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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 given 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 :

GinnyAnswer
To determine the height of the ball at its equilibrium, let's carefully analyze the given equation:

[tex]\[ h = a \sin(b(t - h)) + k \][/tex]

- [tex]\( h \)[/tex]: the height of the ball at any time [tex]\( t \)[/tex].
- [tex]\( a \)[/tex]: the amplitude of the oscillations.
- [tex]\( b \)[/tex]: the angular frequency.
- [tex]\( k \)[/tex]: the equilibrium height, or the mean height around which the ball oscillates.
- [tex]\( t \)[/tex]: the time variable.

The height of the ball at equilibrium is the position where it would rest if it were not oscillating — essentially the average height. In the given equation, this is represented by the term [tex]\( k \)[/tex].

The trigonometric function [tex]\(\sin(b(t - h))\)[/tex] oscillates between -1 and 1. When there is no oscillation, the value of [tex]\(\sin\)[/tex] is zero. At this point, the equation simplifies to [tex]\( h = k \)[/tex].

Thus, the equilibrium height of the ball is [tex]\( k \)[/tex] feet.

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