AC circuit, voltage and frequency problem

[Alice7979]

Homework Statement

According to Equation 20.7, an ac voltage V is given as a function of time t by V = Vo sin 2

ft, where Vo is the peak voltage and f is the frequency (in hertz). For a frequency of 64.7 Hz, what is the smallest value of the time at which the voltage equals one-half of the peak-value?

Homework Equations

V = Vo sin2(3.14)ft

The Attempt at a Solution

1/2Vo = Vo sin(2*3.14*64.7*t)

I know I set this equation wrong with the 1/2Vo but I don't understand how to do that part, the answer I got this way is .0738 V

 

[Charles Link]

You need to know the angle $\theta$ that has $\sin(\theta)=\frac{1}{2}$. And here, with this voltage function as a function of time, we are working in radians, so you need to write that angle $\theta$ in radians. $\\$ For starters, you need to know the angle in degrees that has $\sin(\theta)=\frac{1}{2}$, and then convert that angle to radians.

 

[haruspex]

I know I set this equation wrong

Looks right to me.

the answer I got this way is .0738 V

Please show your working. The answer should be a period of time, not a voltage.

 

[Alice7979]

You need to know the angle $\theta$ that has $\sin(\theta)=\frac{1}{2}$. And here, with this voltage function as a function of time, we are working in radians, so you need to write that angle $\theta$ in radians. $\\$ For starters, you need to know the angle in degrees that has $\sin(\theta)=\frac{1}{2}$, and then convert that angle to radians.

How do you know its in radians?

 

[Alice7979]

You need to know the angle $\theta$ that has $\sin(\theta)=\frac{1}{2}$. And here, with this voltage function as a function of time, we are working in radians, so you need to write that angle $\theta$ in radians. $\\$ For starters, you need to know the angle in degrees that has $\sin(\theta)=\frac{1}{2}$, and then convert that angle to radians.

Never mind, I angular velocity=2pif

 

[haruspex]

How do you know its in radians?

At this level, radians would be standard as input to a trig function. A calculator would generally assume radians unless you tell it otherwise.
In the present case, you have to assume that in the given sin(2

ft) the 2

ft part is in radians. The presence of the "

" just about clinches it.

 

[Alice7979]

Looks right to me.

Please show your working. The answer should be a period of time, not a voltage.

*Vo cancel out*
1/2= sin(406.5t)
1/2= sin(.523599)
t=406.5/.523599 = .0012880664 s

 

[haruspex]

*Vo cancel out*
1/2= sin(406.5t)
1/2= sin(.523599)
t=406.5/.523599 = .0012880664 s

Correct, but don't quote so many digits. The frequency is only specified to three sig figs.

 

[Alice7979]

I entered .001 s and I still have it wrong

 

[Alice7979]

I got it, the answer was supposed to be in scientific notation

 

[Charles Link]

Instead of using $.523599=\sin^{-1}(\frac{1}{2})=\theta$ (radians) from a calculator, could you recognize that $\theta=30^o=\frac{\pi}{6}$ radians? (and then compute $\frac{\pi}{6}$?) Note: $\pi \,(radians)=180^o$. Otherwise, it is correct. $\\$ $\sin(30^o)=\frac{1}{2}$. The angle of $30^o$ is well-known as having the $\sin(30^o)=\frac{1}{2}$. The $\sin(45^o)=\frac{\sqrt{2}}{2}$. Both of these trigonometric values come up time and time again. And also $\sin(90^o)=1$.