Question

What is the period, in seconds, of the simple harmonic motion described by the equation x = 5sin(8πt)?

Answer 1

`\bf \qquad \qquad \qquad \qquad \textit{function transformations}`








`\bf \begin{array}{llll} \bullet \textit{vertical shift by }{{ D}}\\ \qquad if\ {{ D}}\textit{ is negative, downwards}\\ \qquad if\ {{ D}}\textit{ is positive, upwards}\\\\ \bullet \textit{function period}\\ \qquad \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\ \qquad \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta) \end{array}`

now, let's see your equation
`\bf \begin{array}{llccclll} x=&5sin(&8\pi t&+0)&+0\\ &\uparrow &\uparrow &\uparrow &\uparrow \\ &A&B&C&D \end{array}\qquad period\implies \cfrac{2\pi }{B}\iff \cfrac{2\pi }{8\pi }`

Answer 2

t=0.007958x

Hope that helps!