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

In summary: They are extremely important in science and engineering, and are used in many real-world applications.In summary, a sinusoidal voltage is represented by the formula v = 6sin (314t + pi/2). By sketching the graph on the provided graph paper for values of t from 0.0s to 0.04s, we can determine the amplitude and time period of the waveform. The amplitude of the waveform is 6, which is equal to the maximum value of the sine function. The time period can be determined by finding the time t such that the derivative of v with respect to t is equal to 0. Additionally, the frequency of the waveform can be determined from the graph by measuring the number of cycles
  • #1
will.i.am2
14
0
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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  • #2
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)##?
 
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  • #3
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.
 
  • #4
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.
 
  • #5
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
 
  • #6
i think maximum value of sin is 6 but i don't know what's it implying on the amplitude of 6sin.
 
  • #7
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.
 
  • #8
i have done it in rads only. & how can it's be 1?
 
  • #9
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 ?
 
  • #10
will.i.am2 said:
i have done it in rads only. & how can it's be 1?

If you take sin(1.57) in radians, you should get very close to 1. 1.57 is almost pi/2.
 
  • #11
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.
 
  • #12
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.
 
  • #13
This is in the wrong forum. This is neither calculus nor beyond. It's pre-calculus.
 
  • #14
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.
 
  • #15
^^

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
 
  • #16
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.
 

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1. What is trigonometry?

Trigonometry is a branch of mathematics that deals with the relationships between the sides and angles of triangles. It is used to solve problems involving right triangles and can also be applied to other areas of mathematics and science.

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The basic trigonometric functions are sine, cosine, and tangent. These functions are used to calculate the ratios of the sides of a right triangle.

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4. Can you provide an example of a trigonometry problem?

Sure, here's an example: Find the value of x in the right triangle below, given that the sine of angle A is 0.5.
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5. How can I improve my understanding of trigonometry?

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