# Thread: Voltage output sinusoidal function problem

1. ## Voltage output sinusoidal function problem

The voltage output (in volts) of an electrical circuit at time t seconds is given by the following function.

V
(t) = 22 sin(5πt − 3π) + 1

(e) During the first second, determine when the voltage output of the circuit is 3 volts. (Enter your answers as a comma-separated list. Round your answers to four decimal places.)

My approach is to
1. set the function equal to 3
3=22sin(5∏t-3∏)+1

2. take the natural log to get rid of the exponent
ln3=(2sin(5∏t-3∏)+1)ln2
[SUP]
(ln3/ln2)-1=2sin(5∏t-3∏)

[(ln3/ln2)-1]/2=sin(5∏t-3∏)

arcsin ([(ln3/ln2)-1]/2)=5∏t-3∏

t= (arcsin([ln3/ln2)-1]/2) + 3∏) / 5∏

t=0.6189

I don't know if I'm on the right track so far, and I have no idea where to go from here? Any help would be soooooo great!! Thanks.

2. ## Re: Voltage output sinusoidal function problem

Originally Posted by UWstudent
The voltage output (in volts) of an electrical circuit at time t seconds is given by the following function.

V
(t) = 22 sin(5πt − 3π) + 1

(e) During the first second, determine when the voltage output of the circuit is 3 volts. (Enter your answers as a comma-separated list. Round your answers to four decimal places.)

My approach is to
1. set the function equal to 3
3=22sin(5∏t-3∏)+1

2. take the natural log to get rid of the exponent
ln3=(2sin(5∏t-3∏)+1)ln2
[SUP]
(ln3/ln2)-1=2sin(5∏t-3∏)

[(ln3/ln2)-1]/2=sin(5∏t-3∏)

arcsin ([(ln3/ln2)-1]/2)=5∏t-3∏

t= (arcsin([ln3/ln2)-1]/2) + 3∏) / 5∏

t=0.6189

I don't know if I'm on the right track so far, and I have no idea where to go from here? Any help would be soooooo great!! Thanks.
I can suggest you that there is something you miss between these steps:

[(ln3/ln2)-1]/2=sin(5∏t-3∏)

arcsin ([(ln3/ln2)-1]/2)=5∏t-3∏, and you final result will be something like 0.06s , so sounds reasonable since you get a time during the first second

3. ## Re: Voltage output sinusoidal function problem

Hi
As the index is raised to the power 2 you need to take logs to this base.

Then log to base2 of 3 = ln3/ln2

This is a standard result from the laws of logs.