Question 867377: During an experiment, the intensity i(t) of the electric current of a device as a function of time t elapsed since the beginning of the experiment is given by:
i(t)=6sin((pi*t/12)+(2*pi/3)+6
The device emits a sound signal each time the current’s intensity is equal to 9.
The experiment lasts 120 seconds.
How many sound signals does the device emit during the experiment?
After trying to graph it on google, I got 6. I don't fully understand how to get it anyway but the graph way, so I'm grateful for any help.
Thanks in advance!
Found 2 solutions by stanbon, jim_thompson5910:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! During an experiment, the intensity i(t) of the electric current of a device as a function of time t elapsed since the beginning of the experiment is given by:
i(t)=6sin((pi*t/12)+(2*pi/3)+6
The device emits a sound signal each time the current’s intensity is equal to 9.
The experiment lasts 120 seconds.
How many sound signals does the device emit during the experiment?
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Period = (2pi)/(pi/12) = 24 seconds
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Phase shift = -c/b = -[(2pi/3)/(pi/12)] = -8
# of periods in 120 sec = 120/24 = 5
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Since the range of "i" is [0,12] i will equal 9 twice during
each cycle.
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# of alarms = 10
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Cheers,
Stan H.
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Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website! The device emits a signal every time the intensity, i(t), is equal to 9. So we need to find out how many times it equals 9. This is within the interval t = 0 to t = 120 since the experiment lasts 120 seconds.
So that means t is in the interval [0,120]. This is important to selecting the right solutions.
Plug in i(t) = 9 and solve for t
i(t)=6sin((pi*t/12)+(2*pi/3))+6
9=6sin((pi*t/12)+(2*pi/3))+6
9-6=6sin((pi*t/12)+(2*pi/3))
3=6sin((pi*t/12)+(2*pi/3))
6sin((pi*t/12)+(2*pi/3)) = 3
sin((pi*t/12)+(2*pi/3)) = 3/6
sin((pi*t/12)+(2*pi/3)) = 1/2
(pi*t/12)+(2*pi/3) = arcsin(1/2) or (pi*t/12)+(2*pi/3) = pi - arcsin(1/2)
(pi*t/12)+(2*pi/3) = pi/6 or (pi*t/12)+(2*pi/3) = pi - pi/6
(pi*t/12)+(2*pi/3) = pi/6 or (pi*t/12)+(2*pi/3) = 5pi/6
(pi*t/12)+(2*pi/3) = pi/6 + 2pi*n or (pi*t/12)+(2*pi/3) = 5pi/6 + 2pi*n
pi*t/12 = pi/6 + 2pi*n - 2*pi/3 or pi*t/12 = 5pi/6 + 2pi*n - 2*pi/3
t = 12/pi(pi/6 + 2pi*n - 2*pi/3) or t = 12/pi(5pi/6 + 2pi*n - 2*pi/3)
t = 2 + 24n - 8 or t = 10 + 24n - 8
t = 24n - 6 or t = 24n + 2
So from all of that work above, we have determined that the solutions to i(t) = 9 are t = 24n - 6 or t = 24n + 2 where n is an integer.
Now let's plug in integral values for n into the two equations.
Let's pick on the first equation t = 24n - 6
If n = 0, then t is...
t = 24n - 6
t = 24(0) - 6
t = -6 ... out of range of the interval [0,120]
If n = 1, then t is...
t = 24n - 6
t = 24(1) - 6
t = 18
If n = 2, then t is...
t = 24n - 6
t = 24(2) - 6
t = 42
If n = 3, then t is...
t = 24n - 6
t = 24(3) - 6
t = 66
If n = 4, then t is...
t = 24n - 6
t = 24(4) - 6
t = 90
If n = 5, then t is...
t = 24n - 6
t = 24(5) - 6
t = 114
If n = 6, then t is...
t = 24n - 6
t = 24(6) - 6
t = 138 ... out of range of the interval [0,120]
If n = 7, then t is...
t = 24n - 6
t = 24(7) - 6
t = 162 ... out of range of the interval [0,120]
So we see that as n increases, we're getting larger and larger t values that will be outside the interval [0,120]. So we can stop for this equation.
So far, the solutions are: 18, 42, 66, 90, 114
Repeat for the other equation
If n = -1, then t is...
t = 24n + 2
t = 24(-1) + 2
t = -22 ... out of range of the interval [0,120]
If n = 0, then t is...
t = 24n + 2
t = 24(0) + 2
t = 2
If n = 1, then t is...
t = 24n + 2
t = 24(1) + 2
t = 26
If n = 2, then t is...
t = 24n + 2
t = 24(2) + 2
t = 50
If n = 3, then t is...
t = 24n + 2
t = 24(3) + 2
t = 74
If n = 4, then t is...
t = 24n + 2
t = 24(4) + 2
t = 98
If n = 5, then t is...
t = 24n + 2
t = 24(5) + 2
t = 122 ... out of range of the interval [0,120]
If n = 6, then t is...
t = 24n + 2
t = 24(6) + 2
t = 146 ... out of range of the interval [0,120]
The other set of solutions are: 2, 26, 50, 74, 98
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Put them all together to get this final set of solutions: 18, 42, 66, 90, 114, 2, 26, 50, 74, 98
Here is that list sorted: 2, 18, 26, 42, 50, 66, 74, 90, 98, 114
There are 10 solutions listed above, so it emits a signal exactly 10 times.
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