Asked by Daniela
A musician is playing a tuba. If she plays a note and holds it for a significant length of time, the air pressure near the opening of the tuba can be modeled by the following function. pt=−101,325.8 - 0.1 cos163.1t
In this equation, pt represents the air pressure in pascals, and t is the time in seconds.
Find the following. If necessary, round to the nearest hundredth
Period of p :
Frequency of p:
Max air pressure:
In this equation, pt represents the air pressure in pascals, and t is the time in seconds.
Find the following. If necessary, round to the nearest hundredth
Period of p :
Frequency of p:
Max air pressure:
Answers
Answered by
mathhelper
We get the period from
cos163.1t
since period = 2π/k
period = 2π/163.1 = appr .0385 seconds
frequency: 163.1 cycles/second
The max of the cosine is +0.1
so the max of pt is -101325.8 + .1 = -101325.7
(btw, sounds like the musician is playing a third octave E)
cos163.1t
since period = 2π/k
period = 2π/163.1 = appr .0385 seconds
frequency: 163.1 cycles/second
The max of the cosine is +0.1
so the max of pt is -101325.8 + .1 = -101325.7
(btw, sounds like the musician is playing a third octave E)
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