# How do you find the amplitude for y(t) = 1/4e^-t cos6t?

Aug 14, 2018

In this damped oscillation, as in springs, the amplitude
$\left(\frac{1}{4} {e}^{- t}\right) \downarrow$ as $t \uparrow$. See graphs for both the oscillation and the diminishing amplitude.

#### Explanation:

In the amplitude-damping periodic oscillation,

$y = \left(\frac{1}{4} {e}^{- t}\right) \cos 6 t , t \ge 0$,

t-intercepts;$\left(2 k + 1\right) \frac{\pi}{2}$ ( from zeros of $\cos 6 t$ ,

$k = 0 , 1 , 2 , 3 , \ldots$

the amplitude $a \left(t\right) = \frac{1}{4} {e}^{- t}$

is a function of time t that $\downarrow$, as $t \uparrow$.

The period is $\frac{2 \pi}{6} = \frac{\pi}{3}$.

See the second graph, for the exponential decay of amplitude..
graph{y-1/4e^(-x) cos (8x) = 0[0 6 -1.5 1.5]}
Amplitude-decay graph:
graph{(y-1/4e^(-x))(y-0.04) = 0[0 5 0 1]}

See just one-in-many referenceas:

https://deutsch.physics.ucsc.edu/6A/book/harmonic/node18.html