Question
Tue January 10, 2012 By:
Original question is:Derive the time period of sin^2(omega)t.
in the solution they have written,
 sin^2(omega)t=1/2-1/2cos2(omega)t.
I

What are the different conclusions or deductions from following function of time. 1/2-1/2cos2(omega)t.

Expert Reply
Thu January 12, 2012
we know that cos 2?t=1-2sin^2?t
from there we get
 sin^2(omega)t=1/2-1/2cos2(omega)t.
here 1/2 has nothing to do with the timeperiod.1/2-1/2cos2(omega)t will help you to draw the actual graph of the oscillation from where we can have an actual idea about the position of the particle at a particular time.
here time period will be T/as 2(omega)t=2x2?xt/T
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