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

Answer:

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

b) $T(t) = 20 sin (\frac{\pi }{12} (t+15))+50$

Step-by-step 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, $A = \frac{70-30}{2}$

A = 20⁰

Vertical shift, $D = \frac{70 + 30}{2}$

D = 50⁰

Since the low temperature of 30⁰ occurs at 3 AM

t = 3, T(t) = 30

Period, $T = \frac{2\pi }{B}$,   $T = 24 hrs$

$B = \frac{2\pi }{T}$

$B = \frac{\pi }{12}$

$30 = 20 sin (\frac{\pi }{12} (3-C))+50$

$-20 = 20 sin (\frac{\pi }{12} (3-C))$

$-1 = sin (\frac{\pi }{12} (3-C))$

$sin^{-1} (-1)= (\frac{\pi }{12} (3-C))$

$3\pi/2 = (\frac{\pi }{12} (3-C))\\18 = 3 - C\\C = -15$

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

$T(t) = 20 sin (\frac{\pi }{12} (t+15))+50$

Answer 2

Answer:

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

The model of the temperature is  $T(t)= 20 sin (\frac{\pi}{2}(t+15) ) +50$

Step-by-step 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  $T_{avg} = \frac{70+30 }{2} = 50^o$

The Amplitude is  $A = \frac{T_{max} - T_ [min}{2} = \frac{70-30}{2} =20^0$

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 $T = \frac{2 \pi}{b}$ since $T = \frac{1}{frequency}$  

This implies that $b= \frac{2 \pi}{T} = \frac{2 \pi}{24 }=\frac{\pi}{12}$

 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

       $T(3) =20 sin (\frac{\pi}{12}(3 -C) )+50$

=>   $30 = 20 sin (\frac{\pi}{12} (3-c)) +50$

=>    $20 sin (\frac{\pi}{12} (3-c)) = 30 -50= -20$

=>   $sin (\frac{\pi}{12} (3 -c )) = -1$

Since  $-1 = sin \frac{3 \pi}{2}$ we have  

     $sin (\frac{\pi}{12} (3 -c )) = sin(\frac{3 \pi}{2} )$

=>  $3-c = \frac{sin(\frac{\pi}{12} )}{sin (\frac{3 \pi}{2} )}$

=>  $-c = 18 -3$

=>   $c = - 15$

Therefore the sinusoidal model of the temperature is

   $T(t) = 20 sin (\frac{\pi}{12} (t - (-15)) ) +50$

           $T(t)= 20 sin (\frac{\pi}{2}(t+15) ) +50$