Let's suppose that we have a spring which is attached to the ceiling and pulled 12 cm down from equilibrium and released. After 2 seconds, the amplitude has decreased to 7 cm. The spring oscillates 11 times each second. Assume that the amplitude is decreasing exponentially. Now we have to find an equation for the distance, D the end of the spring is below equilibrium in terms of seconds, t.
Since the spring at t=0 is pulled 12 cm down from equilibrium, we know that it starts at a minimum value. So it is a cosine function with a negative amplitude. The frequency is 11 Hz since the spring oscillates 11 times each second, so b (the value we enter in the formula) is $2\pi*11 = 22\pi$ (because the period is $1/11$). The amplitude represents an exponential function, which after 2 seconds decreases to 7 cm. This means that $7 = 12 * g^2 => g = \sqrt\frac{7}{12}$. So the function I end up with is $D(t) = -12 * (\sqrt\frac{7}{12})^t*cos(22\pi t)$ For some reason my answer has been marked as wrong. Can someone please point out my mistake?
The sign of $\operatorname{D}(t)$ is wrong.
It is customary to write: $$\operatorname{D}(t) = \operatorname{A}(t)\cos(2\pi\nu t)$$ $$\operatorname{A}(t) = A_0e^\frac{-t}{T}$$
You are given $\nu=11$, $A_0=12$ and from $\operatorname{A}(2)=7$ you derive $T=\frac{2}{\ln(12) - ln(7)} \approx 3.710599$.
Equivalently, $\operatorname{D}(t)=12\left(\sqrt{7/12}\right)^t\cos(22\pi t)$.