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# A mass is placed on a frictionless, horizontal table. A spring (k = 100 N/m), which can be stretched or compressed, is placed on the table. A 5.00-kg mass is attached to one end of the spring, the other end is anchored to the wall. The equilibrium position is marked at zero. A student moves the mass out to x = 4.0 cm and releases it from rest. The mass oscillates in SHM. (a) Determine the equations of motion. (b) Find the position, velocity, and acceleration of the mass at time t = 3.00 s.

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First, since there are no non-conservative forces acting on the system, the amplitude $A$ of the movement will be equal to the initial displacement $\left(4.0 \text{ cm}\right)$. We can also calculate the angular frequency $\omega$ with the given force constant $k$ and the mass $m$. Once we have calculated $\omega$, we continue to write the equation for the general position $x\left(t\right)$ of the object in simple harmonic motion. Then, we differentiate $x\left(t\right)$ in respect to time to obtain $v\left(t\right)$. Finally, we differentiate $v\left(t\right)$ in respect to time to obtain $a\left(t\right)$. Note that $\phi=0$ because the position is $4.0 \text{ cm}$ when $t=0$.

\begin{align*} A&=4.0 \cdot 10^{-2} \text{ m} \\ \omega&=\sqrt{\dfrac{k}{m}} \\ &=\sqrt{\dfrac{100}{5.00}} \\ &=4.47 \text{ } \dfrac{\text{rad}}{\text{s}} \\ x\left(t\right)&=A \cos\left(\omega t\right) \\ &=\left(4.0 \cdot 10^{-2} \text{ m}\right) \cos\left[\left(4.47 \text{ } \dfrac{\text{rad}}{\text{s}}\right)t\right] \\ v\left(t\right)&=\dfrac{dx}{dt} \\ &=-A \omega \sin\left(\omega t\right) \\ &=-\left(4.0 \cdot 10^{-2} \text{ m}\right) \left(4.47 \text{ } \dfrac{\text{rad}}{\text{s}}\right)\sin\left[\left(4.47 \text{ } \dfrac{\text{rad}}{\text{s}}\right)t\right] \\ &=-\left(1.79 \cdot 10^{-1} \dfrac{\text{m}}{\text{s}}\right)\sin\left[\left(4.47 \text{ } \dfrac{\text{rad}}{\text{s}}\right)t\right] \\ a\left(t\right)&=\dfrac{dv}{dt} \\ &=-A \omega^2 \cos\left(\omega t\right) \\ &=-\left(4.0 \cdot 10^{-2} \text{ m}\right) \left(4.47 \text{ } \dfrac{\text{rad}}{\text{s}}\right)^2\cos\left[\left(4.47 \text{ } \dfrac{\text{rad}}{\text{s}}\right)t\right] \\ &=-\left(7.99 \cdot 10^{-1} \text{ } \dfrac{\text{m}}{\text{s}^2}\right)\cos\left[\left(4.47 \text{ } \dfrac{\text{rad}}{\text{s}}\right)t\right] \end{align*}

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