Question

Joey is running an experiment with a mass on a spring. Joey models the height of the mass above the table he is working on using the equation h(t)=6.85cos(3.42t)+9, where the height is in inches and the time, t, is in seconds since he let the mass free. How many seconds, to the nearest tenth, will it take for the mass to return to its original position?

Answer 1

Answer: 1.8 s

Step-by-step explanation:

Given

The height of the spring mass system above the table is given by

$h(t)=6.85\cos (3.42t)+9$

The mass is performing S.H.M with frequency $\omega =3.42$

and $\omega T=2\pi$

$\therefore T=\dfrac{2\pi }{\omega}$

time when mass returns to its original position

$T=\dfrac{2\pi }{3.42}\\\\T=1.8\ s$