconds = 2 * (π/4) = π/2 radians
Therefore, the phase difference from t=2 sec to t=4 sec will be π/2 radians.