The correct answer is: k feet (Option D)
Explanation:
Let me solve it without using any complex jargon (and formulas).
Given equation is as follows:
[tex]h = a*sin(b(t-h)) + k[/tex] ---- (A)
Here in this case, a represents the distance of the ball from the equilibrium position WHEN the spring is stretched. h inside the sine function represents the horizontal phase shift. Please do not confuse it with the height, and k represents the vertical shift.
Before oscillations, the amplitude is equal to zero. Assume that the ball attached with the spring was pushed down to position a and let go (as shown in the figure below). Now the ball would oscillate between the +a and -a. Why? Because the maximum value of sine function—sin(b(t-h)) (given in the above equation)—is +1, and the minimum value will be -1. When sin(b(t-h)) will be multiplied by a, the maximum value of a*sin(b(t-h)) will be +a, and the minimum will be -a. However, there is a vertical shift, k, involved in the equation as well. Now the maximum value of (a*sin(b(t-h)) + k) will be (+a+k), since the maximum value of a*sin(b(t-h)) is +a. Likewise, the minimum value of (a*sin(b(t-h)) + k) will be (-a+k), since the minimum value of a*sin(b(t-h)).
The equilibrium position lies between (+a+k) and (-a+k). To find the equilibrium position, add both of them and divide it by 2.
At equilibrium position, height, will be:
[tex]\frac{(a+k) + (-a+k)}{2} = \frac{0+2k}{2} = k\thinspace feet[/tex]
Hence, the height of the ball at its equilibrium is k feet (Option D).
