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Worked Example: SHM Solution Formatting
Question: Which of the following functions of time represent (a) simple harmonic motion and (b) periodic motion? Give the period for each case.
Solution:
i) sin Οt - cos Οt = β2[1/β2 sin Οt - 1/β2 cos Οt] = β2 sin(Οt - Ο/4)
This function represents a simple harmonic motion having a period T = 2Ο/Ο and a phase angle (-Ο/4) or (7Ο/4).
ii) sinΒ² Οt = (1-cos(2Οt))/2 = 1/2 - 1/2 cos(2Οt)
The function is periodic having a period T = Ο/Ο. It is not SHM.
Answer:
i) SHM with T = 2Ο/Ο and phase angle (-Ο/4)
ii) Periodic with T = Ο/Ο, not SHM
i) sin Οt - cos Οt = β2[1/β2 sin Οt - 1/β2 cos Οt] = β2 sin(Οt - Ο/4)
This function represents a simple harmonic motion having a period T = 2Ο/Ο and a phase angle (-Ο/4) or (7Ο/4).
ii) sinΒ² Οt = (1-cos(2Οt))/2 = 1/2 - 1/2 cos(2Οt)
The function is periodic having a period T = Ο/Ο. It is not SHM.
Answer:
i) SHM with T = 2Ο/Ο and phase angle (-Ο/4)
ii) Periodic with T = Ο/Ο, not SHM
Problem 1 of 25
Question
Solve for x in the equation: 2x + 5 = 13