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simple harmonic motion , d=8cos((3pi)/(4)t) answer the following questions: Maximum: 8 Frequency: ????
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Let's recall. \[x(t) = A \sin(\omega t)\]where \(\omega\) is the angular velocity. Frequency can be expressed in terms of \(\omega\) as such\[f = {\omega \over 2 \pi}\]
4/2pi (Is that correct?)
I would imagine that \(\omega = 3 \pi /4\)
Ok I got it, I was looking at the wrong thing THANK YOU :)
What about this question: 0=8cos(3pi/4)(t) So far I got: 0=8(-√2/2)(t)
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What are you solving for with this equation?
It says: Determine the least positive value of t which d=0
You can just solve\[\cos({3 \pi \over 4} t ) = 0\]
We know that \(\cos(\pi/2) = 0\) Therefore, \[{\pi \over 2} = {3 \pi \over 4} t\]
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