Question

A weight is attached to to a spring that is fixed to the floor. The equation h=7cos (pi/3 t) models the height, h, in centimeters after t seconds of the weight being stretched and released.

Solve for the solution of t and find the times at which the weight is first at a height of 1 cm, of 3 cm, and of 5 cm above the rest position. Round to the nearest hundreth

Answer 1

lets solve for t in the height equation:

h = 7 cos (pi/3 t)
cos (pi/3 t) = h/7
pi/3 t = cos^-1 (h/7)

t = (3/pi) cos^-1 (h/7)

then we substitute:

t = (3/pi) cos^-1 (1/7)
t = 78.1 s

t = (3/pi) cos^-1 (3/7)
t= 61.71

t = (3/pi) cos^-1 (5/7)
t = 42.41

those are the times in seconds respectively.

Answer 2

`t=(3)/(\pi)\left(\cos ^(-1)(h)/(7)\right)`

Explanation:

The given function is,

`h=7\cos \left((\pi)/(3)t\right)`

`\Rightarrow \cos \left((\pi)/(3)t\right)=(h)/(7)`

`\Rightarrow (\pi)/(3)t=\cos ^(-1)(h)/(7)`

`\Rightarrow t=(3)/(\pi)\left(\cos ^(-1)(h)/(7)\right)`

This is the solution of t.

Now we have to find t when h is 1 cm, 3 cm and 5 cm.

When h=1

`t=(3)/(\pi)\left(\cos ^(-1)(1)/(7)\right)=78.10\ s`

When h=3

`t=(3)/(\pi)\left(\cos ^(-1)(3)/(7)\right)=61.71\ s`

When h=5

`t=(3)/(\pi)\left(\cos ^(-1)(5)/(7)\right)=42.41\ s`