# MATH

posted by SC

Danny is measuring the loudness of a police siren. At t=2 seconds, the siren's loudness is at its maximum of 112 dB for the first time. At t=7 seconds, the siren's loudness is at its minimum of 88 dB for the first time. The loudness is a sinusoidal function of time, t. In the first 13 seconds, how much of the time will the loudness be above 94 dB?

So I know the sinusoidal function is this:
y=Asin(2pi/B(x-C)+D
A=amplitude=12
D=Mean=100
B=period=???
C=phase shift= x-coordinate of max-(B/4)=?????

1. Reiny

your a is correct at 12
from max to min took 5 seconds, so the whole period is 10 seconds
then 10 = 2pi/k
k = pi/5

so the basic curve could be
y = 12sin(pi/5)(x)

this would have a max of 12 but you want the max to be 112 , so let's move it up 100 units
y = 12sin(pi/5)(x) + 100

Right now our max would be at 2.5 seconds, but we want it to be at 2 seconds, so our graph must be shifted to the left .5 units.

Final curve:
y = 12sin(pi/5)(x + .5) + 100

testing:
if t=2, y = 112 , check!
if t=7, y = 88 check!

now let's set y = 94
94 = 12sin(pi/5)(x + .5) + 100
-.5 = sin(pi/5)(x + .5)
(pi/5)(x + .5) = pi+.526599 = 3.66519
x = 5.33333
or
(pi/5)(x + .5) = 2pi-.526599 = 5.75959
x = 8.6667
The next set of answers would be 10 seconds later, (the period is 10), but you wanted it only in the first 13 seconds
So between the times of 5.33 and 8.67 seconds we are below 94 decibels, (looking at our graph)
so for a time period of 8.67-5.33 or 3.34 seconds we are below 94 decibels, which means that in the first 13 seconds, the sound is above
94 decibels for a time of 13-3.34 seconds or 9.66 seconds

another check:
take a time between the above, say t = 5.5
y = 12sin(pi/5)(5.5 + .5) + 100
= 92.9 which is below 94

2. Reiny

got my above and belows mixed up in one line

means that in the first 13 seconds, the sound is below

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