91Ó°ÊÓ

To find the velocity function, we need to take the derivative of the position function with respect to time. Given the position function: \(x(t) = A \cos \omega t\) Now, differentiate \(x(t)\) with respect to time t: \(v_x(t) = \frac{dx(t)}{dt} = \frac{d(A \cos \omega t)}{dt}\) Now, apply the chain rule to differentiate \(A \cos \omega t\): \(v_x(t) = -A \omega \sin \omega t\) Thus, the velocity function is: \(v_x(t) = -\omega A \sin \omega t\)

To verify energy conservation, we need to show that the sum of kinetic energy and potential energy remains constant: Total energy, \(E_{total} = K + U\) where: \(K = \frac{1}{2}mv^2\) (kinetic energy) \(U = \frac{1}{2}kx^2\) (potential energy, considering Hooke's Law for a spring system, where k is the spring constant) Now, let's express the kinetic energy and potential energy in terms of the given expressions for position and velocity: \(K = \frac{1}{2}m(-\omega A \sin \omega t)^2 = \frac{1}{2}m\omega^2A^2\sin^2\omega t\) \(U = \frac{1}{2}k(A\cos\omega t)^2 = \frac{1}{2}kA^2\cos^2\omega t\) Since we are given the position as a function of a simple harmonic oscillator, we know that \(\omega^2 = \frac{k}{m}\). Then, by replacing \(k\) with \(m\omega^2\) in the potential energy formula: \(U = \frac{1}{2}m\omega^2A^2\cos^2\omega t\) Now, let's sum the kinetic energy and potential energy: \(E_{total} = K + U = \frac{1}{2}m\omega^2A^2\sin^2\omega t + \frac{1}{2}m\omega^2A^2\cos^2\omega t\) Factor out \(\frac{1}{2}m\omega^2A^2\): \(E_{total} = \frac{1}{2}m\omega^2A^2(\sin^2\omega t + \cos^2\omega t)\) Using the trigonometric identity \(\sin^2\omega t + \cos^2\omega t = 1\), we get: \(E_{total} = \frac{1}{2}m\omega^2A^2(1)\) Thus, the total energy remains constant, and energy conservation is verified.