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

It would be D at least

Answer 2

Answer:

Option B - 41.4%

Step-by-step explanation:

Given : The formula for determining the frequency, f, of a note on a piano is $f(h)=440(2)^{\frac{h}{12}}$ 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.

To find : How much greater is the frequency of the new note compared with the frequency of the A above middle C?

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

When the note is at initial stage i.e, h=0 frequency is

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

$f(0)=440(1)$

$f(0)=440$

A note is six half-steps away from the A above middle C i.e, h=6

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

$f(6)=440(2)^\frac{1}{2}$

$f(6)=440(1.41)$

$f(6)=622.25$

Initial frequency is 440 hz.

Final frequency is 622.25 hz.

To find change formula is

$=\frac{\text{final} - \text{initial}}{\text{Initial}}$

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

$=0.414$

Frequency change in percentage

$0.414\times 100= 41.4\%$

Therefore, Option B is correct.