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?

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 = 5.0 cm

The initial final amplitude = 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:

In the absence of damping, the angular frequency is:

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

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

the answer is kevlar