A mass \(m=1.00 \mathrm{~kg}\) in a spring-mass system with \(k=\) \(1.00 \mathrm{~N} / \mathrm{m}\) is observed to be moving to the right, past its equi. librium position with a speed of \(1.00 \mathrm{~m} / \mathrm{s}\) at time \(t=0 .\) a) Ignoring all damping, determine the equation of motion. b) Suppose the initial conditions are such that at time \(t=0\), the mass is at \(x=0.500 \mathrm{~m}\) and moving to the right with a speed of \(1.00 \mathrm{~m} / \mathrm{s}\). Determine the new equation of motion. Assume the same spring constant and mass.

The angular frequency, \(\omega\), is related to the spring constant and the mass. It is given by the formula: \(\omega=\sqrt{\frac{k}{m}}\) Given, \(k = 1.00~N/m\) and \(m= 1.00~kg\), we can calculate the angular frequency: \(\omega = \sqrt{\frac{1}{1}}\) \(\omega = 1~rad/s\)

Since the system experiences simple harmonic motion, we can represent the equation of motion in the following form: \(x(t) = A\cos(\omega t+\varphi)\) where \(x(t)\) is the position of the mass at time \(t\), \(A\) is the amplitude, \(\omega\) is the angular frequency (which we calculated in Step 1), and \(\varphi\) is the phase.

For case a), we are given that at time \(t = 0\), the mass is moving to the right with a speed of \(1.00~m/s\) and passing its equilibrium position (\(x = 0\)). Using these initial conditions, we can write: \(x(0) = A\cos(\omega \cdot 0 + \varphi) = 0\) and \(v(0) = -A \omega \sin(\omega \cdot 0 + \varphi) = 1.00~m/s\) From the first equation, we have: \(\cos(\varphi) = 0\) This means that the phase, \(\varphi\), is either \(\frac{\pi}{2}\) or \(-\frac{\pi}{2}\). We can use the second equation to find the amplitude and the correct value of the phase: \(A = \frac{v(0)}{\omega \sin(\varphi)}\) Using \(\varphi = \frac{\pi}{2}\), \(A = \frac{1}{1 \cdot 1} = 1~m\)

Now that we have the amplitude and phase for case a), we can write the equation of motion: \(x(t) = 1 \cdot \cos(1 \cdot t + \frac{\pi}{2})\)

For case b), we are given that at time \(t = 0\), the mass is at \(x = 0.50~m\) and moving to the right with a speed of \(1.00~m/s\). Using these initial conditions, we can write: \(x(0) = A\cos(\omega \cdot 0 + \varphi) = 0.50~m\) and \(v(0) = -A \omega \sin(\omega \cdot 0 + \varphi) = 1.00~m/s\) From the second equation, we can solve for the amplitude and phase: \(A = \frac{v(0)}{\omega \sin(\varphi)}\) Using \(x(0)\) and \(\varphi = \arccos(\frac{x(0)}{A})\), \(A = \sqrt{x(0)^2 + (\frac{v(0)}{\omega})^2}\) \(A = \sqrt{(0.50)^2 + (\frac{1}{1})^2}\) \(A = \sqrt{1.25} \approx 1.12~m\) Now, we can find the phase: \(\varphi = \arccos(\frac{x(0)}{A})\) \(\varphi = \arccos(\frac{0.50}{1.12})\) \(\varphi \approx 1.05~rad\)

Now that we have the amplitude and phase for case b), we can write the equation of motion: \(x(t) \approx 1.12\cos(1 \cdot t + 1.05)\)