need helllllllllllllllllp (1 Viewer)

jimmysmith560 · Jun 9, 2022

Would the following working help?

$x=A\sin \left(nt+K\right)+B$

$x'=An\cos \left(nt+K\right)$

$x''=-An^2\times \sin \left(nt+K\right)$

Maximum speed occurs at $x''=0$ , i.e. $t=\frac{-K}{\:n}$

$x'=An\times \cos \left(0\right)=An$

Thus, max speed is $An$

Time for 1 oscillation is $T=\frac{2\pi }{n}$

$v=\frac{s}{t}\rightarrow v=\frac{4A}{\left(\frac{2\pi }{n}\right)}\rightarrow v=\frac{2An}{\pi }$

At max speed $An\::\:\frac{2An}{\pi }$

$1\::\:\frac{2}{\pi }\rightarrow 2\::\:\pi$

need helllllllllllllllllp (1 Viewer)

Stressedaf0666

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jimmysmith560

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