Question

Sounds are how we perceive pressure waves in the air. The note F-sharp, more than 2 octaves below the note 'middle C', is a sound wave with an ordinary frequency* f - 61.875 Hertz or cycles/sec Model this note with the sine function, assuming that the amplitude is 1 and the phase shift is 0. g(x) The temperature during the day can be modeled by a sinusoid. Answer the following question given that the low temperature of 30 degrees occurs at 3 AM and the high temperature for the day is 70 degrees. Assuming t is the number of hours since midnight, find an equation for the temperature, T, in terms of t. T(t)

Answer 1

a)
`g(x) = sin (123.75\pi t)`

b)
`T(t) = 20 sin ((\pi )/(12) (t+15))+50`

Explanation:

Frequency, f = 61.875 Hz

`g(x) = A sin (wt + \alpha)`...........(1)

Amplitude, A = 1

Since there is no phase shift,
`\alpha = 0`

`w = 2 \pi f`

`w = 2\pi * 61.875`

`w = 123.75 rad/s`

Substituting the values of A and w into equation (1)

`g(x) = sin (123.75\pi t)`

b)

The general equation for a sinusoid

`T(t) = A sin (B(t-C))+D`.................(2)

Lot temperature = 30⁰

High temperature = 70⁰

Amplitude,
`A = (70-30)/(2) \\`

A = 20⁰

Vertical shift,
`D = (70 + 30)/(2)`

D = 50⁰

Since the low temperature of 30⁰ occurs at 3 AM

t = 3, T(t) = 30

Period,
`T = (2\pi )/(B)`,
`T = 24 hrs`

`B = (2\pi )/(T)`

`B = (\pi )/(12)`

`30 = 20 sin ((\pi )/(12) (3-C))+50`

`-20 = 20 sin ((\pi )/(12) (3-C))`

`-1 = sin ((\pi )/(12) (3-C))`

`sin^(-1) (-1)= ((\pi )/(12) (3-C))`

`3\pi/2 = ((\pi )/(12) (3-C))\\18 = 3 - C\\C = -15`

Substituting the values of A, B, C, D into equation (2)

`T(t) = 20 sin ((\pi )/(12) (t+15))+50`

Answer 2

The model of the F-sharp note is
`f(x) = sin(123.75 \pi)`

The model of the temperature is
`T(t)= 20 sin ((\pi)/(2)(t+15) ) +50`

Explanation:

From the question we are told that

The frequency is
`f = 61.875\ Hz`

The Amplitude is A = 1

The phase shift is
`\theta = 0`

The angular velocity of a wave can be given as

`w = 2 \pi f`

Substituting values

`w = 2 \pi *61.875`

`= 123.75\pi`

Now the sine function model of this note is given as

`f(x) = A sin (wt +\theta)`

Substituting values

`f(x) = sin(123.75 \pi)`

The minimum temperature is
`T_(min) = 30^o`

The maximum temperature is
`T_(max) = 70^o`

The average temperature is
`T_(avg) = (70+30 )/(2) = 50^o`

The Amplitude is
`A = (T_(max) - T_ [min)/(2) = (70-30)/(2) =20^0`

The period = 24 This because the function is modeled in such a way that 24 hour is time to complete a full cycle

The General sin equation is mathematically given as

`T(t) = A sin (b (t - c )) +d`

Where A is the Amplitude = 20

b is the angular speed

c is the phase shift

d is the vertical transition

And T(t) is the temperature after time t

And the period is evaluates as
`T = (2 \pi)/(b)` since
`T = (1)/(frequency)`

This implies that
`b= (2 \pi)/(T) = (2 \pi)/(24 )=(\pi)/(12)`

Since average is at 50 then it means that the wave graph has been shifted by 50 unites

d = 50

Now from our question

t = 3 hours when T(t) = 30

This means that

`T(3) =20 sin ((\pi)/(12)(3 -C) )+50`

=>
`30 = 20 sin ((\pi)/(12) (3-c)) +50`

=>
`20 sin ((\pi)/(12) (3-c)) = 30 -50= -20`

=>
`sin ((\pi)/(12) (3 -c )) = -1`

Since
`-1 = sin (3 \pi)/(2)` we have

`sin ((\pi)/(12) (3 -c )) = sin((3 \pi)/(2) )`

=>
`3-c = (sin((\pi)/(12) ))/(sin ((3 \pi)/(2) ))`

=>
`-c = 18 -3`

=>
`c = - 15`

Therefore the sinusoidal model of the temperature is

`T(t) = 20 sin ((\pi)/(12) (t - (-15)) ) +50`

`T(t)= 20 sin ((\pi)/(2)(t+15) ) +50`