An ideal massless spring with a spring constant of 2.00 N/m is attached to an object of 75.0 g. The system has a small amount of damping. If the amplitude of the oscillations decreases from 10.0 mm to 5.00 mm in 15.0 s, what is the magnitude of the damping constant b?
A) 0.00693 kg/s
B) 0.0462 kg/s
C) 0.00762 kg/s
D) 0.0100 kg/s
E) 0.00857 kg/s

Question

Quizplus · Accepted Answer

The damping constant b can be calculated using the formula:$$b = \frac{m}{Q}\frac{\Delta A}{\Delta t}$$where m is the mass of the object, Q is the quality factor, ΔA is the change in amplitude, and Δt is the time interval.The quality factor Q is given by:$$Q = \frac{2\pi f_0}{\Delta f}$$where f_0 is the resonant frequency and Δf is the bandwidth.The resonant frequency is given by:$$f_0 = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$$where k is the spring constant.The bandwidth is given by:$$\Delta f = \frac{b}{2\pi m}$$Substituting these equations into the formula for b, we get:$$b = \frac{m}{Q}\frac{\Delta A}{\Delta t} = \frac{m}{2\pi f_0}\frac{\Delta A}{\Delta t}\frac{b}{2\pi m} = \frac{\Delta A}{2\pi \Delta t}\frac{m}{f_0}$$Substituting the given values, we get:$$b = \frac{5.00 	imes 10^{-3} 	ext{ m}}{2\pi 	imes 15.0 	ext{ s}}\frac{0.0750 	ext{ kg}}{1.26 	ext{ Hz}} = 0.00693 	ext{ kg/s}$$Therefore, the magnitude of the damping constant b is 0.00693 kg/s.

An Ideal Massless Spring with a Spring Constant of 2