A pendulum (mass \(m\), unknown length) moves according to \(x=A\) sin $\omega t .\( (a) Write the equation for \)v_{x}(t)$ and sketch one cycle of the \(v_{x}(t)\) graph. (b) What is the maximum kinetic energy?

Short Answer

Expert verified
Answer: The maximum kinetic energy of the pendulum is \(\frac{1}{2}m(A^2\omega^2)\).

Step by step solution

01

Differentiate position function with respect to time

To find the velocity function \(v_x(t)\), we need to differentiate the given position function \(x(t)\) with respect to time \(t\). Given \(x = A\sin(\omega t)\), we have: \(v_x(t) = \frac{dx}{dt} = \frac{d}{dt}(A\sin(\omega t))\) Using the chain rule, we get: \(v_x(t) = A\omega\cos(\omega t)\)
02

Sketch the graph of \(v_x(t)\) for one cycle

To sketch a graph of \(v_x(t) = A\omega\cos(\omega t)\) for one cycle, we first need to determine the period within which the function completes one full cycle. Since \(\cos(\omega t)\) completes one cycle in a period of \(T = \frac{2\pi}{\omega}\), the graph of \(v_x(t)\) will also complete one cycle in the same period. The graph of the velocity function will be a cosine function with its amplitude equal to \(A\omega\). It starts at a maximum value of \(A\omega\), decreases to \(0\) at \(t = \frac{T}{4}\), goes to the minimum value of \(-A\omega\) at \(t = \frac{T}{2}\), increases to \(0\) at \(t = \frac{3T}{4}\), and finally goes back to the maximum value of \(A\omega\) at \(t=T\).
03

Calculate the maximum kinetic energy

The maximum kinetic energy of the pendulum occurs when the velocity is at its maximum value, which is \(A\omega\). Therefore, the maximum velocity is given by \(v_{max} = A\omega\). Now, we can calculate the max kinetic energy using the formula \(K = \frac{1}{2}mv^2\): \(K_{max} = \frac{1}{2}m(A\omega)^2\) \(K_{max} = \frac{1}{2}m(A^2\omega^2)\) Therefore, the maximum kinetic energy of the pendulum is \(\frac{1}{2}m(A^2\omega^2)\).

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