00:00Hi friends, there is no limit to learning.
00:05The more diligent you are, the more knowledge you will have.
00:11We have a displacement time graph.
00:15Now, we are asked to find the initial phase angle.
00:22As a refresher, consider the following spring ball system.
00:27Both systems have the same amplitude, the same period, and the same frequency.
00:34Why are the motions of the two systems different?
00:39This is the role of the initial phase angle.
00:42That's because the ball was initially displaced from a different point.
00:47So the ball's displacement at the initial time determines the value of the initial phase angle.
00:55Look at the displacement time graph in our problem.
01:00The initial displacement is square root 3 cm, y at 0 seconds is equal to 3 cm.
01:10In addition, we see the amplitude value, a is equal to 2 cm.
01:17Now, write the general equation of simple harmonic motion, y is equal to a sine of omega t plus phi.
01:24If you still remember, the initial deviation angle is measured at the beginning of time.
01:29The value of t is 0.
01:32Y 0 is equal to a sine of phi.
01:38Just substitute the values of a and y 0.
01:43The sine of phi is equal to half the square root of 3.
01:48I think we already know that half the square root of 3 is the value of the sine of 60 degrees.
01:55So phi is equal to 60 degrees.
02:01Is this value really correct?
02:06If we extend the curve to the left, the displacement time graph would be a sine curve.
02:14If so, the initial displacement point is actually in quadrant 2.
02:20The angle phi should be between 90 degrees and 180 degrees.
02:27Besides 60 degrees, 120 degrees also yields a value of half the square root of 3.
02:35This is the answer to this question, phi equals 120 degrees.
02:41In radians, that value is 2 thirds pi radians.
02:47Happy learning, everyone!
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