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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 $t = 55.55$:

$$t = \frac{4}{\pi}\cot^{-1}\left(\dfrac{20\pi}{\sqrt{9+400\pi^{2}}-3}\right)$$

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  • $\begingroup$ The term multiplying the sine function is the amplitude. But it's also time dependent, so you can only sort of talk about instantaneous amplitude. Plotting the graph, indeed you see that the function attains its maximum value at $x\approx 0.97$. Why did you differentiate it twice and not once? I would differentiate it once and numerically investigate the result. $\endgroup$ – Matti P. May 25 '18 at 8:33
  • $\begingroup$ Close to 1 radian is close to 55.55 degrees. That’s not the issue, is it? $\endgroup$ – G Tony Jacobs May 25 '18 at 8:39
  • $\begingroup$ It may be. I'll edit my question and include the answer Wolfram provided. $\endgroup$ – david_10001 May 25 '18 at 8:44
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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.)

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  • $\begingroup$ Thanks very much for this, it's exactly what I was looking for! $\endgroup$ – david_10001 May 25 '18 at 8:59
  • $\begingroup$ @david_10001 : I notice in your title you ask for the minimum as well. Note that this cotangent is periodic: $\cot(\pi t / 2) = \cot(\pi(t+2)/2)$, so the minimum occurs at $t = 2.969627\dots$. $\endgroup$ – Eric Towers May 25 '18 at 9:01
  • $\begingroup$ Yes I was just trying to find where the rates of change was zero but to then find max. $\endgroup$ – david_10001 May 25 '18 at 9:13

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