Question

The formula for determining the frequency, f, of a note on a piano is f=440(2)^h/12 where h is the number of half-steps from the A above middle C on the keyboard. A note is six half-steps away from the A above middle C. The frequency of the A above middle C is 440 Hz. How much greater is the frequency of the new note compared with the frequency of the A above middle C?
A)29.3%
B)41.4%
C)70.7%
D)182.3%

Answer 1

Answer:

i think it is c

Step-by-step explanation:

Answer 2

Answer:

41.4%

Step-by-step explanation:

The formula for determining the frequency: $f(h)=440(2)^{\frac{h}{12}}$  --A

where h is the number of half-steps from the A above middle C on the keyboard.

A note is six half-steps away from the A above middle C.

Now we are supposed to find How much greater is the frequency of the new note compared with the frequency of the A above middle C?

Now initially there is no half steps .

So, substitute h =0

$f(0)=440(2)^{\frac{0}{12}}$

$f(0)=440$

Now we are given that A note is six half-steps away from the A above middle C

So, substitute h =6

$f(6)=440(2)^{\frac{6}{12}}$

$f(6)=622.25$

Now To find change percentage

Formula: $=\frac{\text{final} - \text{initial}}{\text{Initial}} \times 100$

                $=\frac{622.25- 440}{440} \times 100$

               $=0.414 \times 100$

               $=41.4\%$

Hence  the frequency of the new note is 41.4% greater with the frequency of the A above middle C.