A vertical spring with a spring constant of 4.00 N/m,has a 1.00-kg mass attached to it,and moves in a medium with a damping constant 0.025 kg/s.The mass is released from rest at a position of 3.0 cm from equilibrium position.How long will it take for the amplitude to decrease to 1.0 cm?
A)60.5 s
B)87.9 s
C)101 s
D)117 s
E)129 s

Question

Quizplus · Accepted Answer

The amplitude of the spring-mass system is given by $A = A_0e^{-\frac{b}{2m}t}$, where $A_0$ is the initial amplitude, $b$ is the damping constant, $m$ is the mass, and $t$ is time. We are given $A_0 = 3$ cm and $A = 1$ cm, so we need to find the time $t$.

The angular frequency of the system is $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{4}{1}} = 2$ rad/s. The damped angular frequency is $\omega' = \sqrt{\frac{k}{m} - \left(\frac{b}{2m}ight)^2} = \sqrt{\frac{4}{1} - \left(\frac{0.025}{2	imes1}ight)^2} \approx 1.96$ rad/s.

The general solution for the position of the mass as a function of time is $x(t) = Ae^{-\frac{b}{2m}t}\cos(\omega't + \phi)$, where $A$ and $\phi$ are constants determined by the initial conditions. The initial velocity of the mass is zero, so $x'(0) = -A\omega'\sin(\phi) = 0$. Therefore, we must have $\phi = 0$ or $\phi = \pi$.

Using the initial condition $x(0) = A_0 = 3$ cm, we have $A = 3$. Using the condition $x(t_1) = 1$ cm, we have:

$$1 = 3e^{-\frac{0.025}{2	imes1}t_1}\cos(1.96t_1)$$

We can solve for $t_1$ numerically or graphically to get $t_1 \approx 87.9$ s. Therefore, the answer is B.

A Vertical Spring with a Spring Constant of 4