Trig Functions - When wil object be 9cm below 0?

[TN17]

Trig Functions - "When wil object be 9cm below 0?"

Homework Statement

Here is the background information:

A weight hanging from a spring is set in motion by an upward push. It takes 10 s for the weight to complete one cycle from moving 12 cm above 0, then dropping 12 cm below 0, then returning to 0.

Here is the question:

d) In the first 10 s, when will the height of the weight be 9 cm below 0?

Homework Equations

I found the equation to be y=12sin (Pi/5)(x)
Since k=2Pi/10 = Pi/5 and the vertical stretch is 12.

The Attempt at a Solution

I set y=-9 because the weight is below 0, and solved for x, but I didn't know how to continue from there.
-9 = 12sin(Pi/5)(x)

There are 2 answers, 6.3 s and 8.7 s

 

[HallsofIvy]

Homework Statement

Here is the background information:

A weight hanging from a spring is set in motion by an upward push. It takes 10 s for the weight to complete one cycle from moving 12 cm above 0, then dropping 12 cm below 0, then returning to 0.

Here is the question:

d) In the first 10 s, when will the height of the weight be 9 cm below 0?

Homework Equations

I found the equation to be y=12sin (Pi/5)(x)
Since k=2Pi/10 = Pi/5 and the vertical stretch is 12.

The Attempt at a Solution

I set y=-9 because the weight is below 0, and solved for x, but I didn't know how to continue from there.
-9 = 12sin(Pi/5)(x)

Well, solve the equation for x by the usual way- "unpeel" what has been done to x.
First, divide both sides by 12:
[tex]\frac{-9}{12}= -\frac{3}{4}= sin(\pi x/5)[/tex]
Now, use the inverse function, arcsine, to get rid of the sine function:
[tex]\frac{\pi x}{5}= arcsin(-3/4)= -0.8481[/tex]
is the "principal solution" given by a calculator. Of course, we want x to be positive so we use the fact that [itex]sin(\pi- \theta)= sin(\theta)[/itex] and, of course, [itex]sin(2\pi+ \theta)= sin(\theta)[/itex]. [itex]\pi- (-.8481)= 3.1416+ .8481= 3.9896[/itex] and [itex]2\pi+ (-.8481)= 6.2832- .8481= 5.4351[/itex]
From
[tex]\frac{\pi}{5}x= 3.9896[/tex]
and
[tex]\frac{\pi}{5}x= 5.4351[/tex]
we get
[tex]x= (3.9896)\left(\frac{5}{\pi}\right)= 6.3496[/tex]
[tex]x= (5.4251)\left(\frac{5}{\pi}\right)= 8.6502[/tex]
which round to the values you give.

There are 2 answers, 6.3 s and 8.7 s