Question

A mass suspended from a spring is raised a distance of 5 cm above its resting position. The mass is released at time t=0 and allowed to oscillate. After one third of a ​second, it is observed that the mass returns to its highest position, which was 4.5 cm above its resting position. What is the rate of change of the position of the mass at t = 2.1 seconds?

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Rates of change are the change of a quantity over another.

The rate of change of the mass position at 2.1 seconds is -89.6 cm/ s

The given parameters are:

$\mathbf{A = |5|}$ -- the amplitude

$\mathbf{T = \frac 13}$ --- the period

The position of the mass is modeled by:

$\mathbf{y= Acos(wt)}$

Where:

$\mathbf{w = \frac{2\pi }{T}}$

So, we have:

$\mathbf{w = \frac{2\pi}{1/3}}$

$\mathbf{w = 6\pi }$

$\mathbf{y= Acos(wt)}$ becomes

$\mathbf{y = |5|cos(6\pi t)}$

$\mathbf{y = 5cos(6\pi t)}$

Differentiate

$\mathbf{y' = 5 \times -sin(6\pi t) \times (6\pi)}$

$\mathbf{y' = -30\pi sin(6\pi t) }$

When t = 2.1, we have:

$\mathbf{y' = -30\pi sin(6\pi \times 2.1) }$

$\mathbf{y' = -30\pi sin(12.6\pi) }$

$\mathbf{y' = -30\pi \times 0.9510}$

$\mathbf{y' = -89.6296}$

Approximate

$\mathbf{y' = -89.6}$

Hence, the rate of change of the mass position at 2.1 seconds is -89.6 cm/ s

Read more about rates at:

brainly.com/question/15579510