Mastering Trigonometry: Graphing Sinusoidal Voltage with v = 6sin (314t + pi/2)

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A sinusoidal voltage is represented by the formula v = 6sin (314t + pi/2). Sketch the graph on the graph paper provided for values of t from 0.0s to 0.04s and show on the graph the values of
• the amplitude
• the time period

and show how the frequency of the waveform may be determined from the graph.



can i have a small hint on finding the amplitude and the time period?
 

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will.i.am2 said:
A sinusoidal voltage is represented by the formula v = 6sin (314t + pi/2). Sketch the graph on the graph paper provided for values of t from 0.0s to 0.04s and show on the graph the values of
• the amplitude
• the time period

and show how the frequency of the waveform may be determined from the graph.can i have a small hint on finding the amplitude and the time period?
If you know calculus, the amplitude will be the value of v at the time ##t_0## such that ##\left.\frac{dv}{dt}\right|_{t_0}=0##.

If you don't know calculus, what's the maximum value of the sine function and what does that imply about the magnitude of ##6\sin(\xi)##?
 
Last edited:
Mandelbroth said:
If you know calculus, the amplitude will be the time ##t_0## such that ##\left.\frac{dv}{dt}\right|_{t_0}=0##.

No. The amplitude is not a value of ##t##.

If you don't know calculus, what's the maximum value of the sine function and what does that imply about the magnitude of ##6\sin(\xi)##?

In other words, the amplitude.

@will.i.am2: You are required to show an attempt at solving the problem to receive help.
 
LCKurtz said:
No. The amplitude is not a value of ##t##.
Sorry. I misspoke. The amplitude will be the value of v at ##t_0##. I fixed my post.
 
v = 6sin (314t + π/2)
v = 6sin (314(0) + 3.14 /2)
v = 6sin (0 + 1.57)
v = 6sin(1.57)
v = 0.16

v = 6sin (314t + π/2)
v = 6sin (314(0.01) + 3.14 /2)
v = 6sin (3.14 + 1.57)
v = 6sin(4.71)
v = 0.49

v = 6sin (314t + π/2)
v = 6sin (314(0.02) + 3.14 /2)
v = 6sin (6.28 + 1.57)
v = 6sin(7.85)
v = 0.81

v = 6sin (314t + π/2)
v = 6sin (314(0.03) + 3.14 /2)
v = 6sin (9.42 + 1.57)
v = 6sin(10.99)
v = 1.14

v = 6sin (314t + π/2)
v = 6sin (314(0.04) + 3.14 /2)
v = 6sin (12.56 + 1.57)
v = 6sin(14.13)
v = 1.46
 
i think maximum value of sin is 6 but i don't know what's it implying on the amplitude of 6sin.
 
will.i.am2 said:
i think maximum value of sin is 6 but i don't know what's it implying on the amplitude of 6sin.

I think the maximum of sin is 1. Try graphing it. And set your calculator to radians, not degrees.
 
i have done it in rads only. & how can it's be 1?
 
will.i.am2 said:
i have done it in rads only. & how can it's be 1?

What does the value of y = sin(x) look like. What max/min values does it have ? What about y = psin(x) where p is some real number. What are it's max/min values ? How does this relate to the equation you were given ?
 
Dick said:
If you take sin(1.57) in radians, you should get very close to 1. 1.57 is almost pi/2.

why have you taken sin of half pi? I still didn't get it.
 
will.i.am2 said:
v = 6sin (314t + π/2)
v = 6sin (314(0) + 3.14 /2)
v = 6sin (0 + 1.57)
v = 6sin(1.57)
v = 0.16

I'm trying to tell you v=0.16 isn't right because your calculator isn't in radian mode. sin(1.57) should round off to 1.00, so 6*sin(1.57) should be 6.
 
This is in the wrong forum. This is neither calculus nor beyond. It's pre-calculus.
 
will.i.am2 said:
v = 6sin (314t + π/2)
v = 6sin (314(0) + 3.14 /2)
v = 6sin (0 + 1.57)
v = 6sin(1.57)
v = 0.16

How did you arrive at this value... Let's try it

we have

[tex]v = 6sin(314t + \frac{\pi}{2}}[/tex]

verbally stated v = 6 times the sin of 314t + [itex]\frac{\pi}{2}[/itex] radians.

So at time t = 0 we have..

[tex]v = 6sin(314(0) + \frac{\pi}{2})[/tex]

which gives

[tex]v = 6 sin( \frac{\pi}{2})[/tex]

In other words to get v at t=0 we have to multiply the sine of [itex]\frac{\pi}{2}[/itex] radians by 6.

[itex]\frac{\pi}{2}[/itex] radians in degrees is 90 degrees, the sine of which is 1 . So what then happens when you then multiply 1 by 6 to get the final result for v ?

Your calculations are screwed up. As someone else pointed out, you appear to be using a calculator set to degrees but you are providing radians. Either switch the calculator to radians or, convert the radians to degrees and then compute the sine.
 
^^

yes, you guys are absolutely right & my mistake I wasn't doing in rad.

it is equal to 6 then the amplitude would be 6 - (-6) / 2 = 6
 
Given that -1 <= sin(x) <= 1, it follows that -6 <= 6 sin(x) <= 6 (just multiply each part by 6).

PS. It helps to learn all you can about these trig functions.