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Mechanical Oscillations

Problem 4.3

A particle performs harmonic oscillations along the \(x\) axis about the equilibrium position \(x=0\). The oscillation frequency is \(\omega=4.00 \mathrm{~s}^{-1} .\) At a certain moment of time the particle has a coordinate \(x_{0}=25.0 \mathrm{~cm}\) and its velocity is equal to \(v_{x 0}=100 \mathrm{~cm} / \mathrm{s}\) Find the coordinate \(x\) and the velocity \(v_{x}\) of the particle \(t=2.40 \mathrm{~s}\) after that moment.

Reveal Answer
 4.3. x=acos⁡(ωt+α)=−29 cm,vx=−81 cm/s, where a=x02+(vx0/ω)2,α=arctan⁡(−vx0/ωx0)\begin{aligned} &\text { 4.3. } x=a \cos (\omega t+\alpha)=-29 \mathrm{~cm}, \quad v_{x}=-81 \mathrm{~cm} / \mathrm{s}, \text { where }\\ &a=\sqrt{x_{0}^{2}+\left(v_{x 0} / \omega\right)^{2}}, \alpha=\arctan \left(-v_{x 0} / \omega x_{0}\right) \end{aligned}
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