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if you have the function of f(x)=4cos(kt)+15 and the period if 1/60, what would k equal?
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ignoring the constant (+15) which simply raises the graph 15 units, the wave is in the form:\[f(x) = Acos( \omega t)\] w [omega] is the angular frequency. T is period.\[T = \frac{ 2 \pi }{ \omega }\] \[k = \omega = \frac{ 2\pi }{ T } = \frac{ 2\pi }{ \frac{ 1 }{ 60 } } = 120\pi\]
Thank you! @Euler271
glad i could help :)
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