Answer to Question #258321 in Calculus for Bora

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 secondit 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

A mass suspended above its resting position at a distance 5cm

So, amplitude (A)$=|a|=5$

At $t=\frac{1}{3}$ seconds, period$=\frac{1}{3}$

$\\\frac{2\pi}{3}=\frac{1}{3}\implies \omega=6\pi$

The function to model the motion to resting position initially is,

$f(t)=y=Acos(\omega t)\\\implies y=5cos(6\pi t)$

Now state of change of position at t=2.1 seconds.

$y=5cos(6\pi t)\\y'=\frac{dy}{dt}=5[-sin(6\pi t)](6\pi)]\\t=2.1\space seconds$

$y'=-30\pi[sin(6\pi[2.1])]\\y'=-30\pi[0.951]\\y'=-89.63\space cm/sec$