Question

(a) Given that the position of an object is \(x(t)=A\) cos \(\omega t\) show that \(v_{x}(t)=-\omega A\) sin \(\omega t .\) [Hint: Draw the velocity vector for point \(P\) in Fig. \(10.17 \mathrm{b}\) and then find its \(x\) component.] (b) Verify that the expressions for \(x(t)\) and \(v_{x}(t)\) are consistent with energy conservation. [Hint: Use the trigonometric identity $\sin ^{2} \omega t+\cos ^{2} \omega t=1.1$.]

Short answer

Question: Show that the given position function \(x(t) = A \cos \omega t\) has a velocity function \(v_x(t) = -\omega A \sin\omega t\), and verify that these expressions for position and velocity conserve energy in a spring system. Solution: a) Through differentiation, we find that the position function \(x(t) = A \cos \omega t\) has a velocity function \(v_x(t) = -\omega A \sin\omega t\). b) We verified energy conservation by showing that the total energy \(E_{total} = K + U\), which consists of kinetic energy \(K=\frac{1}{2}m\omega^2A^2\sin^2\omega t\) and potential energy \(U=\frac{1}{2}m\omega^2A^2\cos^2\omega t\), remains constant and is equal to \(\frac{1}{2}m\omega^2A^2\).

Step-by-step solution

a) Derive the velocity function

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: \(v_x(t) = -\omega A \sin \omega t\)

b) Verify energy conservation

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.