I have a sinusoidal exponential function that represents the deflection angle of a beam as it vibrates. I am trying to calculate the maximum amplitude of the curve between $t = 0$ and $t = 60$, however Wolfram Alpha is saying the maximum value is at $t = 55.55$, which is not correct. The maximum should be near $t = 1$. I have attempted to differentiate it twice myself and set the $y''$ value to zero to find the min and max, however with the exponential component in the equation I am a bit lost. The equation is:
$$\delta(t)=0.5 \cdot 3.8345e^{-0.075t}\sin\left(\frac{2\pi t}{4}\right)$$
Is there an easier to way to find the maximum amplitudes of this function?
Edit: Wolfram provided this as an answer, which when I calculate comes to $t = 55.55$:
$$t = \frac{4}{\pi}\cot^{-1}\left(\dfrac{20\pi}{\sqrt{9+400\pi^{2}}-3}\right)$$
You shouldn't have trouble getting $$ \delta'(t) = 2.63737 \mathrm{e}^{-0.075 t} \cos(\pi t/2) - 0.125925 \mathrm{e}^{-0.075 t} \sin(\pi t/2) \text{.} $$
This may seem hopeless, but much simplification awaits. Set this equal to $0$ and move the entire sine term to the right and then divide through by it. (We should verify that this divisor's zeroes do not correspond to zeroes of $\delta'$, but you already know this since the divisor is a scaled copy of $\delta$, and the maximum is positive.) \begin{align*} 2.63737 \mathrm{e}^{-0.075 t} \cos(\pi t/2) &- 0.125925 \mathrm{e}^{-0.075 t} \sin(\pi t/2) = 0 \\ 2.63737 \mathrm{e}^{-0.075 t} \cos(\pi t/2) &= 0.125925 \mathrm{e}^{-0.075 t} \sin(\pi t/2) \\ \frac{2.63737 \mathrm{e}^{-0.075 t} \cos(\pi t/2)}{0.125925 \mathrm{e}^{-0.075 t} \sin(\pi t/2)} &= 1 \\ \frac{2.63737}{0.125925}\cot(\pi t/2) &= 1 \text{.} \end{align*}
This is yet another place where your mathematics instruction, preferring symbolics to shoveling piles of digits around, is correct: replacing the constants in your problem with variables clarifies cancellation in the leading constant: \begin{align*} a &= 3.8345 \\ b &= 0.075 \\ \delta(t) &= \frac{1}{2} a \mathrm{e}^{-b t} \sin(2 \pi t/4) \\ \delta'(t) &= \frac{1}{4} \pi a \mathrm{e}^{-b t} \cos(\pi t/2) - \frac{1}{2} a b \mathrm{e}^{-b t} \sin(\pi t/2) \\ \delta'(t) = 0 &\implies \frac{\pi}{2b} \cot(\pi t/2) = 1 \end{align*} So we can find $t$ as $$ t = \frac{2}{\pi} \cot^{-1}\left( \frac{2b}{\pi} \right) = \frac{2}{\pi} \cot^{-1}\left( \frac{0.150}{\pi} \right) = 0.969627\dots \text{.} $$
(What did this symbolic operation show us the original did not? It showed us something that will be obvious in retrospect -- the location of the maximum is independent of the scalar multiple at the front of $\delta$; it only depends on $b$, the decay rate.)