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Question
What is the smallest positive phase constant which is equivalent to 7⋅5 π?
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Solution
Equation of the wave: y = A sin(kx − ωt + Φ)
Here, A is the amplitude, k is the wave number, ω is the angular frequency and Φ is the initial phase.
The argument of the sine is a phase, so the smallest positive phase constant should be
\[\sin\left( 7 . 5\pi \right) = \sin\left( 3 \times 2\pi + 1 . 5\pi \right)\]
\[ = \sin\left( 1 . 5\pi \right)\]
Therefore, the smallest positive phase constant is 1.5π.
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