The height, [tex]h[/tex], in feet of a ball suspended from a spring as a function of time, [tex]t[/tex], in seconds can be modeled by the equation [tex]h = a \sin(b(t - h)) + k[/tex]. What is the height of the ball at its equilibrium?

A. [tex]a[/tex] feet
B. [tex]b[/tex] feet
C. [tex]h[/tex] feet
D. [tex]k[/tex] feet



Answer :

To solve this problem, we need to understand what the height of the ball at its equilibrium position represents.

The equation given is:
[tex]\[ h = \operatorname{asin}(b(t - h)) + k \][/tex]

In this model, [tex]\(h\)[/tex] represents the height of the ball at any time [tex]\(t\)[/tex], the term [tex]\(\operatorname{asin}(b(t-h))\)[/tex] represents the oscillatory component of the ball's motion due to the spring, and [tex]\(k\)[/tex] represents the vertical shift from the origin, or essentially the equilibrium position of the ball.

The equilibrium position is where the ball comes to rest when it is not oscillating. This happens when the oscillatory component [tex]\(\operatorname{asin}(b(t-h))\)[/tex] is zero, as it would be at its mid-point of motion.

So, when the ball is at equilibrium, the term [tex]\(\operatorname{asin}(b(t-h))\)[/tex] becomes zero. This simplifies the equation to:
[tex]\[ h = 0 + k \][/tex]
or simply,
[tex]\[ h = k \][/tex]

Therefore, the height of the ball at its equilibrium position is [tex]\(k\)[/tex] feet.

Thus, the correct answer is:
[tex]\[ \boxed{k} \][/tex] feet.