A dense 1.00-kg mass is attached to a 1.35-m long massless rod that is free to swing from the top.The entire system is placed in a fluid with a small drag force b = 0.120 Ns/m = 0.120 kg/s and the rod is released from an initial displacement angle of 6.00 $\circ$ from vertical.What is the time it takes for the oscillation amplitude to reduce to 3.00 $\circ$ ? You may assume that the damping is small.Also note that since the amplitude of oscillation is small and all the mass of the pendulum is at the end of the rod,the motion of the mass can be treated as strictly linear and we can use the substitution: R$\theta$ (t)= x(t),where R = 1.35 m is the length of the pendulum rod.
A)2.51 s.
B)11.6 s.
C)4.16 s.
D)15.3 s.
E)12.2 s.

Question

Quizplus · Accepted Answer

The time it takes for the oscillation amplitude to reduce to a certain value in a damped harmonic oscillator can be calculated using the formula for the decay of amplitude: $ A(t) = A_0 e^{-\frac{b}{2m}t} $. Solving for $ t $ when $ A(t) = 3^\circ $ and $ A_0 = 6^\circ $, with $ b = 0.120 \, \text{kg/s} $ and $ m = 1.00 \, \text{kg} $, gives approximately 11.6 seconds.

A Dense 100-Kg Mass Is Attached to a 1 ∘\circ∘ from Vertical

A Dense 100-Kg Mass Is Attached to a 1 $\circ$ from Vertical