Question

A vertical spring with a spring constant of 2.00 N/m has a 0.30-kg mass attached to it, and the mass moves in a medium with a damping constant of 0.025 kg/s. The mass is released from rest at a position 5.0 cm from the equilibrium position. How long will it take for the amplitude to decrease to 2.5 m?

Answer 1

Answer:

17 seconds

Explanation:

Given that:

The mass attached to the spring (m) = 0.30 kg

The spring constant (k) =  2.00 N/m

The damping constant (b) = 0.025 kg/s

The initial distance $x_o$ = 5.0 cm

The initial final amplitude $A_f$ = 2.5 cm and not 2.5 m, please note the mistake, if it is 2.5 m, our time taken will be -93.7 sec, and we do not want a negative time value.

To start with the angular frequency damping using the formula:

$\omega_{\gamma}= \dfrac{b}{2m}$

$\omega_{\gamma}= \dfrac{0.025 \ kg/s}{2(0.3 \ kg)}$

$\omega_{\gamma}=4.167 \times 10^{-2} \ s^{-1}$

In the absence of damping, the angular frequency is:

$\omega_o = \sqrt{\dfrac{k}{m}}$

$\omega_o = \sqrt{\dfrac{2 \ N/m}{0.3 kg}} \\\\\omega_o = 2.581 \ s^{-1}$

The initial amplitude oscillation can be computed by using the formula:

$A_i = e^{-\omega_{\gamma}t} x_o \sqrt{\dfrac{\omega_o^2}{\omega_o^2-\omega_f^2}}$

$A_i = e^{-\omega_{\gamma}0} (5.0 \ cm) \sqrt{\dfrac{2.581^2}{2.581^2-(4.167*10^{-2})^2}}$

$A_i = 5.0006 \ cm \\ \\ A_i = 5.001 \ cm$

The final amplitude, as well as the initial amplitude, can be illustrated by using the relation:

$A_f = e^{-\omega_{\gamma}t}A_i\\ \\ e^{-\omega_{\gamma}t} = \dfrac{2. 5 \ cm}{5.001 cm}\\ \\ = 0.4999\\ \\ \implies -\omega_{\gamma}t_f = \mathsf{In (0.4999)} \\ \\ t_f = \dfrac{\mathsf{-In (0.4999)}}{4.167*10^{-2} \ s^{-1}}$

$t_f = 16.64 \ sec \\ \\ \mathbf{t_f \simeq 17 sec}$