Question

If a charged capacitor is connected to a coil by closing a switch, energy

is transferred to the coil and then back to the capacitor in an oscillatory motion. The voltage V (in volts) across the capacitor gradually diminishes to 0 with time t (in seconds).

(a) Graph the function relating V and t: $V\left(t\right)=e^{-\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\cos \left(\mathrm{\pi t}\right),0\le \mathrm{t}\le 3$

(b) At what times t does the graph of V touch the graph of $y=e^{-\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$? When does the graph of V touch the graph of $y=e^{-\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$?

(c) When is the voltage V between $-0.4and0.4$volt?

Short answer

(a)

(b) At $t=0,2$the graph of V will touch the other graph.

(c)$0.35<t<0.675,1.29<t<1.75,2.19<t<3$

Step-by-step solution

Step 1. Given Information   

The given equation is$V\left(t\right)=e^{-\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\cos \left(\mathrm{\pi t}\right),0\le \mathrm{t}\le 3$

Part (a) Step 1. Graph

Since when a charge capacitor is connected to a coil by closing the switch, energy transferred to the coil and then back to the capacitor in an oscillatory motion, so it is simple harmonic motion.

The graph of the given function is

Part (b) Step 1. Explanation

The graph of the function $y=e^{\raisebox{1ex}{$-t$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$is as follows,

The graph of V will touch the graph of $y=e^{\raisebox{1ex}{$-t$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$when both will preserve the same value.

$$\begin{gathered} V=e^{\raisebox{1ex}{$-t$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \\ {e}^{\raisebox{1ex}{$-t$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\cos \left(\mathrm{\pi t}\right)={\mathrm{e}}^{\raisebox{1ex}{$-\mathrm{t}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.};0\le \mathrm{t}\le 3 \\ \cos \left(\mathrm{\pi t}\right)=1;0\le \mathrm{t}\le 3 \\ \mathrm{t}=0,2\mathrm{as}\cos \left(0\right)=1=\cos \left(2\mathrm{\pi }\right) \\ \mathrm{At}\mathrm{t}=0,\mathrm{V}\left(0\right)=\mathrm{y}=1 \\ \mathrm{At}\mathrm{t}=2,\mathrm{V}\left(2\right)=\mathrm{y}={\mathrm{e}}^{\raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\approx 0.513 \end{gathered}$$

This can be verified from figure -1 and figure -2

The graph of the function $y=-e^{\raisebox{1ex}{$-t$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$is as follows,

The graph of V will touch the graph of $y=-e^{\raisebox{1ex}{$-t$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$when both will preserve the same value.

$$\begin{gathered} V=-e^{\raisebox{1ex}{$-t$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \\ {e}^{\raisebox{1ex}{$-t$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\cos \left(\mathrm{\pi t}\right)=-{\mathrm{e}}^{\raisebox{1ex}{$-\mathrm{t}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.};0\le \mathrm{t}\le 3 \\ \cos \left(\mathrm{\pi t}\right)=-1;0\le \mathrm{t}\le 3 \\ \mathrm{t}=1,3\mathrm{as}\cos \left(\mathrm{\pi }\right)=-1=\cos \left(3\mathrm{\pi }\right) \\ \mathrm{At}\mathrm{t}=1,\mathrm{V}\left(1\right)=\mathrm{y}=-e^{\frac{-1}{3}}\approx 0.716 \\ \mathrm{At}\mathrm{t}=3,\mathrm{V}\left(3\right)=\mathrm{y}=-{\mathrm{e}}^{-1}\approx -0.368 \end{gathered}$$

This can be verified from figure -1 and figure -3

Part (c) Step 1. Explanation

From figure -1 we can say that the voltage is between $-0.4and0.4$volt for three intervals $\left[0,1\right],\left[1,2\right]and\left[2,3\right]$of time $0\le t\le 3$.

The graph of the function in the above three intervals respectively and $\left[-0.4,0.4\right]$are as follows,

And

Thus, from figure 4, 5 and 6, it is clear that for $-0.4<V<0.4$, we have,

$0.35<t<0.675,1.29<t<1.75and2.19<t<3$