Trigonometry word problem help!

[Guest]

I am on the last question of my trig assignment and i need help!

The voltage of an alternating current flowing in an electrical circuit is given by V = Vo sin (ωt + a) where:

V is the voltage
Vo , ω and α are constants
T is the time in seconds since measurements of the voltage were begun

The maximum voltage in an electric circuit is 312 volts. The time between maximums is 0.02 seconds. The initial voltage measured was 200 volts.

- Find the value of the constants Vo, ω and α and hence find an equation for V. (All values to 3 significant figures)
- What will the voltage be after 1.2 seconds? (All times to 3 significant figures)
- What are the first 3 times that the voltage will be 240 volts? (All times to 4 significant figures)

Any help at all would be amazing

[Guest]

The maximum value [tex]\sin[/tex] attains is 1. So the maximum voltage attained by [tex]V = V_0\sin(\omega t+\alpha)[/tex] is simply [tex]V_0[/tex] (assuming [tex]V_0[/tex] is positive which for the purpose of this question we are free to do). This means [tex]V_0=312[/tex] volts.

(If we assume [tex]V_0[/tex] is negative then [tex]V_0=-312[/tex] volts and the value of [tex]\alpha[/tex] we calculate later should be replaced by [tex]\alpha+180^\circ[/tex]).

The maximum peaks in [tex]\sin[/tex] occur at intervals of [tex]360^\circ[/tex] (or [tex]2\pi[/tex] radians). Let [tex]t_1[/tex] be the time of the first peak and let [tex]t_2[/tex] be the time of the second peak. We know that [tex](\omega t_2 +\alpha) - (\omega t_1+\alpha) = 360^\circ[/tex] (assuming [tex]\omega[/tex] is positive which for the purpose of this question we are free to do) which simplifies to [tex]\omega(t_2-t_1)=360^\circ[/tex] which in turn tells us that [tex]\omega = 360^\circ/(t_2-t_1)[/tex]. The question tells us that [tex]t_2-t_1=0.02[/tex] seconds, so that means [tex]\omega = 18000[/tex] degrees/second (or [tex]100\pi[/tex] radians/second).

(If we assume [tex]\omega[/tex] is negative then [tex]\omega = -18000[/tex] degrees/second.)

At [tex]t=0[/tex] the question tells us that [tex]V=200[/tex] volts. Substituting this into [tex]V = V_0\sin(\omega t+\alpha)[/tex] gives [tex]200 = 312\sin\alpha[/tex]. So [tex]\sin\alpha = 200/312 = 25/39[/tex] and therefore [tex]\alpha = \sin^{-1} (25/39) = 39.86...^\circ[/tex] or [tex]140.13...^\circ[/tex] plus or minus multiples of [tex]360^\circ[/tex] which for the purposes of this question make no difference, (or equivalently 0.6958... or 2.4457... radians plus or minus multiples of [tex]2\pi[/tex]).

Unfortunately the question is badly written and there is no way to determine which of the two solutions for [tex]\alpha[/tex] we should take.

"What will the voltage be after 1.2 seconds?"
Substituting this into the formula gives [tex]V = 312\sin (18000\times 1.2+\alpha) = 312\sin(216000+\alpha) = 312\sin(60\times 360+\alpha) = 312\sin\alpha = 312\times \frac{25}{39} = 200[/tex] volts.
In 1.2 seconds the voltage goes through 60 full oscillations to end up back where it started which is at 200 volts.

"What are the first 3 times that the voltage will be 240 volts?"
This depends on which of the two solutions for [tex]\alpha[/tex] we take.

[tex]240 = 312\sin(18000t+\alpha)[/tex]
which implies [tex]18000t+\alpha = \sin^{-1}(240/312) = 50.28...+360n[/tex] or [tex]129.71...+360n[/tex] for some integer [tex]n[/tex].
This rearranges to [tex]t = (50.28-\alpha)/18000+0.02n[/tex] or [tex](129.71-\alpha)/18000+0.02n[/tex].

If [tex]\alpha = 39.86...[/tex] we get
[tex]t = ..., -0.0194213, 0.000578695, 0.020578695, 0.040578695, ...[/tex] or [tex]..., -0.0150085 ,0.00499148, 0.02499148, 0.04499148, ...[/tex]
So the first three times are: [tex]0.000578695, 0.00499148, 0.020578695[/tex].

If [tex]\alpha = 140.13...[/tex] we get
[tex]t = ..., -0.00499148, 0.0150085, 0.0350085, 0.0550085, ...[/tex] or [tex]..., -0.000578695, 0.0194213, 0.0394213, 0.0594213 ...[/tex]
So the first three times are: [tex]0.0150085, 0.0194213, 0.0350085[/tex].

Hope this helped,

R. Baber.