91Ó°ÊÓ

To find the velocity as a function of time, we need to differentiate the given position function x(t) with respect to time t. Given position function: \(x(t) = x_{m}\cos(\omega t)\) Differentiating x(t) with respect to t: \(v_{x}(t) = \frac{dx(t)}{dt} = -\omega x_{m} \sin(\omega t)\) So, the velocity function is: \(v_{x}(t) = -\omega x_{m} \sin(\omega t)\)

Here, we are given \(\Phi(t)=\Phi_{0} \sin \omega t\), and we need to find the slope of its graph at t=0. To do this, we first find the first derivative of the function with respect to t and then evaluate it at t=0. Differentiating Φ(t) with respect to t: \(\frac{d\Phi(t)}{dt} = \omega \Phi_{0} \cos(\omega t)\) Now, substitute t=0 in the above expression to find the slope of the graph at t=0: \(\frac{d\Phi(t)}{dt} \Big|_{t=0}=\omega \Phi_{0} \cos(\omega \cdot 0) = \omega \Phi_{0} \cos(0)\) Since cos(0)=1, we get: \(\frac{d\Phi(t)}{dt} \Big|_{t=0}=\omega \Phi_{0}\) Now, let's check if our result agrees with the given value of ΔΦ/Δt=ωΦ₀cos(ωt) at t=0: \(\Delta \Phi / \Delta t=\omega \Phi_{0} \cos(\omega \cdot 0)=\omega \Phi_{0} \cos(0)=\omega \Phi_{0}\) As both the calculated slope and given value for ΔΦ/Δt at t=0 are equal, we can conclude that our result agrees with the provided expression. Therefore, the slope of the graph of Φ(t) at t=0 is ωΦ₀, and it agrees with the provided value of ΔΦ/Δt at t=0.