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

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Homework Help Overview

The discussion revolves around the sinusoidal voltage represented by the formula v = 6sin(314t + π/2). Participants are tasked with sketching the graph for values of t from 0.0s to 0.04s, identifying the amplitude, time period, and frequency from the graph.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the concept of amplitude and time period, with some suggesting that the amplitude relates to the maximum value of the sine function. Questions arise regarding the implications of the sine function's maximum value on the amplitude of the sinusoidal equation.

Discussion Status

There is ongoing clarification regarding the calculations of the amplitude and the correct use of radians versus degrees in trigonometric functions. Some participants have provided hints and corrections, while others are still grappling with the concepts.

Contextual Notes

Participants are required to show attempts at solving the problem to receive help, which has led to discussions about the correct interpretation of the sine function and its maximum values.

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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 ?
 
  • #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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