24P3Q03

3)

$\displaystyle \frac{2\pi }{T}=15.7\text{ }\Rightarrow \text{ }T=0.400\text{ s}$

$\displaystyle {{v}_{\max }}=\omega {{x}_{0}}=(15.7)(0.160)=2.51\text{ m }{{\text{s}}^{-1}}$

$\displaystyle {{a}_{\max }}={{\omega }^{2}}{{x}_{0}}={{(15.7)}^{2}}(0.160)=39.4\text{ m }{{\text{s}}^{-2}}$

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