Find the time when the spring force reaches max magnitude

[SirChris93]

Homework Statement

Show how you use calculus to find the time when the magnitude of the Spring Force reaches its maximum. Then, when you found that time show how you calculate the Spring force at that time as well.

m = 1.125kg
v_i = .8 m/s
k = 2250 N/m
x = 0m

Homework Equations

F_s (t) = [-mv_i sqrt(k/m)] sin (sqrt(k/m)t)

The Attempt at a Solution

So the homework provided the Time and we had to find the Spring Force and put them in a table, which I did. After viewing the graph I figured, that the magnitude reaches its maximum between .03 and .04, but I can't figure out how to find that.

I didn't know how to make a table on here, but I tried the best I could

time (t) | Spring Force (F)
 0            0
.01     |       -17.41
.02     |      -31.39
.03     |     -39.2
.04     |    -39.3
.05     |    -31.67
.06     |   -17.81
.07     |   -.45

 

[RedDelicious]

The problem statement tells you to "use calculus."

How do you usually go about finding the max/min of a function in calculus?

 

[SirChris93]

The problem statement tells you to "use calculus."

How do you usually go about finding the max/min of a function in calculus?

My teacher said not to find the derivative, so that’s out of the question. The only other way I can think of is t=F_s(t)/[(-mvsqrt(k/m)]sin sqrt(k/m)), but it won’t work. Unless I would the integral of the equation...

 

[RedDelicious]

My teacher said not to find the derivative, so that’s out of the question. The only other way I can think of is t=F_s(t)/[(-mvsqrt(k/m)]sin sqrt(k/m)), but it won’t work. Unless I would the integral of the equation...

That's fine. It's not necessary to take the derivative anyway.

The only time depenendent part of your function is the sine. And so your function will have its max value when the sine function has its max value.

Is sin(x) ever going to be greater than 1? For what x does it have a max value? Think of the unit circle.