MHB What is the smallest time from the data given

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The discussion revolves around determining the smallest time from the equation \(\sin(120\pi t)=\frac{2}{5}\). It is clarified that the smallest time corresponds to one cycle, and multiple solutions for \(t\) exist based on the equation derived. The expression for \(t\) is given as \(t=\frac{n}{120}+\frac{(-1)^n}{120\pi} \sin^{-1}\frac{2}{5}\), where \(n\) is a non-negative integer. Substituting \(n=0\) yields the smallest value for \(t\), though further proof is suggested to confirm this minimum. The conversation emphasizes the need for clarity in the definition of "smallest time" in the context of the problem.
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https://www.physicsforums.com/attachments/1729

before I proceed with b, c, and d. what really is meant by the smallest time. does this mean the time of one cycle. there are multiple answers if it calculated for t. the steps I used give one but that may not be the correct process. no ans given
 
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karush said:
https://www.physicsforums.com/attachments/1729

before I proceed with b, c, and d. what really is meant by the smallest time. does this mean the time of one cycle. there are multiple answers if it calculated for t. the steps I used give one but that may not be the correct process. no ans given

Hi karush, :)

You have obtained,

\[\sin(120\pi t)=\frac{2}{5}\]

Generally, the values that satisfies this equation are given by,

\[120\pi t=n\pi+(-1)^n \sin^{-1}\frac{2}{5}\]

where \(n\in \mathbb{Z}^{+}\cup\{0\}\). Therefore we have,

\[ t=\frac{n}{120}+\frac{(-1)^n}{120\pi} \sin^{-1}\frac{2}{5}\]

So by substituting \(n=0,\,1,\,2\,\cdots\) and so on we can see that at \(n=0\), \(t\) attains it's smallest value. That's not a proof though, I guess to show that \(t(n)=\frac{n}{120}+\frac{(-1)^n}{120\pi} \sin^{-1}\frac{2}{5}\) has a minimum at \(n=0\) in a more abstract manner will need more work. I can't think of any simple method. :confused:
 
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