A 4.24 m long,1.27 kg rope under a tension of 475 N oscillates with a frequency of 11.2 Hz.If the oscillation amplitude is 6.32 cm,how much energy is required to keep the rope oscillating for 5.52 s?

Question

Quizplus · Accepted Answer

The energy required to keep the rope oscillating is equal to the energy dissipated due to damping. The damping force is proportional to the velocity of the rope, so the power dissipated is proportional to the square of the velocity. The average power dissipated over one cycle is given by:$$P_{avg}=\frac{1}{2}bv^2$$where b is the damping constant and v is the velocity of the rope. The damping constant can be determined from the frequency and amplitude of the oscillation:$$b=\frac{2\pi f}{Q}$$where f is the frequency and Q is the quality factor. The quality factor is a measure of the damping of the system, and it is given by:$$Q=\frac{2\pi f_0}{\Delta f}$$where f_0 is the resonant frequency and Δf is the bandwidth of the oscillation. The resonant frequency is the frequency at which the system oscillates with the maximum amplitude, and the bandwidth is the difference between the upper and lower frequencies at which the amplitude is half of the maximum amplitude.In this problem, we are given the frequency, amplitude, and tension of the rope, but we are not given the damping constant or the quality factor. However, we can use the fact that the rope is oscillating with a constant amplitude to conclude that the damping is small. This means that the quality factor is large, and we can approximate it as:$$Q\approx\frac{2\pi f_0}{\Delta f}\approx\frac{2\pi f}{0}=∞$$With this approximation, the damping constant becomes:$$b=\frac{2\pi f}{Q}\approx0$$This means that the power dissipated is negligible, and the energy required to keep the rope oscillating is zero.Therefore, the correct answer is A.

A 424 M Long,1